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PHYS 220 Exam 1 Study Guide

by: Sean Anderson

PHYS 220 Exam 1 Study Guide PHYS 220

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Sean Anderson
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These notes cover all the material that we have learned in class thus far, and includes all of the important formulas and concepts we will need to know for the exam.
General Physics I
Dr. Matthew Jones
Study Guide
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This 27 page Study Guide was uploaded by Sean Anderson on Sunday October 2, 2016. The Study Guide belongs to PHYS 220 at Purdue University taught by Dr. Matthew Jones in Fall 2016. Since its upload, it has received 2 views.


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Date Created: 10/02/16
Sean Anderson Professor Physics 220 10/03/16 Physics 220 Exam #1 Study Guide General Physics   Metric system o Newton (N) = (kg*m)/s 2 o Joule (J) = (Kg*m )/s o N*s = (Kg*m)/s o Hertz = 1/s o Watt = joule/sec Motion in One Dimension   Kinematics is the study of how objects move  Objects in motion o You need to identify both the object of interest and the observer to describe the motion of the object  What is motion? o Motion is the change in an object’s position relative to a given observer during a certain time interval o Without identifying the observer, you can’t say whether the object is moving o A reference frame provides:  An object (or point on an object) of reference  A coordinate system with a scale for measuring distance  A clock for measuring time  Linear motion o We will assume we can choose a reference frame and tell where the object is at any time o To describe linear motion, we need to define:  Time and time interval  Position, displacement, distance, and path length  Scalar components of displacement for motion along on axis  Position, displacement, etc… o Position is an objects location with respect to a particular coordinate system o Displacement is a vector that starts from an object’s initial position and ends at its final position o Distance is the magnitude (length) of the displacement vector o Path length is how far the object moved as it traveled from its initial position to its final position (imagine lying a string along the path the object took, the length of the sting is the path length) o Example:  The initial position and the origin of a coordinate system are not necessarily the same points  The displacement for the whole trip is a vector that points from the starting position at X ti the final position at X f  dx= X –fX i  The distance for the trip is the magnitude of the displacement (always positive)  d = I Xf– X i  The path length is the distance from X to 0 pius the distance from 0 to X (fotal distance traveled, always positive)—path length does not equal the distance o Example:  X 1+3m, X =+4m2 X =-1m, X 3+5m 4  X2– X 1 4-3 = 1m  X3–X 2= -1-4 = -5m  X4–X 3 5 -(-1) = 6m  Total path length: l = (1m) + (5m) + (6m) = 12m  Kinematic graphs o Time (t) is usually the independent variable (x-axis) o The position (x), is the dependent variable (y-axis)  Linear motion o A straight line graph can be described by the equation:  X(t) = kt + b  b is the y-intercept (value of y when t=0)  The slope is k = ∆x/∆t (m/s)  Velocity and speed o Velocity is the slope of the position vs time graph  V x ∆x/∆t  If slope is (-) object moves in –X direction  If slope is (+) object moves in +X direction o Velocity has both magnitude and direction o The magnitude of the slope (always +) is the speed of the object  Constant velocity linear motion o Position equation for constant velocity linear motion:  X(t) = X +0V t x o X(t) means that position, X, is a function of time, t o The initial position at t=0 is 0 o The velocity, V xs the slope of the position vs time graph  Graphs of velocity o The velocity is always positive o The velocity is increasing with time o A horizontal line on the V vs t graph means the velocity is constant  Displacement from a velocity graph o If velocity is positive, the displacement will be positive o If velocity is negative, the displacement will be negative  When velocity is not constant o The instantaneous velocity is the velocity of an object at a particular time o The average velocity is the ratio of the change in the position and the time interval over which the change occurred  Acceleration o The simplest type of linear motion with changing velocity occurs when the velocity changes at a constant rate o It increases or decreases by the same amount, ∆V in exch equal time interval, ∆t  Finding acceleration from a V vs t graph o Acceleration is the slope of the velocