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General Physics for Life Science Oriented Majors

by: Miss Lyla Rolfson

General Physics for Life Science Oriented Majors PHYS 2414

Marketplace > University of Oklahoma > Physics 2 > PHYS 2414 > General Physics for Life Science Oriented Majors
Miss Lyla Rolfson
GPA 3.7

Michael Strauss

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Michael Strauss
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This 127 page Class Notes was uploaded by Miss Lyla Rolfson on Monday October 26, 2015. The Class Notes belongs to PHYS 2414 at University of Oklahoma taught by Michael Strauss in Fall. Since its upload, it has received 6 views. For similar materials see /class/229268/phys-2414-university-of-oklahoma in Physics 2 at University of Oklahoma.


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Date Created: 10/26/15
Physics 2414 Strauss Chapter 3 Lecture Notes Main Ideas 1 Vectors 2 Projectile Motion 3 Relative Velocity 1 Vectors 1 1 VECTORS AND SCALARS In our study of physics we come across two kinds of information The first kind is something that can be described by a single number like the area of a circle or speed or distance This is called a scalar We will also come across quantities that have two pieces of information a magnitude and a direction Velocity is such a quantity It has a magnitude say 50 mileshour and a direction say Northwest We write scalars simply as numbers because that is what they are We write vectors in print as bold face A or we write them in handwriting with an arrow over them 11 Every vector quantity must have a magnitude and a direction If you are asked to give an answer for a vector quantity on a problem both the magnitude and the direction must be given or the problem is only half correct Vectors will be used in this class throughout the whole year so it is very important that you learn this material and become comfortable with it We draw vector quantities with an arrow The magnitude of the vector is proportional to its length The direction of the arrow points in direction of vector Every time we encounter a vector we must use its direction and it magnitude Vectors do not have a particular location I can move a vector which points up and is 2 units long anywhere I want as long as it is the same length and points in the same direction The magnitude of a vector is a scalar and is written with a regular font The magnitude of V is v 1 2 MULTIPLICATION BY CONSTANT If I multiply a vector by a constant that changes the magnitude of the vector For example draw 2V and V 1 3 ADDITION AND SUBTRACTION How do we add vectors There is a way to draw vectors which adds them We put the tail of one vector on the tip of the other and draw the resultant If in the same direction simply a linear addition If I walk north for a distance then north again I can just add the two distances to see how far I am from where I started But if I walk north for a while then east I can t just add the distances to see how far I am from where I started But if you draw the lines very accurately then you can actually measure the result To add three 5 Note that you must always put the tail of the one vector on the tip of the other However this is not always accurate enough so there is another way of adding vectors using components The components are the part of the vector along different axes usually chosen to be the x and the y aXis Every vector can be written as the sum of its components A Ax Ay This is vector addition in Ax How long are the parts of the vector along the x aXis and along the y aXis A and Ay To answer that we use trigonometric relations Recall that sin 6 2 side oppos1te 2 COS 6 2 side adjacent g a tan 6 2 side opposite 2 hypotenuse h hypotenuse h side adjacent a So look at the triangle We see that cos0 AA where A and A are the magnitude of the vectors AK and A Then we get the length A A cos0 In a similar manner sin0 AyA gt A A sin0 So there are two ways to specify a vector We can give its magnitude A and its direction as an angle or compass direction or something or we can give its x and y components Suppose we have A and Ay then what is the magnitude of the vector From the pythagorean thereom A2 A2 Ay2 Problem Find the components of the following vectors A and B and draw the components 30 A Magnitude of A is 500 and B magnitude ofB is 700 40 How do we add the vectors using components It is very easy once you have found the components You can see this by first drawing the addition of A and B graphically as already discussed Then it is easy to see that to find the vector C we just use C A B and Cy Ay By In order to solve problems with vectors you must always follow these steps If you don t you will almost always get the wrong answer Solving Problems with Vectors 1 Determine which direction is the positive x direction and which is the positive y direction Draw and label the axes Draw and label all the vectors Determine the individual components of each vector 0 The components of the vector must be along the x and y aXis and add up as vectors to equal the original vector The length of each component may not be greater than the original vector Draw and label the components Use trigonometric functions to determine the length of each component The sign of the component is given by the direction of the component s arrow 4 Do calculations in the x direction and in the y direction separately using only the x and y components of the vectors respectively 5 Combine the results from the x and y directions into one vector WN Problem What is A B Problem A hiker walks 250 km due southeast then 400 km in a direction 60 north of east What are the component of her displacement vectors and how far is she from where she started Problem A skier skis down a hill which is 770 m long The vertical drop is 230 m What is the angle of the hill relative to at ground You should know how to use sin391 on calculator 2 Projectile Motion We have previously discussed motion in a constant gravitational field where the acceleration is constant If I define the positive y direction as being up and g 98 msz then we know that writing the equations in the vertical y direction vy v yo gt y yo Vyot 1219t 2 yfyotVDW V Vyo 290 39yo 17y vy0 vy2 What about the motion in the x direction Is there any acceleration If we throw a ball across the room we might get some idea of what is going on Is the particle changing velocity in the x direction Amazingly it is not Galileo first showed this The velocity and acceleration in the horizontal direction is independent of what is going on in the vertical direction This is always true The motion in each direction is independent of the motion in the other direction So we have general equations to use when the acceleration is constant but not zero and equations when there is no acceleration in the x direction which is the usual case for objects moving in a gravitational field General Equations Equations for x acceleration 0 vxzvx0axt VXZVXO x x0 vx0t 12 cat 2 x x0 vx0t xxOUDmvt v2vx022axx x0 x7x vx0vx2 If the positive x direction is up away from the center of the earth then for objects in free fall we find a g See Problem Solving boxes on pages 56 and 61 in your textbook Let s use these equations to solve some problems Problem A plane must drop a package from 100 m above the ground The plane is travelling at 400 ms a Where does the package strike the ground b What are the horizontal and vertical components of the velocity just before the package strikes the ground Problem A golfer hits a ball with an initial speed of 403 ms at an angle of 320 from the horizontal a How far does the ball go and how long is it in the air b What is its speed when it hits the ground Remember to draw a diagram The final velocity is the same as the initial This is true when the final position is at the same level as the initial position And of course there is no horizontal acceleration and no air resistance Problem a What angle should we launch something in order to make it go its maximum distance on level ground No air resistance b What is the maximum distance the golf ball can go Problem A baseball player hits a home run and the ball lands in the left field seats 76 m above the point at which the ball was hit The ball lands with a velocity of 49 ms at an angle of 31 to the horizontal What is the initial velocity of the ball when the ball leaves the bat Problem A plane is ying horizontally at an altitude of 420 km with a speed of 225 ms When the plane is directly overhead a projectile is fired at an angle 0 with a speed of 389 ms The projectile hits the plane What is the angle0 Sometimes problems can be a little harder if the initial velocity is not given Let s try a problem where no initial velocity is given Problem A golfer hits a ball 180 m on level ground at an angle of 60 above the horizontal What was the initial speed of the ball The speed of the ball is the magnitude of the velocity O We know v vx0 v0 cos60 and x xo vxt We also havevyo v0 sin60 y y0 0 and a 980 msz So we can write two equations with two unknowns The two unknowns are t and v0 DemonstrationProblem Monkey Gun A hunter aims his gun at a monkey hanging on a branch Just as the hunter shoots the monkey drops from the branch to avoid the bullet What happens 3 Relative Velocities We use vectors to solve problems involving the two velocities which are relative to each other The book uses subscripts to help you remember what velocity is relative to another For instance if a plane is ying through the wind we have the velocity of the plane relative to the ground vpg the velocity of the plane relative to the wind va and the velocity of the wind relative to the ground vwg Then when you add the relative velocities always put the letters that are the same next to each other in the addition vpg va vwg Problem A person looking out of a window of a stationary train notices that the drops are falling at a rate of 50 ms When the train is moving the drops are making an angle of 25 with the vertical How fast is the train moving 5 VRT VRG 50 ms VGT Whenl try to ride my bicycle it always seems like I have a headwind and seldom a tailwind Let s investigate this situation using addition of vectors This doesn t really work because we are neglecting the friction of the tires on the road but it giveS some idea of why it always seems like I have a headwind Problem Supposel can pedal 18 mph when there is no wind velocity of the bike with respect to the wind VBW I am travelling east on highway 9 and there is a 10 mph cross wind from the south velocity of the wind with respect to the ground VWG How fast am I travelling velocity of the bike with respect to the ground VBG Neglect friction with respect to the road Wind resistance is the major force making bicycling difficult You might think that a cross wind wouldn t make it harder to pedal But my velocity is a vector and the cross wind is a vector so let s look at the problem like that We will use VBG VBW VWG The velocity of the bike with respect to the wind VBW will always stay the same and will be 18 mph Before starting the problem we note that it is clear that a headwind will certainly make you go slower Can you draw vectors to show this is true Now for the problem let s investigate what happens when we have a cross wind VBG VBw VWG This problem isn t really accurate because we have neglected the friction holding my tire to the road However it is true that depending on the magnitude of the wind cross winds and even partial tail winds can make it harder to pedal a bicycle Since any kind of head winds cross winds and even partial tail winds can make it harder to pedal it is much more probable that any wind will make riding more difficult Problem A lightweight plane can y 19 ms in still air The plane is trying to y east but there is a wind blowing 20 east of north at 15 ms a What direction must the plane fly in order to go east b How fast is the plane going relative to the ground Note again that VPG VPW VWG which means VPWx VWGx VPGx VPWy VWGy VPGy In this case the wind is a partial tail wind but the plane is still going slower than it would if there were no wind Chapter 2 An object goes from one point in space to another After it arrives at its destination its displacement is A either greater than or equal to B always greater than C always equal to D either smaller than or equal to E always smaller than the distance it traveled A car travels in a straight line covering a total distance of 900 miles in 600 minutes Which statement concerning this situation is true A The velocity of the car is constant B The acceleration of the car must be nonzero C The rst 45 miles must have been covered in 30 minutes D The speed of the car must be 90 miles per hour throughout the entire trip E The average velocity of the car is 90 miles per hour in the direction of motion Chapter 2 When is the average velocity of an object equal to the instantaneous velocity A This is always true B This is never true C This is the case when the velocity is constant D This is the case only when the velocity is increasing at a constant rate Which physical quantity is not correctly paired with its SI unit and dimension Quantity Unit Dimension A velocity ms L T B path length m L C speed ms L T D displacement ms2 L T 2 E speed gtlt time m L Chapter 2 Suppose that an object is moving with constant acceleration Which of the following is an accurate statement concerning its motion A In equal times its speed increases by equal amounts B In equal times it velocity changes by equal amounts C In equal times it moves equal distances D None of the above is true Starting from rest a particle that is con ned to move along a straight line is accelerated at a rate of 4 ms2 After 10 seconds how far will the particle have traveled A 20 m C 100 m E 400 m B 40 m D 200 m Chapter 2 A car starts from rest and accelerates at a constant rate in a straight line In the rst second the car covers a distance of 20 meters How much additional distance will the car cover in the second second A 20m C 60m E 125m B 40m D 80m Ball A is dropped from a window At the same instant ball B is thrown downward and ball C is thrown upward om the same window Which statement concerning the balls is necessarily true if air resistance is neglected A At one instant the acceleration of ball C is zero B All three balls strike the ground at the same time C All three balls have the same velocity at any instant D All three balls have the same acceleration at any instant E All three balls reach the ground with the same velocity Chapter 2 If you drop a brick from a building in the absence of air resistance it accelerates downward at 98 ms2 If instead you throw it downward its downward acceleration after release is A less than 98 ms2 B 98 ms2 C more than 98 ms2 D impossible to determine with the information given A ball is released from rest and falls 100 feet near the surface of the earth Neglecting air resistance how long will it take for the ball to fall the 100 feet A 250 s C 450 s E 100 s B 312 s D 625 s Chapter 2 A ball is thrown straight up from the surface of the earth with an initial speed of 196 ms Neglecting air resistance what maximum height will the ball reach before it begins to fall downward A 980m C 196m E588m B 147 m D 245 m Two balls are thrown straight up The rst one takes twice as long to return to earth as the second one Ignore air resistance How much faster was the rst ball thrown A 2 times as fast B Twice as fast C Three times as fast D Four times as fast E Impossible to tell without knowing the exact times Chapter 2 Two balls are thrown straight up The rst is thrown with twice the initial speed of the second Ignore air resistance How much higher will the rst ball rise A 2 times as high B Twice as high C Three times as high D Four times as high E Eight times as high Two rocks are dropped into two different deep wells The rst one takes three times as long to hit bottom as the second one Ignore air resistance How much deeper is the rst well than the second A 3 times as deep B Three times as deep C Four and a half times as deep D Six times as deep E Nine times as deep Chapter 2 A person standing at the edge of a cliff throws one ball straight up and another ball straight down at the same initial speed Neglecting air resistance the ball that hits the ground below the cliff with the greater speed is the one initially thrown A upward B downward C neither they both hit at the same speed D It is impossible to tell with the information given Ball A is dropped from the top of a building One second later ball B is dropped from the same building Neglecting air resistance as time progresses the difference in their speeds A increases B remains constant C decreases D depends on the size of the balls Chapter 2 Ball A is dropped from the top of a building One second later ball B is dropped from the same building Neglecting air resistance as time progresses the distance between them A increases B remains constant C decreases D depends on the size of the balls An object is moving along a straight ne The graph at the right shows its position E from the starting point as a function 0 l 2 3 4 5 6 of time I seconds In what section of the graph does the object have the fastest speed A AB C CD E AB and CD B BC D DE Chapter 2 An object is 40 moving along a 30 straight ne The a 20 graph at the right B shows its position E from the starting point as a function 0 l 2 3 4 5 6 of time I seconds What was the instantaneous velocity of the object at I 4 seconds A 60 ms C 100ms E 40 ms B 80 ms D 133 ms An object is 20 moving along a Q 15 straight ne The Q 10 graph at the right gt 5 shows its velocity as a function Of 0 0 1 2 3 4 5 6 time I seconds During which intervals of the graph does the object travel equal distances in equal times AOt02s DOt02sand3st05s B2sto3s E0t02s3st05sand5sto6s C3st05s Chapter 2 The graph shows position as a function of time for two trains running on parallel tracks Which is true A At time IE both trains have the same velocity B Both trains speed up all the time C Both trains have the same velocity at some time before 3 D Somewhere on the graph both trains have the same acceleration Chapter 7 A rubber ball and a lump of putty have equal mass They are thrown with equal speed against a wall The ball bounces back with nearly the same speed with which it hit The putty sticks to the wall Which object experiences the greater momentum change A The ball B The putty C Both experience the same momentum change D Cannot be determined from the information given A person attempts to knock down a large wooden bowling pin by throwing a ball at it The person has two balls of equal size and mass one made of rubber and the other of putty The rubber ball bounces back while the ball of putty sticks to the pin Which ball is most likely to topple the bowling pin A the rubber ball B the ball of putty C makes no difference D need more information Chapter 7 Suppose a pingpong ball and a bowling ball are rolling toward you Both have the same momentum and you exert the same force to stop each How do the time intervals to stop them compare A It takes less time to stop the pingpong ball B Both take the same time C It takes more time to stop the pingpong ball Consider two carts of masses m and 2m at rest on an air track If you push rst one cart for 3 s and then the other for the same length of time exerting equal force on each the momentum of the light cart is A four times B twice C equal to D onehalf E onequarter the momentum of the heavy cart Chapter 7 Consider two carts of masses m and 2m at rest on an air track If you push rst one cart for 3 s and then the other for the same length of time exerting equal force on each the kinetic energy of the light cart is A larger than B equal to C smaller than the kinetic energy of the heavy cart A 65 g tennis ball moving to the right with a speed of 15 ms is struck by a tennis racket causing it to move to the left with a speed of 15 ms If the ball remains in contact with the racquet for 002 s what is the magnitude of the force experienced by the ball A zero B 975N C 163N D 163 X105 N E 98 gtlt104 N Chapter 7 A small car meshes with a large truck in a headon collision Which of the following statements concerning the magnitude of the collision force is correct A The truck experiences the greater average force B The small car experiences the greater average force C The small car and the truck experience the same average force D It is impossible to tell since the masses and velocities are not given A compact car and a large truck collide head on and stick together Which undergoes the larger momentum change A car B truck C the momentum change is the same for both D you can t tell without knowing the nal velocity and combined mass Chapter 7 A compact car and a large truck collide head on and stick together Which undergoes the larger acceleration during the collision A car B truck Cboth experience the same acceleration D you can t tell without knowing the nal velocity and combined mass Suppose the entire population of the world gathered together at one spot and everyone jumps up at the same time While all the people are in the air does the earth gain momentum in the opposite direction A No the inertial mass of the earth is so large that the planet s change in motion is imperceptible B Yes because of its much larger inertial mass however the change in momentum of the earth is much less than that of all the jumping people C Yes the earth recoils like a ri e ring a bullet with a change in momentum equal and opposite to that of the people Chapter 7 Suppose the entire population of the world gathered together at one spot and everyone jumps up at the same time When the 5 billion people land back on the ground the earth s momentum is A the same as what is was before the people jumped B different from what is was before the people jumped A 30 kg cart moving to the right with a speed of 10 ms has a head on collision with a 50 kg cart that is initially moving to the left with a speed of 2 ms After the collision the 30 kg cart is moving to the left with a speed of 1 ms What is the nal velocity of the 50 kg cart A zero B 08 ms to the right C 08 ms to the left D 20 ms to the right E 20 ms to the left Chapter 7 Suppose you are on a cart initially at rest on atrack with no friction You throw balls at a partition that is rigidly mounted on the cart If the balls bounce straight back as shown is the cart put in motion A Yes it moves to the right B Yes it moves to the left C No it remains in place A 50 g lump of clay moving horizontally at 12 ms strikes and Sticks to a statiome 100 g cart whichcan move on a frictionless air track Determine the speed of the cart and the clay after the collision A 2 ms B 4 ms C 6 ms D 8 ms E 12 ms Chapter 7 A 1000 kg car traveling east at 20 ms collides with a 1500 kg car traveling west at 10 ms The cars stick together after the collision What is their common velocity after the collision A 1 ms west B 2 ms east C 4 ms east D 6 ms west E 16 ms east While driving on a one way street you notice an identical car coming at you at the same speed as you are going You can either hit the car head on or swerve and hit a massive concrete wall also head on In the split second before impact you decide to A hit the other car B hit the wall C hit either one it makes no difference D consult your lecture notes Chapter 7 Tightrope walkers walk with a long exible rod in order to A increase their total weight B allow both hands to hold onto something C lower their center of mass D move faster along the rope E lower their potential energy A plane ying horizontally releases a bomb which explodes before hitting the ground Neglecting air resistance the center of mass of the bomb fragments just after the explosion A is zero B moves horizontally C moves vertically D moves along a parabolic path E not enough information to determine the path Chapter 4 Lecture Notes Physics 2414 Strauss Formulas IF ma FG mg FSF S usN FKF KN Main Ideas 1 Newton s Three laws 2 Weight 3 Solving Problems 4 Friction In this chapter we will study Newton s laws of motion These are some of the most fundamental and important principles in physics 1 Newton s First Law Every object continues in its state of rest or of uniform speed in a straight line unless it is compelled to change that state by a net force acting on the object First note what is not stated or implied by Newton s first law 0 This doesn t say that every moving object has a force acting on it o This doesn t say that a stationary object has no forces acting on it This says that if an object is stationary it will stay stationary and if it is moving it will stay moving in a straight line unless a net force acts on it Therefore if an object is stationary or moving in a straight line at a constant velocity you can be sure that the net force acting on the object is zero A net force is the vector sum of all the forces on an object A nonzero net force means that the vector sum of all the forces is not zero A book sitting on a table has a net force equal to zero because it is in a state of rest The table is exerting a force upward and gravity is exerting a force downward so that the vector sum of the forces is zero Note that the object exerting a force does not have to be moving but can be stationary like the table Note also that an object that is at rest or an object that is maintaining a constant velocity are equivalent in that both need a force to change them and that in both cases the net force on the object is zero A net force on an object is not needed to maintain the velocity only to change it Let s look at an example of what Newton s first law implies about the motion of objects Suppose you were to push a book and let it slide across a table after brie y pushing it Why does it stop According to Newton s first law there must be a net force slowing it down which in this case is caused by friction But what if there was no friction Then what would keep the book going The answer is that nothing keeps the book going That s the point Once the object starts in motion it will continue in motion unless it experiences a net force It does not take any magic force for the object to continue in motion The push from your hand only acts on the book during the brief time your hand is touching the book It is not the push from your hand that keeps the book moving but the fact that once the book is set in motion it will continue in a straight line at the same speed unless a net force acts on it The force required to change a body s state of motion is a measure of the inertia of the body That SI unit of inertia is called mass It is harder to change the motion of something with more mass whether or not there is air resistance gravity etc So even in the vacuum of space it is harder to change the motion of a spaceship than a wrench There is the force of gravity in space just not air resistance More about that later 2 Newton s Second Law In chapters one through three we discussed how an object moves when it is experiencing a constant acceleration However we never discussed what causes an object to accelerate Newton s second law gives an explanation for what causes an object to undergo an acceleration It is the net force applied to the object The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass The direction of the acceleration is in the direction of the applied net force a ZFm usually written IF ma You should understand what this formula means Remember that the formulas in physics have a physical meaning They describe the physical universe This formula indicates that if the vector sum of all the forces on an is zero the object will not accelerate If the vector sum of all the forces on the object is not zero the object will accelerate To use this formula we draw a free body diagram showing all of the forces acting on a certain object The sum of the forces on this object and the mass of this object will determine its acceleration We do not consider the forces that this object exerts on other objects but only the forces exerted on the object in question If all the forces add up using the rules for adding vectors to be zero then there is no acceleration If they do not then there is an acceleration in the direction of the net force The converse of this is true as well If an object is accelerating the net force on it must not be zero On the right hand side of the equation always use the mass of the object in question the object which has the forces acting on it which will cause an acceleration of that object The SI unit of force is kgms2 Newtons N One Newton is the force required to accelerate one kg one meter per second per second The British unit of force is the pound Note that pounds and kg are not a measure of the same quantity Pounds is a measure of force and kilograms is a mass The pound is the unit of weight when the acceleration is g 321734 st s2 The unit of mass in the British system is the slug which is the mass that undergoes an acceleration of 1 stZ when a force of 1 pound is applied to it In an environment where gravity is different like on the moon the weight of an object changes Newtons or pounds but the mass stays the same kg or slugs lbs slugs X g Using measures of force we find that 1 lb E 448 N Note that the Newton s first law is a special case of the Newton s second law When IF 0 a 0 and so AV 0 because a AVAt Because the acceleration of an object is one variable in Newton s second law and also in the kinematic equations of chapters 2 and 3 we will often use the kinematic equations to find the acceleration which will be used in Newton s second law or Vice versa that is find the acceleration from Newton s second law and use it to solve the kinematic equations Problem What is the average net force exerted by a shot putter on a 70 kg shot if the shot is moved through a distance of 28 m and is released with a speed of 13 ms Also note that the second law is a vector equation This means that for problems involving two dimensions Newton s second law actually implies two separate equations When working with Newton s law in more than one dimension you must always break down all vectors into their components along different directions and use two equations separately for example ZFy may and ZFX max Do not combine forces or accelerations in one direction with forces or accelerations in another direction Problem Two forces F1450 N and F2250 N act on a 500 kg block sitting on a table as shown a What is the horizontal acceleration magnitude and direction of the block b What happens in the vertical direction 629quot F2 3 Newton s Third Law Whenever one object exerts a force on a second object the second object exerts an equal and opposite force on the first It is important to realize that the equal and opposite forces described by Newton s third law are not exerted on the same body For instance when a nail is hammered the hammer exerts a force on the nail which drives the nail into the wood and the nail exerts an equal and opposite force on the hammer which stops the hammer when it hits the nail When determining if something will move like will the nail go into the wood we must always look only at the forces acting on that object That is if you want to determine the motion of the nail you must consider the force of the hammer on the nail but you can not consider the force of the nail on the hammer Newton s third law always deals with two different objects the force object A exerts on object B is equal and opposite to the force object B exerts on object A Problem An astronaut who is walking in space pushes on a spaceship with a force of 36 N The astronaut has a mass of 92 kg and the spaceship has a mass of 11000 kg What happens Problem Suppose I throw a baseball My body exerts a force on the baseball and the baseball exerts a force back on my body How does the force of the ball on my body compare with the force of my body on the ball Is my body accelerated back If not why not Note the difference between Newton s second law and Newton s third law Newton s Second law has to deal with all of the forces acting on a single object Newton s third law has to do with the forces acting and reacting on two different objects 4 Force of Gravity and the Normal Force We know that an object near the surface of the earth will undergo an acceleration g 98 msz Using Newton s second equation we see that for a body near the earth the force of gravity is Fmamg since the acceleration of an object near the surface of the earth is g This implies that the force acting on an object of mass m near the surface of the earth is equal to mg and is directed toward the center of the earth This is what is often called the object s weight The weight is the force of the earth or other celestial body on the particular object Remember that the British unit pounds is a force but the SI unit kilograms is a mass So an object of 1 kg near the surface of the earth has a weight of F mg lkg 98 ms2 98 N which is 22 pounds On the moon where the gravity is 16 of the earth s the mass will still be 1 kg Mass never changes no matter what the gravity However the weight will now be F lkgl6 ms2 16 N which is 37 pounds The weight which is a force depends on the acceleration of gravity but the mass is constant no matter what the force of gravity is even when there is no gravity at all Remember that even if there is no force of gravity it still takes more force to accelerate a spaceship of 11000 kg than to accelerate a man of 92 kg at the same rate Now suppose an object is placed on a table What are the forces acting on this object Certainly gravity is acting on it But if there was only gravity then the object would accelerate down by Newton s second law IF m 3 There must be a force pushing up on the book or it would accelerate down This is the force of the table It pushes up against the book It is equal and opposite to the force of the earth pulling on the book It is perpendicular to the table s surface so we call it the normal force Normal means perpendicular This is not the equal and opposite force from Newton s third law because we are dealing with two different forces acting on the same object the book What is the equal and opposite force of the earth pulling on the book It is the book pulling on the earth As we will see in chapter 5 the book really does pull up on the earth Problem What is the normal force for a book with a mass of 080 kg Is the normal force always equal to the weight Problem Suppose I tie a string to the book and lift it with a force of 52 N What is the normal force then Problem If I push down with 52 N what is the normal force 5 Solving Problem 5 There are some basic steps in solving problems using Newton s Laws see pages 87 and 100 Draw a diagram Consider only the forces acting on one object at a time and draw a free body diagram 3 Determine the components of each of the forces and set 2F ma in each direction If there is no acceleration along a particular direction then 2F 0 along that direction 4 Determine if there are other equations that need to be used eg kinematic equations to get the acceleration 5 Solve the equations Jp A Problem A man is pulling a wagon at a constant rate of 2 ms with a force of 52 N with the handle at an angle of 30 What is the force of friction holding the wagon back FP52N Ff 4 Problem If the wagon has a mass of 12 kg what is the normal force When working with more than one object Draw free body diagrams of all the objects separately write equations from each from body diagram and look for relationships between the different objects PROBLEM Consider a block on a frictionless table with a mass hanging down as shown a What is acceleration and tension in the cord b How long will it take the block on top to move 070 m Look at block falling down draw a free body 2g The tensions must be the same and the accelerations must be the same so for block 1 tension and acceleration are in the same direction Problem A train is given an acceleration of 5 msz What is the tension between the two cars with a mass of 2000 kg 6 Friction and Inclines In most cases we need to include friction For many surfaces the force of sliding friction is proportional to the normal force That is as the normal force increases the force of friction increases linearly When this is true we write Ffr ocFN with the proportionality constant called Mk the coefficient of kinetic friction This is the friction when the object is moving FKF lLLkFN39 If I push a table and it doesn t move there must also be a force pushing back That force comes from friction as well It is the friction when the object is not moving or static called the coefficient of static friction It is always greater than or equal to the coefficient of kinetic friction It is often harder to start something sliding than to keep it sliding due to the difference in these coefficients F SF SI F N Problem A 20 kg sled is being pulled across a horizontal surface at a constant velocity The pulling force has a magnitude of 800 N and is directed at an angle or 300 above the horizontal Determine the coefficient of kinetic friction We know that FfVLk FN or ch Ff FN so to solve the problem we need to find what F f and FN are Many problems involving friction also involve surfaces at an incline In such a case it is almost always a good strategy to set one aXis parallel to the surface and one aXis perpendicular to the surface Why Because the acceleration is along the surface Along the direction perpendicular to the plane then 2 F ma 0 and all of the acceleration is along the surface parallel to the plane ProblemDemonstration Calculate the acceleration of gravity using an inclined plane and an air track Consider an object on a inclined plane tilted at an angle of 0 Note that the angle between the normal and the vertical is also given by 0 We draw a free body diagram for the case when there is no friction between the object and the surface Inclined plane problems are usually easier to solve if we chose the x aXis along the incline In general always chose the direction of acceleration as one of the axes N y Wy mgsinQ Since the acceleration is along this x aXis let s look at the forces in that direction 2F ma mgsinQ ma If we know a and 0 we can use this to measure g Using kinematic equations in that x direction gives xx 0 vt l2atz or with v00 a 2xx0t2 Plugging this into 2F ma from above gives mg sin0 m 2xx0t2 g 2x x0t2sin0 Problem An 180 kg box is released on a 370 incline and accelerates down the incline at 0270 msz Find the coefficient of friction and the frictional force Friction is not always a force that will slow something down Some times friction is the force that causes things to go Think of an example like a car driving along the road or a dog standing up in the back of a pickup truck 7 Solving Some Harder Problems Problem An electric motor is lowering a 452 kg crate with an acceleration of 160 msz What is the tension in the cord The only tricky part of this is to realize that all of the cords have the same tension so the free body diagram looks like this and with tension in the opposite direction from the acceleration Problem Given the following situation with uk 050 what is the acceleration of m1 and m2 Problem A 30 kg mass hangs at one end of a rope that is attached to a support on a railroad car When the car accelerates to the right the rope makes an angle of 40 with the vertical Find the acceleration of the car bC Chapter 9 Lecture Notes Physics 2414 Strauss Formulas Stress ForceArea FA Strain ALL0 E FLOAAL StressStrain G FL0 AAL StressStrain B AP V0 AV F kAL 1 Static Equilibrium When the forces on an object are balanced so that the object does not have any acceleration or angular acceleration we say the object is in equilibrium The object is either at rest or in a state of constant velocity andor angular velocity So for an object to be in a state of equilibrium the net force and the net torque must be zero If there were a net force or a net torque applied to the object it would have an acceleration or an angular acceleration respectively This means that three equations must be satisfied for an object to be in equilibrium 2F 0le7y 0210 The torque can be measured relative to any axis of rotation We must be careful to look only at the forces and torques acting on the object and not any of the forces or torques exerted by the object Remember that we can consider the force of gravity acting on an object as completely acting at the point of its center of mass Problem A uniform 400N board supports two children weighing 500 N and 350 N The support is under the center of gravity of the board and the 500N child is 150 m from the center a What is the force which the support exerts b Where should the 350 N child sit to balance the board N SOON mg40N 350N Problem A traffic light hangs from a structure The pole is L 75 m long and has a mass of 80 kg The mass of the light is 110 kg Determine the tension in the horizontal cable and the vertical and horizontal components of the force on the pivot 38m I Problem A 50 N weight is held in a person s hand with the forearm horizontal The bicep muscle is attached 00300 m from the joint The weight is 035 m from the joint The mass of the forearm is 13 kg with the center of mass located 17 m from the joint What is the force of the humerous and the bicep on the arm Problem A rod is hanging from two cords What is the tension in the first string right after the second string is cut 2 Balance The rest of this chapter consists of concepts you should be familiar with which may have medical applications such as human balance for physical therapy and bone fracture but are not fundamental concepts in physics Consequently we will cover them brie y but not in detail Something that is not moving or moving at a constant velocity is said to be in equilibrium because the net vector force acting on the body is zero If I change an object s position slightly it can do one of three things 1 Return to its original position called stable equilibrium 2 Stay in its new position called neutral equilibrium 3 Move farther away from its original position unstable equilibrium Examples of each are l tipping a chair over slightly 2 sliding a chair along the oor 3 a pencil balanced on its tip An object will be in stable equilibrium when a line projected downward from the center of mass of the object falls within the base of the support This is why a person shifts their weight when carrying a heavy object 3 Stress Strain and Fracture l Definitions Stress ForceArea FA Pressure applied to an object Strain ALL0 Percentage that something has deformed 2 Results of Stress Elastic Limit The relations below hold and Hooke s law is valid F kAL Plastic Limit The relations below do not hold and the material is permantly deformed Fracture Forces exceed maximum limits derived from the relations below and material breaks 3 Types of Stress Appropriate Constant o Tensile Stress or Young s Modulus Compressive Stress E FLOAAL Stress Strain o Shear Stress Shear Modulus G FL0 AAL StressStrain o All directions Bulk Modulus B AP V0 AV For all situations we have two definitions we need to know The first is the stress on an object defined as the force per unit area Stress FA Next is the strain defined as how much the length of the object has been changed compared to its original length Strain ALLO When I exert forces on an object there are three things that can happen 1 The object can go back to its original form If the object does that then I have only strained the object within its elastic limit 2 The object can retain its new shape If this happens we say that I have strained the object within its plastic limit 3 it can break The object has been strained too much and fractures When 1 occurs the material obeys Hooke s law F kAL The material will retain its original shape when the forces are removed How much it moves from its original position when the forces are on it depend on the size and shape of the body and how the force is applied The force can be applied in three ways tension compression and shear The change in shape for each of these three ways is determined from the elastic modulus for tension and compression and from the shear modulus for shear If the pressure is distributed on all parts of the object the volume of the material changes and the bulk modulus is used See table 91 for various values Finally if the stress exceeds some maximum the object will break The maximum stress depends on whether there is a tensile compressive or shear force Maximum strengths are given in table 92 Problem A nylon tennis string on a racquet is under tension of 250 N If it has a diameter of 100 mm by how much is it lengthened from its untensioned length of 300 cm Young s modulus for nylon is 5 X 109 Nmz Problem What is the maximum tension possible in a 100 mm diameter nylon tennis racquet string Tensile strength for nylon is 500 X 106 Nmz Chapter 7 Lecture Notes Physics 2414 Strauss Formulas p mv 2F ApAt FAt Ap Zpi pr xCM me 2m yCM Zmy 2m Main Ideas Momentum and Impulse Conservation of Momentum Elastic Collisions Inelastic Collisions Center of Mass and motion LIIbLIJlJ A 1 Momentum and Impulse Last chapter we talked about the fundamental law of the conservation of energy Energy is always conserved in any process This chapter will introduce another conservation law which is not applicable in all circumstances but in many important ones That is the conservation of linear momentum First we have to define what linear momentum is Linear momentum p is defined as the mass of an object times the velocity of that object It is a vector having the same direction as the velocity of the object We use momentum in this way in everyday language A 300 lb lineman has more momentum than a 180 lb anker when they are both moving at the same velocity because the lineman has more mass A car going 60 mph has more momentum than one going 10 mph because it has a higher velocity Now consider Newton s second law IF ma mAVAt mv2 v1At mv2 mv1At p2 p1At So Newton s second law can be restated as the rate of change of momentum of a body is proportional to the net force applied to it In fact this way of stating Newton s law as the change in momentum is how Newton originally formulated his second law I can rewrite this equation as ZFAI Ap In physics you should understand that the equations mean something This equation says that the net force acting on an object over a period of time At produces a change in momentum Ap Often when the momentum of an object is changed it is because there was a very large force acting on the object for a very brief time or the force of impact is in a different direction from other forces acting on an object This happens when you hit a ball or bounce a ball or hammer a nail or in many collisions In such cases we neglect all forces except the major force changing the momentum and write dropping the summation where the quantity F is the major force changing the momentum and FAt is called the impulse It is the quantity which gives the impetus necessary to change the momentum Problem A 50g golf ball is struck with a club The ball is deformed by about 20 cm during the time of collision and the ball leaves the club face with a velocity of 44 ms a What is the impulse during collision b How long is the collision c What is the average force during the collision Problem A 100g ball is dropped from 200 m above the ground It rebounds to a height of 150 m What was the average force exerted by the floor if the ball was in contact with the floor for 100 X 10392 s 2 Conservation of Momentum One of the most important concepts in understanding conservation of momentum is to understand what we mean by a system A system will be defined by the person trying to solve the problem or look at the situation Suppose we look at a collection of objects that we want to define as a system If we have two billiard balls that could be the system A ball bouncing off the oor could consist of the ball and the oor as the system or I could define just the ball itself as the system Let s define two classes of forces The first are forces internal to the system These are forces between only objects that are in our defined system For example if our system consists only of two balls the only forces internal to the system would be the equal and opposite force between the two balls when they collide or the force of gravity between the two balls Other forces like the force of gravity from the earth or the normal force of the table on the balls are external to the system That is they require agents which are not a part of the system They require the earth to make gravity or the table to impart the forces to the system If there are no external forces we say that the system is isolated In many cases where a collision occurs the internal forces are much greater than the external forces If I hit a baseball or two cars collide the internal forces between the ball and the bat or between the two cars are much greater over the very brief period of time that the collision took place than the external forces like gravity In such a case the system acts as if it is isolated during the collision So we approximate collisions as isolated systems and treat them as isolated systems A major question then is what happens in an isolated or nearly isolated collision Look at two balls colliding Before collision After collision 0 lt0 lt0 Ogt mlvli mzvzi mlvlf mzvzf Looking at ball 1 and at ball 2 FlAt Apl mlvlf 39mivii FzAt Apz mzvzf 39mzvzi where F1 and F2 are the force of ball 2 on ball 1 and the force of ball 1 on ball 2 respectively during the collision From Newton s third law we know that the force of ball 2 on ball 1 F 1 is equal to the force of ball 1 on ball 2 Fz and the time of the collision is the same for each ball so we set the two equations above equal to each other with a minus sign EN F2At This leads to two important results First when two objects collide the magnitude of the change in momentum for each object is exactly the same Apl 39Apz Second when we set the two equations equal to each other we get mlvlf 39mivii 39mzvzf 39mzvzi or mlvli m2V2i mlvlf mzvzf which generalizes for any number of objects to be Zpi pr which is the law of the conservation of momentum It states The total momentum of an isolated system of bodies remains constant Problem A 90kg fullback attempts to dive over the goal line with a velocity of 600 ms He is met at the goal line by a 110kg linebacker moving at 400 ms in the opposite direction The linebacker holds on to the fullback Does the fullback cross the goal line Since pi pf is a vector equation if we have a collision in two dimensions we must conserve momentum in each direction So the conservation of momentum can be written as two equations pxi pr and pyi p f We must make the momentum initially and finally equal in both the x direction and the y direction Problem A firecracker weighing 100 g initially at rest explodes into 3 parts One part with a mass of 25 g moves along the x aXis at 75 ms One part with mass of 34 g moves along the y aXis at 52 ms What is the velocity of the third part 3 Elastic Collisions In a certain class of collisions both the momentum and the kinetic energy are conserved That is the total momentum and the total kinetic energy are the same before and after the collision The only collisions that conserve kinetic energy are elastic collisions All collisions that can be considered isolated conserve momentum Remember that momentum is a vector so the total momentum must be summed like a vector Kinetic energy is a scalar so the total energy is summed like regular numbers An example of a nearly elastic collision is a super ball Problem A ball with a mass of 12 kg moving to the right at 20 ms collides with a ball of mass 18 kg moving at 15 ms to the left If the collision is an elastic collision what are the velocities of the balls after the collision 4 Inelastic Collisions An inelastic collision is one in which kinetic energy is not conserved A collision is called a completely inelastic collision if the objects stick together after colliding In an inelastic collision kinetic energy is not conserved However linear momentum is conserved even in inelastic collisions DemonstrationProblem The ballistic pendulum is used to determine the velocity of a projectile Suppose we have a projectile with mass m and velocity v1 and a pendulum with mass M The projectile hits the pendulum and sticks to it so that the projectile and pendulum have a mass m M and final velocity v The pendulum then rises a distance h from its original position From conservation of momentum during the collision we know that mv1 mMv After the collision then mechanical energy is conserved and we have Ki Uf l2mM v2 mMgh 12 mzvlzmM mMgh v1 Zgh mMm Now h is not easy to measure but it is given by hl R R IR cos0 gt Z Rcos0 h R R cos0 v1 2gR lcos0 mMm If the gun is horizontal then from kinematic equations we know y vyit l2gt2 y l2gt2 t V 2yg x vlt 2x yR lcos0 mMm 5 Center of Mass We have talked about the motion of an object a lot in this class We have usually used simple objects and described their movement But sometimes the movement of an object is not so simple like a spinning wobbling tossed frisbee How do we describe the motion of that object The motion of most of the points on the object is quite complicated However there is one point of the object which behaves in the same way as a single particle would move subject to the same forces That point of the body is called the center of mass Even with rotating and spinning the center of mass moves in the same way that a single particle would move That is the same way as all of We find the center of mass of an object along a certain aXis by using the equation XCM 2mx 2m or yCM Zmyy 2m Problem If the mass distribution of a person sitting down with his legs outstretched can be approximated by the two rectangles where is the person s center of mass 10 cm 0 65 k 0 X 25 kg 45 cm Q 6 cm 4 N b 20 cm 40 cm Note a few things about this example 1 The CM may be outside the body 2 The CM depends on the shape of the body If this person stands up the center of mass will change location relative to his body In fact it will then be somewhere in his torso The high jumper described in the book can actually clear the bar without the center of mass clearing the bar A basketball player or gymnast in ight has their center of mass follow the same path that a single particle would follow parabolic near the earth if we neglect air resistance although their extremities may be changing direction and they may be twisting and tumbling This means that even if there are internal force like muscles pulling the motion of the center of mass of the system is governed by the external forces only Suppose I shoot fireworks up in the air and they explode What is the motion of the center of mass of all the particles What is the motion of each particle Look at the rocket example in the book Case where mII mI Problem Suppose that mII 3m Where would mI land mI would still land straight down and now the CM must be at 2D so Note that every one of the individual objects as well as the center of mass of all the objects follow a parabolic path A straight line down is also a type of parabola Chapter 8 Steve S and his brother Mark M are riding on a merrygoround as shown A They have the same speed but different angular velocities B They have the same speed and the same angular velocities C They have different speeds and different angular velocities D They have different speeds and the same angular velocity The record playing on the turntable is rotating clockwise as seen from above After turning it off the turntable is slowing down but hasn t stopped yet The direction of the acceleration at point P is A B 39 P C gt ml ET Chapter 8 The diagram shows a door that is 2 m wide A force of 10 N is applied at the point shown What is the magnitude of the torque on the door with respect to the hinge 15 m a 30 10 N Hinge L 47 A zero B 13 Nm C 75 Nm D 10 Nm E 17 Nm The diagram show the top View of a door that is 2 m wide Two force are applied to the door as indicated in the diagram What is the magnitude of the net torque on the door with respect to the hinge 10 m ION O Hinge 60 60 A zero B 100 Nm 10 N C 40 Nm D 260 Nm E 87 Nm Chapter 8 A force F is applied to a dumbbell for a time interval At first as in a and then as in b In which case does the dumbbell acquire the greater centerofmass speed F F a gt b A a B b 0 C no difference A force F is applied to a dumbbell for a time interval At first as in a and then as in b In which case does the dumbbell acquire the greater energy F gt F a b 3 A a B b 0 C no difference Chapter 8 A hollow cylinder of mass M and radius R rolls down an inclined plane A block of mass M slides down an identical inclined plane If both objects are released at the same time A the block will reach the bottom rst B the cylinder will reach the bottom rst C the block will reach the bottom with greater kinetic energy D the cylinder will reach the bottom with greater kinetic energy E both the block and the cylinder will reach the bottom at the same time A solid sphere S a thin hoop H and a solid disk D all with the same radius are allowed to roll down an inclined plane without slipping In which order will they arrive at the bottom The st one down listed rst A HDS B HSD C SDH D SHD E DHS Chapter 8 In what circumstances can the angular velocity of a system of particles change without any change in the system s angular momentum A This cannot happen under any circumstances B This can happen if a net extemal force acts on the center of mass C This can happen if the only forces acting are internal to the system D This can happen if an external net torque is applied properly to the system E This can happen if there are only conservative forces acting on the system An ice skater performs a pirouette by pulling her outstretched arms close to her body What happens to her moment of inertia about the axis of rotation A It does not change B It increases C It decreases D It changes but it is impossible to tell which way Chapter 8 An ice skater performs a pirouette by pulling her outstretched arms close to her body What happens to her angular momentum about the axis of rotation A It does not change B It increases C It decreases D It changes but it is impossible to tell which way A ball on a string is rotating in a circle The string is shortened by pulling it through the axis of rotation What happens to the angular velocity and the tangential velocity of the ball angular veloci tangential velocig A increases decreases B increases stays the same C increases increases D stays the same stays the same E stays the same increases Chapter 8 An ice skater performs a pirouette by pulling her outstretched arms close to her body What happens to her rotational kinetic energy about the axis of rotation A It does not change B It increases C It decreases D It changes but it is impossible to tell which way Physics 2414 Strauss Chapter 1 Lecture Notes 1 WHAT Is PHYSICS 11 Qualitative Description The goal of physics is to understand and explain the physical universe Physicists observe the physical world around them and try to categorize and understand the phenomena they observe 12 Quantitative Reasoning In physics we don t just want a general idea or description of how things work we want a precise understanding of physical phenomena To demonstrate that understanding we need to be able to describe events quantitatively For instance if I drop a ball from a certain height the ball will hit the ground after a short period of time Now I know that if I drop the ball from a higher height then the ball will take longer to hit the ground But to demonstrate that I really understand the phenomena of the ball dropping I have to be able to answer quantitative questions If I drop the ball from twice the height how much longer will it take for the ball to hit the ground The answer is found in an equation we will encounter in chapter 2 y vot l2gt 2 We will discuss this equation in detail in the coming weeks but basically it says that if I double the height the ball will take about 14 times longer to reach the floor Without the quantitative equation I can t really make precise statements about the physical phenomena The goal of physics is to predict the future That is given a known set of circumstances what will occur as a result of the applicable physical phenomena As a physicist I only will say that I understand something if I can make quantitative predictions based on known laws and theories 13 Interpreting Formulas Some students think physics is difficult because of the mathematics used Yet it is the mathematics itself which allows us to understand the world and make predictions about how things will behave It is very important then that you understand the equations and formulas They have a meaning For instance in chapter 2 we will encounter the equation Ax vt This equation has a meaning It says that if I am moving at a constant velocity v then the distance I move Ax is equal to that velocity times the time If I were to move for twice the time I would cover twice the distance If I double the velocity I double the distance If we cut the time in half we cut the distance in half It is quantitative If I double my speed from 30 mileshour to 60 miles per hour then it only takes 12 the time to go 20 miles Also the symbol means something As a student you should always know what the symbol means or you can t use the equation It doesn t matter which symbol is used as long as you know what it means It makes sense to use I for time but one could use anything 14 Deriving Formulas There are a few very fundamental laws in physics Much of the rest of physics can be derived from these few fundamental laws Occasionally I or the book will actually derive one equation from a set of other equations Why do we do this To show the relationships between the more fundamental laws and the derived laws 2 UNITS Every measurement or quantitative statement requires a unit If I say I am driving my car at a speed of 30 that doesn t mean anything Am I driving it 30 mileshour 30 kmhour or 30 stec 30 only means something whenI attach a unit to it What is the speed of light in a vacuum 186000 milessec or 3gtlt108 msec The number depends on the units 21 SI Units We will most use the International System of Units Systeme International SI units These consist of the meter length the second time and the kilogram mass Each of these have prefixes and suffixes which you have encountered and should be familiar with For instance centi means 102 So a centimeter is 1gtlt10392 meters 002 meters A kilogram is 1gtlt103 grams 1000 grams There is a table in the front cover of the book of suffixes and prefixes 22 Changing Units Occasionally you might have to change to a different set of units Units behave like any algebraic quantity and cancel when multiplication is performed Problem You are travelling 65 mileshour How fast is this in ftsecond meterssecond 23 Consistency of Units When working with formulas and solving problem you must make sure that the units are always consistent Problem You are travelling 30 meterssecond How far do you travel in one hour 3 SOLVING PROBLEMS IN PHYSICS 31 Significant Figures Notice in the above problem I wrote 11gtlt105 meters rather than 108gtlt105 meters Why did I do that BecauseI can only know my answer to a specific accuracy Every number has a number of significant figures That is what is the precision of the number The number 25 has two significant figures If a table is 25 inches wide that means it is more than about 245 and less than about 255 inches However if the table is 250 inches wide then I now have 3 significant figures The table is more than about 249 inches wide and less than about 251 inches wide 3100 has four significant digits but what about 310 Does it have two or three significant digits One reason we use scientific notation is to clarify the number of significant digits 31gtlt102 has two significant digits while 310gtlt 105 has three significant digits Suppose I have a rectangular garden which is 103 by 42 meters What is the area of the garden My calculator says it is 4326 square meters But that is way too accurate How could I know the area to 01 square meters when I only know the length to 1 meters The rules for using the correct number of significant figures are as follows 1 When multiplying or dividing numbers the answer has only as many digits as the number with the least number of significant digits So 32 X 563 18 Only two significant digits because the 32 has only two significant digits 2 When adding or subtracting the number of decimal places in the answer should match the number with the smallest number of decimal places So 326 43 73 32 Dimensional Analysis This topic is covered in Appendix B of the textbook Dimensional analysis helps you solve problems as well as check whether your solution is correct Since most values have some kind of units we can use these units to our advantage Suppose you want to calculate how far a car will travel that is going a certain speed for a certain time You can t remember if the equations for calculating this is x 12vt2 or x 12vt We write the dimensions of the quantities in square brackets using L for length and T for time L LTT 2 LT or L LTT L so the correct equation must not have the t2 term This still doesn t guarantee that the relation is correct only that it is not incorrect For instance the constant 12 doesn t have any dimensions so we don t really know what the constant is In fact it should be 1 here not 12 You can also use dimensional analysis to help check your answer For instance speed should be L T If it is not then there is a problem with your answer 33 Rapid Estimating Order of Magnitude Many times it is important to get an idea of what an answer may be without actually working out the exact number Often the answer you get this way will be perfectly adequate for the question being asked Even if it is not a rapid estimate will help you know if the exact answer you calculate is reasonable Problem How long would it take to drive a car around the world Problem How many quarters would it take to fill up Owen field Chapter 8 Lecture Notes Physics 2414 Strauss Formulas v Al At rA0At ra aT Av At rAa At ra ac v2 r a 2r a 600 at 0 wot 12oct2 0 12a0at 602 woz2a0 1 rF sine I Zmr2 I Mk2 L 03 217 cc ALAt K l2Ico2 Main Ideas Angular Quantities Rotational Kinematics Torque and Rotational Inertial Rotational Kinetic Energy Angular Momentum LIIbLIJlJ A 1 Angular Quantities This chapter will again deal with rotation like chapter 5 did We will look rigid bodies that have purely rotational motion Rigid bodies are objects that have a definite shape Purely rotational motion means that all points in the object rotate in circles They all rotate around a fixed axis of rotation Finally we will combine rotational motion with translational motion The concepts in this chapter are very similar to the concepts we learned in chapter 2 through 7 regarding kinematic motion in one and two dimensions Newton s Laws the conservation of mechanical energy and the conservation of momentum As we develop concepts in this chapter you should try to relate the concepts to those in previous chapters It will help you understand this chapter 1 1 ANGULAR DISPLACEMENT When an object moves along a straight path we describe how far it has moved by its displacement When an object rotates we describe how far it has rotated by its angular displacement 0 The mathematics of circular motion is much PointP at t2 simpler if we measure the angle in radians r rather than degrees One radian as defined as 0 Point P at t1 an angle whose arc length is equal to its radius or in general 0 lr where r is the radius of the circle and l is the arc length subtended by the angle 0 If the angle does a complete revolution then I 275V which is the circumference So 360 21 radians and l radian is about 573 Note that as the object rotates every point on the object undergoes the exact same angular displacement The angular displacement 0 will play the same role in angular kinematics that the displacement x played in linear kinematics It is customary to set counterclockwise rotations to have a positive angular displacement and clockwise rotations to have a negative angular displacement 1 2 AN GULAR VELOCITY The angular velocity is defined in relationship to the angular displacement in the same way that the linear velocity was defined in relationship to the linear displacement The average angular velocity is given by the Greek letter omega w and is defined as the rate of change of the angular displacement wA0At The instantaneous angular velocity is given by A9 d9 60 lim AHO A d 1 3 ANGULAR ACCELERATION Likewise the average angular acceleration is defined as the rate of change of the angular velocity and is given by the Greek letter alpha a 07 Am m and the instantaneous angular acceleration is given by Aw da 06 lim AHO A d 1 4 RELATIONSHIP BETWEEN ANGULAR AND LINEAR QUANTITIES Before using these angular quantities let s look at the relationship between them and the linear quantities we have used We know that velocity is defined as the change in distance divided by the change in time In this case the velocity is in the direction tangent to the circumference of the circle the tangential velocity VT so the distance is given by the arc length l lvT AlAt rA0At rw Also the tangential component of the acceleration is laTAvTAt rAwAt ra Recall the definition of centripetal acceleration with total acceleration and since aT and aC are at right angles to each other the magnitude of the total acceleration is given by a liarz aczi 1 5 OTHER USEFUL DEFINITIONS Frequency is defined as the number of revolutions per unit time To change from frequency to angular frequency I must multiply frequency by 275 since one revolution is 275 radians l revs 275 radiansrev gives radianssec which is the units of angular velocity The period is the time required to make one complete revolution so Problem How fast is the outer edge of a CD at 60 cm moving when it is rotating at its top speed of 220 rads 2 Rotational Kinematics The definitions of the angular quantities are analogous to the definitions of linear quantities with 0 playing the role of x a playing the role of v and 05 playing the role of 51 Consequently the kinematic equations derived in chapter 2 are valid for rotational motion with constant angular acceleration using the same derivations done in chapter 2 We get a w0at 602 w02 2050 0 wot 12oct2 0 12a0 60t Problem A compact disk player starts from rest and accelerates to its final velocity of 350 revs in 150 s What is the disks average angular acceleration Problem How many rotations does the CD from the previous problem make while coming up to speed 3 Torque and Rotational Inertia We have talked about the description of angular motion Now we talk about the causes or dynamics of angular motion What causes something to rotate For translational motion we found that a force caused an acceleration That is a force caused an object to move For rotational motion it also takes a force to start something rotating However there is more to it than that The location thatI apply the force is important If I want to start a wheel rotating I can not apply the force at the aXle of the wheel It will not rotate To start something rotating I must apply a force that is some distance from the aXis of rotation The perpendicular distance from the aXis of rotation is called the lever arm The torque is defined as the force times the lever arm In general we write torque as 1 rFsinQ where 0 is the angle between the direction of the force and a line drawn from the aXis of rotation to the force If I were to push along the direction of F then there would be no rotation around the hinge since the force is directed right through the aXis of rotation If I were to push along F i then there would be a rotation Consequently the only component of the force F which causes a rotation is the component perpendicular to the lever arm V which is given by F sin0 So what produces a greater torque A force applied at some distance from the axis of rotation at some angle other than directly through the axis of rotation If I increase the lever arm I get a greater torque I need to do this to take off the drain plug from my car If I increase the force the torque will increase I can also maximize the torque by making the angle 0 90 Because torque has a distance in its definition the lever arm the torque is only defined around a certain axis of rotation The torque around two different axes of rotation may be very different Problem A crane picks up a heavy steel beam that is 100 m long The tensions in the cable is 200 X 104 N What is the torque around point 0 Problem Calculate the torque around an axis perpendicular to the paper through a point 0 and b point C 2cm 2cm Recall in our study of linear motion that we said that the property of a body that resisted a change in velocity was called mass Even if there was no gravity the mass of a body would resist a change in velocity That is what Newton s second law says IF ma There is also a property of a body which resists a change in angular velocity It is called the moment of inertia If a body has a large moment of inertia then it is difficult to change its angular velocity If it has a small moment of inertia it is easier to change its angular velocity The moment of inertia for any object depends on a number of factors including the object s mass its shape and the axis of rotation Let s first calculate the moment of inertia for a simple object like a sphere at the end of the string Since the sphere has much more mass than the string we will neglect the mass of the string We look at the tangential forces and accelerations FTmaT FT mra rFT mrza 1 mrza In this case the quantity mr2 is called the moment of inertia and given the symbol I So this becomes 1 I a If we have more then one source of torque we get 21 Ia which is the rotational equivalent of Newton s second law The reason we write I rather than mr2 is because I depends on different factors and although it is mr2 for a sphere on a string it is not mr2 for every object In general we can write the moment of inertia for a collection of particles as I Zmr2 This can be used to calculate the moment of inertia of a baton for instance but not for many other extended objects In that case calculus is needed Page 204 lists the moment of inertia for a few shapes In any problem you have to do you will be given the equation for the moment of inertia Problem Suppose a baton is 10 m long with each end weighing 03 kg Neglect the mass of the bar What is the moment of inertia for a baton a spinning around its center b spinning around one end of the baton Note that the moment of inertia depends on where the aXis of rotation is Another quantity is the radius of gyration It is given the symbol k and is defined as the location on the object that would have the same moment of inertia as the original object if all the mass where located at k The relationship between the moment of inertia and the radius of gyration is always given by I Mk2 Let s now solve some problems using Newton s second law in this form for rotational motion We use it the same way as we used Newton s second law for purely translational motion We must draw a free body diagram determine the torques and the accelerations and solve for the unknown quantities Up until now we have not been concerned about where a force acted on an object But because torques produce rotation around some aXis that depends on the distance from where the force is applied it is important to look at where a force is applied on an object For the force of gravity we assume the force is applied at the center of gravity of an object Other forces like tension friction and normal forces are applied where the objects contact Problem A cylindrical 300 kg pulley with a radius of R0400 m is used to lower a 200 kg bucket into a well The bucket starts from rest and falls for 300 s a What is the linear acceleration of the falling bucket b How far does it drop c What is the angular acceleration of the cylinder So these problems are solved just like other problems using Newton s second law Draw force diagrams Set 2F ma and 21 I a Break vectors into components if you need to Solve for the unknown quantities Remember that translational motions are produced by forces and rotational motions are produced by torques Sometime objects are constrained to rotate around a certain aXis because they are attached by hinges or something similar Sometimes objects are free to rotate about any aXis because the have no constraints like when you throw an object in the air An object free to rotate about any aXis will always rotate around its center of mass 4 Rolling Motion When an object is rolling we can observe some aspects about its motion Suppose the object is moving to the right with a speed of v Look at the linear speed of various points on the object Where the object touches the ground its linear speed is zero At the center of the wheel the linear speed is v and at the top of the wheel the velocity is 2v The angular velocity then is given by V260quot where v is the linear speed of the wheel What causes the wheel to rotate It is the force of static friction on the bottom It is static friction because the place where the wheel touches does not move sideways with respect to the ground The static friction produces a torque around the middle point of the wheel That is the aXis of rotation 5 Rotational Kinetic Energy We have learned that when something is moving it has a kinetic energy which is equal to l2mv2 When something is rotating it has a kinetic energy which is equal to 12 602 If it has both translational and rotational motion then it has both forms of kinetic energy as well Problem Two bicycles roll down a hill which is 20 m high Both bicycles have a total mass of 12 kg and 700 mm diameter wheels r350 m The first bicycle has wheels that weigh 6 kg each and the second bicycle has wheels that weigh 3 kg each Neglecting air resistance which bicycle has the faster speed at the bottom of the hill Consider the wheels to be thin hoops The only friction is static friction so there are no nonconservative forces Static friction involves no motion and since work is defined as W Fd when there is no distance involved there is no workenergy used Why does the bike with the lighter wheels have the faster speed Because each bike starts and ends with the same potential energy so they also must have the same total kinetic energy at the end More of the potential energy goes into rotational kinetic energy and less into translational kinetic energy when the wheel has more mass Because the translational kinetic energy is less the speed is smaller 6 Angular Momentum In the same way that linear momentum is defined as p mv so the angular momentum of a rigid body rotating is defined as L Ia Similarly Newton s law for rotational motion can be written as 21 Ia ALAt and just as linear momentum is conserved if there is no net external force so angular momentum is conserved it does not change if there is no net external torque The law of the conservation of angular moment says The total angular momentum of a rotating body remains constant if the net external torque acting on it is zero Some examples of this include an ice skater who is rotating and pulls her arms in or a diver who goes into a tucked position Why do they rotate faster 2Li ZLf 2160i 2160f Zmr izwi Zmr fzwf So as the distance from the aXis of rotation decreases the angular velocity 60 must increase forL to be the same Problem A student is sitting on a swivel seat and is holding a 20 kg weight in each hand If he is rotating at l revs 628 rads when the weights are held in outstretched arms 75 m from the aXis of rotation how fast is he rotating when he pulls the weights in to the aXis of rotation The rest of his body can be approximated as a cylinder with mass of 72 kg and radius of 25 m 7 Vector Nature of Angular Quantities The angular equivalent of all of the linear quantities which are vectors are also vectors This includes all of the angular quantities we have discussed except moment of inertia The direction is chosen arbitrarily by the right hand rule Your fingers rotate in the direction of the rotational motion and your thumb points in the direction of the vector Note however that the relationship between angular and linear quantities like v ra aT ra and ac 60 2r are not vector equations The linear quantities point in the direction of motion which changes as the object rotates but the angular quantities point in the direction given by the right hand rule 8 Inertial and Non Inertial Frames The Coriolis Force Any frame of reference where Newton s laws appear to work is called an inertial frame of reference Sometimes it may appear that his laws do not work For instance in a rotor carnival ride we may think that we are being thrown outward However there is no force which is throwing us outward instead we are being accelerated inward by the centripetal acceleration provided by the wall of the ride This reference frame where we think there is a force but there really is not one is called a noninertial reference frame and the force we think we feel is called a psuedoforce The Coriolis force is one such force Suppose I have a plate rotating around its center Points on the inside of a plate have a lower tangential velocity than those on the outside of the plate Consequently something thrown outward appears to curve in the opposite direction of rotation This is an important psuedoforce in understanding weather and in understanding the path of ballistic shells Ships at sea used to carry tables of values that they could use to estimate the Coriolis force depending on latitude and direction they wanted the shell to travel Then they could correct for the Coriolis force Chapter 6 Lecture Notes Physics 2414 Strauss Formulas W FHd Fd cos0 K E l2mv2 Wnet WNC WC 2 AK UG mgh WG AUG F kx UE l2kx2 WNc AK AU EKUl2mvzU EfEi P Wt F v Main Ideas 1 Definition of Work 2 Kinetic Energy and the WorkEnergy Theorem 3 Potential Energy 4 Conservative and Nonconservative Forces 5 Conservation of Mechanical Energy 6 Nonconservative Forces and Conservation of Energy 7 Power This chapter will deal with one of the most important concepts introduced this semester It is the concept of energy We will find that energy is always conserved in every reaction 1 Definition of Work In physics the term work has a very specific definition Work is de ned as the product of the component of force along the direction of displacement times the magnitude of the distance Note we use distance and not displacement We write this mathematically as W FHdchos0 where F H means the force parallel to the direction of motion 0 is the angle between the direction of the force and the direction of the motion and d is the distance traveled The SI unit of work is Nm which is given the name Joule l Nm l J Note that to do work you need to move some distance as well as a force acting with some component in the direction of motion There will always be an object or objects doing work on another object It is always necessam to determine what entity is doing the work on which object Work is only defined in relation to one thing doing work on another object To solve problems you must first determine this or you will not correctly solve for the work which is being asked for Problem Suppose I pull a crate with a force of 98 N at an angle of 55 above the horizontal for a distance of 62 m What is the total work done by me on the crate Note that F cos0 is the component of the force along the direction of motion Problem Suppose I now pull a crate with a force of 98 N at an angle of 55 above the horizontal for a distance of 62 m in the other direction What is the total work done by me on the crate in both directions If there is no force in the direction of motion then there is no work done on that object When we carry a package at a constant velocity a long distance we may say that we have done a lot of work but using the definition of work here we have done no work because the force of us holding the package up did not have any vector component parallel to the displacement The angle between the direction of force and the displacement was 90 and the cosine of 90 is 0 so WFd cos0 0 If I push on a wall and the wall does not move the work is 0 because the distance the object moved is 0 Work is a scalar but the sign of the work is important It is very important to ask who or what is doing the work We can either ask how much work is done on an object all the forces on the object or how much work does a certain object do For instance Problem A box weighing 23 kg slides down an incline of 25 at a constant velocity The box slides 15 m a What is the work done by the normal by gravity and by friction b What is the total work done on the box To solve a problem with work rst determine the force or forces that are doing the work Look at the question carefully Is it asking how much work a certain object is doing or how much total work is being done on an object Then determine the force or forces doing the work and the angle between the displacement and the force 2 Kinetic Energy and the WorkEnergy Theorem Energy is one of the most important concepts in physics Yet it is not an easy concept to describe simply There are many different kinds of energy Any one of the individual kinds of energy can be described but it is difficult to give an encompassing definition of energy In any process we nd that the total energy of the system never changes It is conserved However energy can change from one form to another When two cars collide the energy they had because of their motion called kinetic energy is changed primarily into heat In this chapter we will look at a number of types of energy The first will be the energy of motion called kinetic energy We look first at the kinematic equations Newton s second law and the definition of work We use vf2 v2 Zad gt a va vi22d and 2F ma Wm ZFd mad m va vi22dd 02me2 l2mv2 Kf KI The term l2mv2 is called the kinetic energy K E l2mv2 We say the net work done on an object is equal to its change in kinetic energy Note that this is the net work done on an object from all the forces acting on it This is known as the workenergy theorem The workenergy theorem only applies to all of the forces acting on a single object It does not apply to any single force Keep in mind when solving problems that kinetic energy is related to the speed squared of an object Problem How much work does it take to stop a 1000 kg car traveling at 28 ms Problem A 58kg skier is coasting down a slope inclined at 25 above the horizontal A kinetic frictional force of 70 N opposes her motion Near the top of the slope the skier s speed is 36 ms What is her speed at a point which is 57 m downhill 3 Potential Energy Another form of energy other than kinetic energy is potential energy Potential energy is the potential an object has to do work 31 GRAVITATIONAL The most common example of potential energy is the gravitational potential energy Suppose I raise a ball to a height h yf yi How much work has been done on the ball by me by gravity and in total WME Fd cos9 mg yfyi mgh The change in kinetic energy of the ball is 0 so the net work must also be 0 by the work energy theorem Because only me and gravity are doing work on the ball and the sum of the work done on the ball is zero we know that the work done by gravity must be equal but opposite in to the work I have done WG mg 2 2 mgyi mgyf U Uf where U is called the potential energy The book uses PE for potential energy but I will usually use U Note that this is initial height minus final height so we say WG mg Viyf Ui 39 Uf WG AU since AU Uf Ui Look at the definition of gravitational potential energy U mgy That means that the larger y is the higher something is above the ground the larger the potential energy is The book uses the symbol h to mean the object s height and then defines the potential energy of gravity or the gravitational potential energy as The gravitational potential energy is the potential to do work due to gravity If I raise a book above the ground it has the potential to do work If I drop it I could drive a nail I could attach it to a toy and drop it and accelerate the toy The object has the potential to do work due to gravity One important point to note is that there is no absolute scale for potential energy We always measure the height with respect to some reference point For instance if I define the gravitational potential energy to be 0 on the table then I raise the book it has a potential energy of mgh relative to the table Something which is higher has a higher potential energy than something that is lower It has more ability to do work SupposeI take two different paths to raise the book 1 meter One is straight up and the other is very circuitous Compare the ability to do work with this book from the two different methods It is exactly the same This illustrates one of the most important principles regarding gravitational potential energy GPE GPE is always the same at the same height regardless of the path taken to reach that height 32 ELASTIC SPRING Another form of potential energy is elastic potential energy EPE One example of this is the potential energy stored in a spring If I compress a spring it has the potential to do work Watches that wind up use this principle The amount of energy stored in a spring depends on how much it is compressed In many cases the force necessary to compress a spring is proportional to the distance it is compressed x The force of the spring is then given by F kx where k is a constant that depends on the construction of the spring The minus sign means that the spring wants to push in the opposite direction from the direction it was compressed Now suppose I compress the spring a distance x When I first start to compress it takes almost no force When it is very compressed it takes a force of kx to compress it The average force it takes is 12F initial Ffinal 12 kx 12 So the work it takes to compress the spring is W Fx 12 kx2 which is also called the elastic potential energy U 12kx2 where x is the distance the spring has been compressed or stretched This is the energy stored in a spring I can use it to do work So we have talked about two kinds of potential energy gravitational mgh and elastic l2kx2 4 Conservative and Nonconservative Forces As we said the work done to move an object in a gravitational field does not depend on the path taken Such a field is called a conservative field and the force here the force of gravity is called a conservative force If work done by a force in moving an object between two positions is independent of the path of motion the force is called a conservative force Kinetic friction is not such a conservative force because W Fd cos0 If I take a longer distance the work increases because d is always in the opposite of F for the case when friction is opposing the motion For gravity d can either be in the opposite direction of F or in the same direction and this contributes to the force being conservative Because potential energy is the energy associated with an object s location potential energy can only be defined for a conservative field We can extend the work energy theorem to include nonconservative forces as well In that case Wm Wc WNC AK recall that the net work done is the change in K where WC is the work done by conservative forces and WC is the work done by nonconservative forces For gravity we saw that WC AU and this is true in general for conservative forces so AU WNC AK WNc AK AU This is probably the most important equation to know from this chapter for it is a mathematical statement that energy is always conserved When we use this along with W Fd cosQ often for the nonconservative force and AK Wm we can solve many problems Also note that when there are no nonconservative forces WNC 0 and AK AU 0 which is the crucial equations derived in the next section If I drop something which smashes when it hits the ground what is happening to the energy in this case Initially there is a lot of potential energy then a lot of kinetic energy Finally the energy changes to nonconservative forms heat and deformation 5 Conservation of Mechanical Energy The equation above be rewritten as WNC AK AU WNC 02me2 l2mv2 sz U l2mvf2 Uf l2mv UI Ef E where E l2mv2 U is the defined as the total mechanical energy So the work done by nonconservative forces is equal to the change in mechanical energy If there are no nonconservative forces no friction heat etc so that the only external forces are gravitational and elastic then WNC 0 and we find that mechanical energy is conserved That is EfE If only conservative forces are acting the total mechanical energy of a system neither increases or decreases in any process It stays constant It is conserved This principle is not new but simply a restatement of WNC AK AU when there are no nonconservative forces However this principle is very important and can be used to solve many problems in physics where there are no nonconservative forces andor all nonconservative forces are acting perpendicular to the direction of motion Problem A child and sled with mass of 50 kg slide down a frictionless hill If the sled starts from rest and has a speed of 300 ms at the bottom how high is the hill We can t do this with kinematic equations unless we know the angle of the hill or at least some more information However we can do it with conservation of mechanical energy Problem A motorcycle rider leaps across a canyon with an initial speed of 380 ms from a height of 700 m He lands at a height of 350 Hi What is his final velocity Problem A 05 kg block is used to compresses a spring with a spring constant of 800 Nm a distance of 20 cm 02 in When the spring is released what is the final speed of the block Problem A 70 kg person bungee jumps off of a 50 in bridge with his ankles attached to a 15 m long bungee cord Assuming he stops just at the edge of the water and he is 20 m tall what is the spring constant of the bungee cord 6 Nonconservative Forces and Conservation of Energy So far in our study of conservation of energy we have neglected friction and other nonconservative forces When there are no nonconservative forces present then the total mechanical energy is conserved as we have seen However there is a more fundamental law of the conservation of energy which includes nonconservative forces It states The total energy in any process is not increased or decreased Energy can be transformed from one form to another and transferred from one body to another but the total energy remains constant This is ALWAYS true For example when we compress a spring then use it to accelerate a block the potential energy of the spring is changed into the kinetic energy of the block What about if I slide something across the oor and it comes to rest It started with a lot of kinetic energy and at the end it has no greater potential energy What happened to it The kinetic energy changed into thermal energy or heat and the book and floor heated up The forms of forces like friction which reduce the mechanical energy of a system are called dissipative forces So when dissipative forces are present the total energy remains constant but the mechanical energy decreases Energy can be transformed from one form to another There are many forms of energy electrical chemical nuclear thermal gravitational elastic kinetic Although energy can be transformed from one of these forms to another it can not ever be created or destroyed We use this fact in solving problems where the mechanical energy is not conserved Recall the equation we derived which included nonconservative forces WNc AK AU l2mvf2 l2mv2 Uf UI If there are no nonconservative forces then we use this equation by setting WNC 0 and the mechanical is conserved If there are nonconservative forces then the loss in mechanical energy goes into work done due to these forces For instance we did a problem with a child on a sled who weighed 50 kg on a hill that was 046 m high and had a final velocity of 30 ms Suppose instead the final velocity is 26 ms on the same hill Problem A child and sled with mass of 50 kg slide down a hill with a height of 046 m If the sled starts from rest and has a speed of 26 ms at the bottom how much thermal energy is lost due to friction ie what is the work that friction does If the hill has an angle of 20 above the horizontal what was the frictional force So if the mechanical energy is not conserved that means that work must have been done by nonconservative forces That is energy was transformed from mechanical energy to other forms of energy No energy was created or destroyed 7 Power Sometimes it is more important to know not just the work that has been done but rather the rate at which work has been done For instance I may have two identical cars except for the engines The two cars my require the same amount of work to go from 0 to 60 mph but one car can do it in 5 seconds while one car can do it in 12 seconds The one that can do it is a more powerful car This illustrates the concept of power Power is defined as the rate at which work is done P Wt The SI unit of power is watts W l Js l W Sometimes we measure power in horsepower One horsepower 550 ftlbs and is equal to 746 W Power can also be written as P Fdt Fdt Fv Whenever you want to determine power you must first determine the force and the velocity or the work being done and the time Problem A deep sea observation chamber is raised from the bottom of the ocean 1700 m below the surface by means of a steel cable The chamber moves at constant velocity and takes 500 minutes to read the surface The cable has a constant tension of 8900 N How much power is required to pull the cable in Give the answer in horsepower Problem A 1000 kg elevator carries an 800 kg load A constant frictional force of 4000 N retards its motion What minimum power must the motor deliver to lift the elevator at a constant rate of 300 ms Chapter 10 When atmospheric pressure increases what happens to the absolute pressure at the bottom of a pool A It does not change B It increases by an amount less than the atmospheric change C It increases by an amount equal to the atmospheric change D It increases by an amount greater than the atmospheric change E It decreases Three drinking glasses all have the same area base and are all lled to the same level Glass A is cylindrical Glass B is larger at the top and holds more water Glass C is smaller at the top and holds less water Which glass has the greatest liquid pressure at the bottom A Glass A B Glass B C Glass C D All have the same pressure Chapter 10 Consider two fish tanks that are the same height and the same width TankA is 3 feet long and tank B is 6 feet long S A is the force on the side of tank A and SB is force on the side of tank B BA is the force on the bottom of tank A and BB is force on the bottom of tank B Which statement below is true A SASB and BABB BSAZSB and BABB C ZSASB and 2BABB D SASB and 2BABB E SAZSB and BA2BB A column of water 70 cm high supports a column of an unknown liquid as shown in the gure Assume that both liquids are at rest and that the density of water is 1000 kgm3 Neglect the small difference in air pressure at the top of the two liquids What is the density of the unknown liquid A 1000 kgm3 70 cm B 27701000 kgm3 C 70200000 kgm3 27 cm D 27701000 kgm3 E None of the above Chapter 10 A Ushaped tube is open at both ends and lled with water Oil is then poured into the right arm and forms a column of liquid as shown in the gure on the right At what height is the pressure in the right arm equal to the pressure in the left arm Before After A A A B B B C C C 39 39 39 D A and C E A B and C A container lled with oil is tted with pistons on both ends The area of the left piston is 10 mm2 and that of the right piston is 100002 m What force must be exerted on the left piston to move the 10000 N car upward at a constant velocity A 10N B 100N C 10000N D 106N E 108N Chapter 10 50 cm3 of wood is oating on water and 50 cm3 of iron is totally submerged Which has the greater buoyant force on it A The wood B The iron C Both have the same buoyant force D It is impossible to tell from the information given A 10 kg piece of aluminum sits at the bottom of a lake right next to a 10 kg piece of lead Which has the greater buoyant force on it Aluminum is less dense than lead A The aluminum B The lead C Both have the same buoyant force D It is impossible to determine without knowing their volumes Chapter 10 An object oats with half its volume beneath the surface of the water The weight of the displaced water is 2000 N What is the weight of the object A 1000 N B 2000 N C 4000 N D Impossible to determine without more information Salt water is more dense than fresh water A ship oats in both fresh water and salt water Would the ship displace a larger amount of fresh water salt water or neither A Fresh water B Salt water C Neither D Cannot be determined from the information given Chapter 10 A beaker of water is sitting on a balanced scale If you put your nger in the water without touching any surface of the beaker what happens to the side of the scale with the beaker on it A A It goes down B It goes up C It stays in the same place A boat carrying a large boulder is oating on a lake The boulder is thrown overboard and sinks The water level in the lake with respect to the shore A rises B drops C remains the same Chapter 10 A piece of wood is oating in a bathtub A second piece of wood sits on top of the rst piece and does not touch the water If the top piece is taken off and placed in the water what happens to the water level in the tub A It goes up B It goes down C It stays the same D It depends on whether the two pieces of wood are the same kind of wood and have the same density Consider an object that oats in water but sinks in oil When the object oats in water half of it is submerged If we slowly pour oil on top of the water so it completely covers the object the object A moves up B stays in the same place C moves down Chapter 10 Water enters a pipe of radius r with a certain velocity The water encounters a location in the pipe where its velocity is increased to 4 times its initial velocity What is the diameter of this portion of the pipe A rl6 cm B r4 cm C r2 cm D 2r cm E 4r cm Blood ows through a coronary artery that is partially blocked by deposits along the artery wall Through which part of the artery is the ow rate ux largest A The narrow part B The wide part C The ow rate is the same in both parts Chapter 10 Blood ows through a coronary artery that is partially blocked by deposits along the artery wall Through which part of the artery is the ow speed largest A The narrow part B The wide part C The speed is the same in both parts A blood platelet drifts along with the ow of blood through an artery that is partially blocked by deposits As the platelet moves from the narrow region to the wider region it experiences A an increase in pressure B no change in pressure C a decrease in pressure Chapter 10 The next three questions all have to do with the following situation A glass tube has three different cross sectional areas with the values indicated in the gure below A piston at the left exerts pressure so the mercury in the tube ows out the right end with a speed of 8 ms 2 in cm 56 cm2 60 cm2 8 ms Piston A B C Atmospheric Pressure is 101 x 105 Nm2 Density of Mercury is 13600 kg m3 2 12 cm 56 cm2 60 cm2 8 ms A B C Atmospheric Pressure 101 x 105 Nm2 Density of Mercury 13600 kg m3 At what speed is mercury owing past the point labeled A A2ms C 8ms E 16ms B 4 ms D 12 ms Chapter 10 2 12 cm 56 cm2 60 cm2 8 Ms A B C Atmospheric Pressure 101 x 105 Nm2 Density of Mercury 13600 kg m3 What is the total pressure at point C A 101gtlt 105 Nm2 D 644 x 105 Nm2 B 326 X 105 Nm2 E 745 x 105 Nm2 C 366 x105 Nm2 2 12 cm 56 cm2 60 cm2 8 Ms A B C Atmospheric Pressure 101 x 105 Nm2 Density of Mercury 13600 kg m3 What is the total pressure at point A A 101gtlt105 Nm2 D 326 x105 Nm2 B 202 X 105 Nm2 E 427 x 105 Nm2 C 225 x 105 Nm2 Chapter 10 When a tube of diameter d is placed in water the water rises to a height k If the diameter were half as great how high would the water rise A 112 B h C 2h D 4h E Not enough information to know Two Styrofoam balls of radii R and 2R are released simultaneously from a tall tower Which will reach the ground rst Do not neglect air resistance and assume the air ow is laminar A The large one B The small one C Both will reach the ground simultaneously D It depends on the atmospheric pressure Chapter 6 I lift a barbell with a mass of 50 kg up a distance of 070 in Then I let the barbell come back down to where I started How much net work did I do on the barbell A 340 J B 0 J C 35 J D 340 J E 690 J You lift a 10 N physics book up in the air a distance of 1 meter at a constant velocity of 05 ms The work done by gravity is A 10 J B 10 J C 5 J D 5 J E zero Chapter 6 In which of the following situations will there be an increase in kinetic energy A A projectile approaches its maximum height B A box is pulled across a oor at a constant speed C A child is pushing a merrygoround causing it to rotate faster D A satellite travels in a circular orbit around a planet at a xed altitude E A stone at the end of a string is whirled in a horizontal circle at a constant speed Two marbles one twice as heavy as the other are dropped to the ground from the roof of a building Just before hitting the ground the heavier marble has A as much kinetic energy as the lighter one B twice as much kinetic energy as the lighter one C half as much kinetic energy as the lighter one D four times as much kinetic energy as the lighter one E impossible to tell Chapter 6 A 40 N book is carried up a ight of stairs which is 25 m long and 8 m high What is the change in potential energy of the book A OJ B 5 J C 40J D 320J E 1000J At a bowling alley the ball feeder must exert a force to push a 50 kg bowling ball up a 10 m long ramp The ramp raises the ball 05 m above the base of the ramp Approximately how much force must be exerted on the bowling ball 9 F 05 m A200N B 50N C25N D 50 N E Not enough info to tell Chapter 6 Suppose you wanted to ride your mountain bike up a steep hill Two paths lead from the base to the top one twice as long as the other Compared to the average force you would exert if you took the short path the average force you exert along the longer path is A four times as small B three times as small C half as small D the same E it depends on the time taken In which system is there a decrease in potential energy A A boy stretches a spring B A child slides down a sliding board C A crate rests at the bottom of an inclined plane D A car ascends up a steep hill E Water is forced upward through a pipe Chapter 6 A ball drops some distance and gains 30 J of kinetic energy Do not ignore air resistance How much gravitational potential energy did the ball lose A More than 30 J B Exactly 30 J C Less than 30 J D It depends on the mass of the ball E More information is needed to determine the answer A stone is thrown upward into the air In addition to the force of gravity the stone is subject to air resistance The time the stone takes to reach the top of its ight is A larger than B equal to C smaller than the time it takes to return from the top to its original position Chapter 6 A 30 kg block falls a distance of 6 m in a tube with no air resistance near the surface of the earth What is its speed after it has covered the 6 in distance A 8 ms B 11 ms C 13 ms D 118 ms E 176 ms A woman stands on the edge of a cliff She throws a stone vertically downward with an initial speed of 10 ms The instant before the stone hits the ground below it has 450 J of kinetic energy If she were to throw the stone horizontally outward from the chff with the same initial speed of 10 ms how much kinetic energy would it have just before it hits the ground A 50 J B 100 J C 450 J D 800 J E Not enough information was given to answer the question Chapter 6 A roller coaster travels along a straight path at a speed of 20 ms What would its speed be after climbing a 15 m hill if friction is ignored v20 ms A 3 ms B 5 ms C 10 ms D 14 ms E 17 ms Suppose you wanted to ride your mountain bike down a steep hill Two paths lead from the top to the base one twice as long as the other Neglect friction and air resistance Compared to the maximum speed you would reach if you took the short path the maximum speed you will reach along the longer path is A one quarter as fast B twice as fast C the same speed D half as fast E four times as fast Chapter 6 A spring loaded toy dart gun shoots a dart straight up in the air and the dart reaches a maximum height of 24 m The same dart is shot straight up a second time but this time the spring is compressed only half as far How far up does the dart go this time neglecting friction A96m B48m C24m D12m E6m How much energy is dissipated in braking a 1000 kg car to a stop from an initial speed of 30 ms A 30000 J B 200000 J C 450000 J D 850000 J E 900000 J Chapter 6 A cart on an air track is moving at 05 ms when the air is suddenly turned off The cart comes to rest in l m The experiment is repeated with the cart moving at 1 ms when the air is turned off How far does the cart travel before coming to rest A l m B 2 m C 3 m D 4 m E impossible to determine A 4 kg mass with a speed of 2 ms and a 2 kg mass with a speed of 4 ms are gliding over a horizontal frictionless surface Both objects encounter the same horizontal force which opposes their motion and are brought to rest Which statement best describes their stopping distances A The 4 kg mass travels twice as far as the 2 kg mass before stopping B The 2 kg mass travels twice as far as the 4 kg mass before stopping C Both mass have the same stopping distance D the 2 kg mass travels farther but not necessarily twice as far Chapter 6 A 50 kg woman runs up a ight of stairs in 5 s Her net upward displacement is 5 m What power did the woman exert while she was running A 250 w B 750 w C 05 kW D 10 kW E 5 kW A car accelerates from 0 to 30 mph in 15 s How long does it take for it to accelerate from 0 to 60 mph assuming the power of the engine to be independent of velocity and neglecting iction A 20 s B 30 s C 45 s D 60 s E 90 s Physics 2414 Strauss Chapter 2 Lecture Notes Formulas v v 0 at x x0 v0t 12at 2 x x0 12v0 v t v2v022ax x0 17 2 v0 v 2 Constants g 980 ms2 Main Ideas 1 Define various physical quantities relating to motion 2 Understand uniform accelerated motion including falling bodies 3 Learn how to solve problems 4 Understand graphical analysis of motion 1 Definitions 1 1 DISTANCE AND DISPLACEMENT Distance is the total length that an object has moved If I walk 2 km north then 1 km south the distance I have moved is 3 km On the other hand displacement is the net change in position My displacement is only 1 km If I go 3 km west and 4 km north my distance 7 km and displaceLment 5 km For displacement Ax x xi The displacement and distance can be drawn on an axis 1 2 SPEED AND VELOCITY Average speed is defined as distancetime E d t Problem If ajogger jogs for 15 hours at an average speed of 222 ms how far does he go Velocity is defined as a quantity which has a magnitude or value as well as a direction Such a quantity is called a vector In print a vector is usually designated by using bold face print For instance velocity might be written using the symbol V Average velocity is defined as displacemenUtime xf xi Ax v tf ti At If it takes me 45 minutes to walk 20 km west then 10 km east my average speed is 30075 40 kmhour while my average velocity is 1075 13 kmhour Problem The land speed record was set by the car ThrustSSC in 1997 and is 341 ms 763 mileshour This was the first land vehicle to travel faster than the speed of sound For this record the car is first driven one direction then the other and the two speeds are averaged Suppose the car covered 1609 m in one direction then turned around and covered the same distance If the first pass is done in 474 s and the second in 469 s what is his average velocity on each pass Instantaneous velocity is the velocity at a single instant It is defined as the average velocity over a very short period of time as the period of time approaches an infinitesimal limit AX V lim AHO A d Note that the magnitude of the instantaneous velocity is the instantaneous speed Your car speedometer gives your instantaneous speed but not instantaneous velocity Why 13 ACCELERATION Average acceleration is defined as Vf Vi tf ti At a Note that an object is accelerating if it changes speed or if changes direction since acceleration and velocity are vectors Acceleration is the rate at which velocity changes while velocity is the rate at which position changes Problem A sports car accelerates from 0 to 60 mph in 60 seconds What is its average acceleration Problem You are driving 20 ms when a dog runs across your path You slam on your brakes and in 2 seconds slow to 5 ms What was your average acceleration Instantaneous acceleration is given by AV dv a lim AHO At dt 2 Uniform Acceleration We now use these simple definitions to derive equations for the special case when acceleration is a constant The equations we will derive can only be used when the acceleration is a constant That means when the rate of change of the velocity is constant It doesn t change The velocity itself will change but the rate of the change will always be the same We start with the definition of acceleration and set the initial time ti 0 the final time If t the initial velocity V v0 and the final velocity vf v a ltv w 2 Remember mathematical equations have a meaning What does this one mean Settingt00weget7x x0t gt xx0 17t orwith 7v0 v2 gt xx0 vv0t Since v v0 at we substitute this in the previous equation to get x x0 12 v0 at v0t x x0v0t l2at2 Finally from substituting I from first boxed into second boxed we get x x 0 l2v v0 v v0 a x x 0 l2v2 v02a v2 v022ax x0 Writing all the equations together gives o xx0 a lt v v0at x x0 l2v0vt x x0 v0t l2at2 v2v022ax x0 4646 as as lt Another equation that is useful to know is 17 2 v0 v 2 and recall that x x 0 is the distance traveled so often we set x 0 0 and just use x for the distance These only work when acceleration is a constant Magnitude and Direction If you are given three known quantities and one unknown you can chose one of these equations If you are given two known quantities and two unknown ones you will have to use two of these equations See Section 26 Solving Problems Solving problems will be a big part of this class 1 Read the Problem Carefully 2 Draw a Diagram Write down what is known or given and what you want to know Think about the physics principles and make sure the equations are valid Do the calculation Think about the answer Is it reasonable Order of magnitude Check the units OUIJgtUJ Problem Previously we had seen that our sports car could accelerate at 45 msz This is a constant acceleration After 8 seconds how far has it gone Problem What is the speed of the car after 10 seconds NOTE We often use the word deceleration to mean that an object is slowing down What is deceleration Is it the same as negative acceleration NO Negative acceleration means the sign of the acceleration is negative Deceleration is when the direction of acceleration is opposite the direction of motion Suppose a car is travelling in the negative direction To decelerate it must have an acceleration in the positive direction On the other hand it will gain speed when the acceleration is in the negative direction Problem A car is traveling in the negative direction at 32 ms The driver applies his brakes and stops in 73 seconds What was the car s acceleration Problem Suppose a spacecraft is traveling with a speed of 3250 ms and it slows down by firing its retro rockets so that a 10 msz What is the velocity of the spacecraft after it has traveled 215 km Problem Suppose a slow Texas lineman picks up a fumbled football on the 20 yard line and runs toward the end zone at 73 ms The safety is standing on the 23 yard line and needs to catch up to the lineman before he scores a touchdown If the safety can accelerate at a constant rate what must be his minimum acceleration to catch the lineman What will be the safety s final velocity 3 Free Falling Bodies One of the most important cases of uniform accelerated motion is the case of objects which are near to earth allowed to free fall At first it might seem that different objects accelerate at different rates near the earth depending on their weight That is what people thought before Galileo did his experiments in the late 1500 s However Galileo showed that objects of different weights dropped at the same rate The reason that some things drop slower in the air is because air resistance pushes against the moving object If there was no air resistance then even the open paper would drop at the same rate On the moon where there is no air a feather and a hammer fell at the same rate When an object is dropped its velocity increases by 980 ms every second It continues to go faster If there was no air resistance this would continue every second Because there is air resistance the object eventually reaches a terminal velocity and doesn t go any faster But for many applications we can negelect air resistance Then everything near the surface of the earth accelerates toward the center of the earth with a consant acceleration Therefore all of the equations we have derived for constant acceleration apply to an object in free fall neglecting air resistance All objects fall with a constant acceleration of about 980 ms2 which we call g the acceleration due to gravity Problem A stone is dropped from a very tall building What is its position and velocity after 300 seconds Problem A boy throws a ball upward from the top of a building with an initial velocity of 200 ms The building is 50 meters high Determine a the time needed for the stone to reach its maximum height b the maximum height c the time needed for the stone to reach the level of the thrower d the velocity of the stone at this instant e the velocity and position of the stone after 500 s f the velocity of the stone when it reaches the bottom of the building 4 Graphical Analysis of Linear Motion We can analyze motion on a graph The most important thing in doing this is to look at how the axis is labeled Look at What is constant and What is changing linearly at a constant rate Also determine what the slope represents Slope is defined as the change in the vertical axis divided by the change in the horizontal axis If the vertical axis is displacement x and the horizontal axis is the time t then the slope is AxAt v the slope is the velocity If the vertical axis is velocity v and the horizontal axis is time t then the slope is given by AvAt a the slope is the acceleration Compare the next two graphs of the same motion Position no speed x nstant speed stant speed Time I Velocity V I I Time I Compare the next three graphs as they analyze another series of movements Velocity V Time Position Time Acceleratio r1 l I l l Time Chapter 5 A 1500 kg car travels at a constant speed of 22 ms around a circular track which has a radius of 80 m Which statement is true concerning this car A The velocity of the car is changing B The car is characterized by constant velocity C The car is characterized by constant acceleration D The car has a velocity vector that points along the radius of the circle E The car has an acceleration vector that is tangent to the circle at all times A rock is twirled on a string at a constant speed The direction of its acceleration at point P is m Bgt C D at Chapter 5 A girl attaches a rock to a string which she then swings clockwise in a horizontal circle The string breaks at point P on the sketch which shows a view from above What path will the rock follow D B C A boy is whirling a stone around his head by means of a string The string makes one revlution every second and the tension in the string is T The boy then speeds up the stone keeping the radius of the circle unchanged so that the string makes two complete revolutions every second What happens to the tension in the string A It remains unchanged B It is reduced to half of its original value C It is increased to twice its original value D It is reduced to onefourth of its original value E It is increased to four times its original value Chapter 5 A rider in an amusement park Q ride the barrel of fun nds herself stuck with her back to the wall Which diagram correctly shows the forces C acting on her A B C D E A cart of massM travels along a straight horizontal track As suggested in the gure the track then bends into a vertical circle of radius R Which expression determines the minimum speed that the car must have at the top of the track if it is to remain in contact with the track M R gt A vMgR B v2 2gR C v gR D V 2gR E v2 gR Chapter 5 A plane is travelling at 200 ms following the arc of a vertical circle of radius R At the top of its path the passengers experience weightlessness To one signi cant gure what is the value of R v gt A 200 m B 1000 m R C 2000 m D 4000 m E 40000 m A car is speeding up as it enters the freeway on a circular entrance ramp as shown in the gure at right What is the direction of the acceleration of the car when it is at the point ramp indicated A B CT D E gt car freeway Chapter 5 A spaceship is traveling to the moon At what point is it beyond the pull of the earth s gravity The mas of the moon is 180 the mass of the earth and the surface gravity of the moon is 16 that of the earth A When it gets out of the atmosphere B When it is halfway there C When it is 56 of the way there D When it is 7980 of the way there E It is never beyond the pull of earth s gravity The Moon does not fall to Earth because A It is in Earth s gravitational eld B The net force on it is zero C It is beyond the main pull of Earth s gravity D It is being pulled by the Sun and planets as well as by Earth E None of the above Chapter 5 Two satellitesA andB of the same mass are going around the earth in concentric orbits The distance of satellite B from the earth is twice that of satellite A What is the ratio of the centripetal force acting on B to that acting on A A 18 B 14 C 12 D WE E 1 Two satellites A andB of the same mass are going around the earth in concentric orbits The distance of satellite B from the earth is twice that of satellite A What is the ratio of the tangential speed of B to that of A A 12 B 1N5 C 1 DH E 2 Chapter 5 The earth exerts the necessary centripital force on an orbiting satellite to keep it moving in a circle at constant speed Which statement best explains why the speed of the satellite does not change even though there is a net force exerted on it A The acceleration of the satellite is zero B The centripital force has a magnitude of var C The centripital force is canceled out by the reaction force D The centripital force is always perpendicular to the velocity E The satellite is in equilibrium which means there is no net force acting on it An astronaut is inside a space shuttle in orbit around the earth She is able to oat inside the shuttle because A her weight is zero and her shuttle s weight is zero B the force of gravity is very small on the astronaut C she and her shuttle move with the same constant velocity D she and her shuttle move with the same centripetal acceleration E The force of the earth on the spaceship and the force of the spacehip on the earth cancel because they are equal in magnitude and opposite in direction Chapter 5 A satellite encircles Mars at a distance above its surface equal to 3 times the radius of Mars The acceleration of gravity of the satellite as compared to the acceleration of gravity on the surface of Mars is A zero B the same C onethird as much D oneninth as much E onesixteenth as much Chapter 11 Lecture Notes Physics 2414 Strauss Formulas F kx f lT E 12mv2 02ch2 12mv02 12kA2 v imp12 xzA2 v02 kmA2 T2rc E k x Acoswt Acos2nft Acos27ttT v aA sinwt a 602A coswt T27rZ g vlTlf v F T mL ln2Ln n FT 2L fl nV fn ZWmL n fmm quot s1n02s1n01 vzv1 Main Ideas 1 Simple Harmonic Motion OOOOOIUJOONOOOO Energy Description Kinematic Description Relationship with Circular Motion Applied to a Pendulum Other Periodic Motion Damped Motion Forced Vibrations and Resonance Wave Motion Types of Waves Description of Waves Superposition and Re ection Standing Waves Resonant Frequencies Refraction and Diffraction 1 Simple Harmonic Motion Vibrations and waves are an important part of life Every sound you hear is a result of something first Vibrating then a sound ane traveling through the air as the air molecules Vibrate then your eardrum Vibrating and the brain interpreting that as sound The simplest Vibrational motion to understand is called simple harmonic motion SHM SHM occurs when the I mOVe an object from its equilibrium position and the force that tries to restore the object back to its equilibrium position is equal to the distance from the equilibrium position In other words F kx This is exactly Hooke s law for springs So ideal springs exhibit SHM The motion of the object at the end of a spring repeats itself after a period of time It is periodic with a period T the amount of time it takes to complete one cycle The frequency is the number of cycles per unit time so f UT The frequency is measured in cyclessecond l cyclesecond l Hertz Hz The maximum distance from the equilibrium point is called the amplitude A and the distance from the equilibrium point at any time is the displacement Problem A 035kg mass attached to a spring with spring constant 130 Nm is free to mOVe on a frictionless horizontal surface If the mass is released from rest at x010 in find the force on it and its acceleration at a x010 m b x0050 m c x0 m and d x 0050 m Let s look at Various aspects of simple harmonic motion including energy motion relationship with circular motion and relationship with pendulum motion 1 1 ENERGY OF SIMPLE HARMONIC MOTION The simple harmonic oscillator is an example of conserVation of mechanical energy When the spring is stretched it has only potential energy U l2kx2 12ch2 where A is the maximum amplitude When the spring is unstretched it has only kinetic energy K l2mv2 l2mv02 where v0 is the maximum Velocity which occurs when the spring in unstretched At any point in the motion of the object it has a total energy equal to its potential energy plus its kinetic energy E 12mv2 12kx2 12mv02 12kA2 Problem A 24kg mass is attached to a horizontal spring that has a spring constant of 86 Nm The spring is initially stretched to 023 m If there is no friction so that the spring oscillates with SHM how much energy is kinetic and potential when the spring is at a x 10 m b x 0 m Let s suppose that the spring is not horizontal but is instead in a vertical position The motion is basically the same as a horizontal spring except for where the equilibrium position is located The equilibrium position is found by looking at all the forces on the mass Let x0 be the new equilibrium position kx0mg gt x0mgk kx0 If you now move the spring an additional distance of x the forces on the mass are given by mg F kx0 x mg kmgk x mg kx so even a vertical spring behaves as if it was horizontal with the restoring force equal to kx Also the energy of compressing a vertical spring is the same as a horizontal spring So a vertical spring acts exactly like a horizontal spring only the equilibrium position is displaced due to gravity 1 2 KINEMATICS OF SIMPLE HARMONIC MOTION So how does a simple harmonic oscillator move We have seen that when the spring is displaced by its maximum amount in a particular situation then the potential energy is a maximum and the kinetic energy velocity is zero When the spring is at its equilibrium position after being displaced then it has a maximum velocity and therefore a maximum kinetic energy If I use the energy equations from above to solve for velocity I get 12mv2 02ch2 12kA2 v ikmAz x2 And if I set the maximum kinetic energy equal to the maximum potential energy I get l2mv02 l2kA2 gt Vol km AZ so plugging this into the above equation gives v imp12 xZA2 This tells the velocity at any position as a function of the maximum velocity and of the maximum displacement amplitude Look at what it says When the position is the same as the amplitude x A the velocity is zero When the position is the equilibrium position x 0 the velocity is a maximum v0 1 3 RELATIONSHIP OF SHM TO CIRCULAR MOTION Consider an object rotating around in a circle If I look at one point of that object from the side I will see that the object appears to be following simple harmonic motion We will look at the projection of the motion along the x axis See figure 116 in the book The object always has a speed of v0 However the x component of the velocity the part of the velocity which is viewed from our observer changes Because the triangles are similar all three angles are the same vvo x A2 x2A v imp12 xzA2 which is the equation for a simple harmonic oscillator If the equations are the same then the motion is the same Since we have already dealt with uniform circular motion it is sometimes easier to understand SHM using this idea of a reference circle For instance the speed of the ball going around the circle is given by distance divided by time v0 272AT or T 27mm where T is the period andA is the amplitude and v0 is the maximum velocity If I now use v02 kmAz I get T Z JE or for the frequency I get 2 k 27r m So we see that frequency does not depend on amplitude but it does depend on the spring constant and the mass If we know the location of the mass and the amplitude of a the oscillation then we know the velocity from v iv0 12 xZA2 But suppose I wanted to know the location as a function of the time Where will the mass be after 1 second or after 100 seconds From our figure we see that x Acos0 and since the mass is rotating with angular velocity 60 we see that 0 wt and from a 27tf 27tT we get x Acoswt Acos2nft Acos2739ctT Now we see that the motion is periodic After a time where t T we get the mass is at a position of cos27c cos0 cos2nrc where n is any integer The mass keeps coming back to the same position The motion is also sinusoidal as a function of time That is if I plot position versus time I will get a sine or cosine curve Similarly the velocity is sinusoidal v v012 xzA2 v0lz Acoswt2A2 v0lz coszwt v0 sinzwt vosinwt aA sinwt since v0 60A and the acceleration is sinusoidal We can see this from Newton s second law a Fm kxm kAm coswt volA coswt 602A coswt To use these keep in mind that v0 60A and v02 kmAz Whether it is a sine or cosine function for x and acceleration just depends on whether the mass starts at the equilibrium position or at the maXimum Displacement and acceleration are the same sine or cosine but velocity is the opposite We say velocity is out of phase with displacement Problem Suppose I have a spring which oscillates according to the following equation What is the amplitude the frequency the period of the oscillation x 025 m cos7z t80 1 4 SHM AND SIMPLE PENDULUMS A simple pendulum acts like a harmonic oscillator if the displacement is small We can see this from looking at the forces on a pendulum F mg sine To be a simple harmonic oscillator the force must be proportional to the distance the pendulum bob has moved Recall that the arclength l is given by 0 l L in radians and that sin0 xL from the figure Now if the angle is small then x zl and 0 sin0 xL or F mg0 mgxL which looks just like F kx if k mgL So for small angles a pendulum acts like a simple harmonic oscillator with a spring constant of mgL Remember if the equations are the same then the motion is the same The period is given by T27rE27r m 27 k mgL g So the period or frequency does not depend on the mass of the pendulum only its length Problem A man wants to know the height of a building which has a pendulum hanging from its ceiling He notices that in one minute the pendulum oscillates 8 times a What is the height of the building b If the length were cut in half what would the new frequency be 2 Other Periodic Motion 2 1 DAMPED OSCILLATIONS Most oscillations do not continue on forever but eventually stop due to some kind of nonconservative force like friction or air resistance Some vibrations are purposely stopped Your shock absorbers in your car are made to stop the vibrations set up by the road All these vibrations which are eventually stopped are called damped vibrations or if the harmonic motion is stopped it is called damped harmonic motion When the motion is damped mechanical energy is not conserved 2 2 DRIVEN OSCILLATIONS The opposite of damped Vibrations are Vibrations which are driven or forced I can drive an oscillation at any frequency but if I drive it at its natural frequency ie resonance frequency then its amplitude rises without any limits Everything has a natural frequency like musical instruments children s swings and bridges Foot soldiers don t march in step across a bridge so that they won t accidentally set up a resonance in the bridge and it oscillates and collapses Oscillating air at the resonant frequency of a glass is what causes it to shatter 3 Wave Motion Oscillations can cause waves So it is natural to use the concepts and terminology developed in our discussion of harmonic motion to begin to talk about waves The world is full of waves For example there are sound waves waves on a string earthquake waves and electromagnetic waves Waves are oscillations which carry energy from one place to another yet matter is not carried with the wave Setting up dominoes and knocking them down illustrates this Each domino moves very little but the energy may move over a long distance If you watch a leaf or a cork oating on a lake as waves move by it the leaf or cork moves very little but the wave moves a long ways Waves may move in one two or three dimensions A wave pulse on a rope basically moves in one dimension Water waves on a lake move in two dimensions and spoken sound waves spread out in three dimensions In this chapter we will talk about waves which require some kind of medium to travel in like sound waves earthquake waves and waves on a string Next semester we will talk about electromagnetic waves which do not require any medium to travel in We will find that there are many similarities in all wave motion 3 1 TYPES OF WAVES There are two basic types of waves transverse waves and longitudinal waves Each type of wave describes the motion of a small segment of the wave In a transverse wave each segment moves perpendicular or transverse to the direction of motion of the wave This is like a rope with a pulse on it In a longitudinal wave each segment moves along the direction of the motion of the wave This is like a spring or slinky which is quickly compressed then returned to its original position 32 DESCRIPTION OF WAVES The shape of a wave on a string gives a good picture of a wave The maximum height of the wave is called the amplitude The distance from one point on the wave to the same point on the next wave like crest to crest or trough to trough is one wavelength denoted by the Greek letter lambda 2 The speed with which the wave moves is called the wave velocity v That is what is the distance an individual crest travels in a certain amount of time The wave velocity is not the velocity of the individual segments of the wave a part of the rope or a single domino It is the velocity that the wave as a whole moves If I stay at one location and count the number of waves that go by in a given amount of time that is called the frequency f The period T is the time it takes for one complete oscillation of the wave Since velocity is distance divided by time we see that the wave velocity is given by The speed of a wave in a certain medium is determined by the properties of the medium We find that for a stretched string the speed with which the wave moves is given by FT V mL where FT is the tension in the rope L is the length of the rope and m is the mass of the rope The quantity u Lm is called the linear density of the rope Sound travels by creating longitudinal waves in the air Some kind of medium is required for sound waves to travel In the vacuum of space sound waves do not travel and no one can hear you scream In general the velocity of a longitudinal wave traveling in a liquid is given by v Bp where B is the bulk modulus see table 91 page 240 and p is the density of the material The density is defined as the mass per unit volume pm V kgm3 and many values of densities for various materials are given in Table 101 on page 260 This equation also works approximately though not precisely for a gas Problem A string has a mass of 0300 kg and a length of 600 m and is set up as shown in the picture at the right What is the speed of a pulse on this string Problem How fast does sound travel in air and in water 3 3 SUPERPOSITION AND REFLECTION An important aspect of waves is the principle of the superposition of waves This principle states If two or more traveling waves are moving through a medium the resultant wave is found by adding together the displacements of the individual waves point by point So if there are two waves approaching each other on the same string they will interfere with each other If they both have a positive amplitude the will interfere constructively and the resultant wave will have a larger amplitude If one wave has a positive amplitude and one has a negative amplitude they will interfere destructively and the resultant wave will have a smaller amplitude See figure 1129 in the book If the waves are moving in the same direction then there will be constructive or destructive interference only if the waves also have the same wavelength If the waves have their greatest positive amplitudes at the same point we say that the waves are in phase and they demonstrate constructive interference If one wave has its greatest positive amplitude where the other wave has its greatest negative amplitude we say the waves are out of phase and they demonstrate destructive interference One way of creating a wave travelling in the opposite direction is to have the wave re ect off of an object If a rope is attached to a wall and a pulse is generated on the rope it will re ect when it reaches the end If the rope is firmly attached the reflection will actually invert the amplitude of the pulse If the rope is attached with a ring it is free to move then the amplitude will not be inverted If a thick rope is attached to a thin rope then part will be re ected and part will be transmitted 34 STANDING WAVES RESONANT FREQUENCIES One of the most important concepts regarding waves is the concept of standing waves and resonant frequencies These ideas explain many aspects of sound including why different instruments or different shaped volumes create different sounds A standing wave is a wave in which every point on the wave oscillates with the same frequency and each point has a maximum amplitude which does not change although every point may have a different maximum amplitude Any standing wave will have certain parts of the wave which do not move at all and are called nodes and other parts of the wave which have maximum movement and are called antinodes For any given geometric situation a standing wave can only be created for certain frequencies which are called natural frequencies or resonant frequencies A standing wave is really a set of waves moving in opposite directions with the same velocities That is why you can create standing waves by attaching a rope at one end and vibrating the other end The waves are re ected off of each end and the two waves interfere to produce a standing wave This will only work if the rope is vibrated at the certain resonant frequencies A standing wave does appear to be standing in place and a travelling wave does appear to be moving Yet a standing wave is really the superposition of travelling waves moving in opposite directions If I have a rope that is tied down at each end and I want to set up standing waves in the rope I have a constraint The constraint is that each end of the rope can not move So each end of the rope is a node Then there are only certain ways I can set the rope vibrating These different ways are called harmonics See figure 1133 in the book If the only nodes are at the end we call this the first harmonic or it is also called the fundamental frequency or mode When I have one other node this is called the second harmonic or the first overtone and when I have two more nodes we call it the third harmonic or the second overtone and so forth This is true not only for ropes but also for standing sound waves in an air column So the harmonics of an instrument are determined by the length of the air column in the instrument In fact every object can be set to vibrating at different harmonic frequencies which are determined by the size and shape of the object u i 39 39 quot 39 Fundamental or First Harmonic Second Harmonic or First Overture Third Harmonic or Second Overture How do we determine the wavelength or frequency of the harmonics We note that the must be some fraction of the total length of the string or air cavity for sound waves For the first harmonic L 121 For the second harmonic L 12 For the third harmonic 32L 13 and so on So in general L min2 or in 2Ln for the nth harmonic The standing wave is equivalent to two travelling waves moving in opposite directions so we can still calculate the velocity of the waves moving on the string From the equations v l T if and v FLm we find that f nvZL fn 2 L FT 2L mL The last two equations are very useful Although you may need them on a test they are not necessarily on the equation sheet because they are easily derived F from the more fundamental equations in 2Ln v MT if andv j In You should know how to use these three equations in appropriate combinations What happens when you play a stringed instrument and you press down on the strings in a certain point You are changing which harmonic oscillates Unfingered Fingered at a low harmonic I quot Fingered at a high harmonic or no harmonic or with a fret that keeps the string from Vibrating The mass per unit length stays the same mL and the tension stays the same so the velocity of the travelling waves which make up the standing waves also stays F the same sincev However the frequency and the wavelength change in m such a way that v 1 If the wavelength is reduced by a factor of 2 the frequency increases by a factor of 2 so that v stays the same Problem A violin string of length 33 cm is under a tension of 55 N The fundamental frequency of the string is 196 Hz a At what speed do the waves travel on the string b What is the mass of the string c How far from one end of the string would you have to press the string in order that the remainder of the string have a fundamental frequency of 300 Hz F Here is a trick to solving these problems Since v i then any string with a m given tension and mass per unit length has the same velocity So v an anLn meLm or n fmm where n is for one harmonic and m is for another harmonic If I set m l I get This formula can be very useful for solving many problems Problem A string has its fundamental at 200 Hz What is the difference in the frequency between the first two overtones 3 5 REFRACTION AND DIFFRACTION When a wave encounters an obstacle they bend around the obstacle This is called diffraction The amount of bending depends on the size of the wavelength and the size of the object It is diffraction which allows us to hear sounds even when the source is not directly in line with us For instance if you go around a corner and talk I can still here you because the waves are bent around the obstacle which is the wall We will study diffraction quite a bit more next semester when we encounter electromagnetic waves When a wave travels from one medium to another like from air to water part of the wave is re ected and part continues into the medium and is transmitted The part that is transmitted will often move in a different direction from the original wave This phenomena is called refraction Refraction will particularly important when we discuss light in the next semester The wave is refracted because it changes velocity in the new medium See figure 1136 which shows how wave fronts will change direction if they move at a different velocity We find that the change of angle is related to the velocity of the wave in the different media so SinQZSine1 vzv1 where 01 is the angle of incidence 02 is the angle of refraction v2 is the velocity in the medium of refraction and v1 is the velocity in the initial medium Problem A longitudinal earthquake wave strikes a boundary between two types of rock at a 25 angle As it crosses the boundary the specific gravity of the rock changes from 37 to 28 Assuming that the elastic modulus is the same for both types of rock determine the angle of refraction The specific gravity is the ratio of the density of a substance to the density of water We use the ratio of the sines Refraction and diffraction occur for all types of waves They are very important concepts when we talk about electromagnetic waves next semester


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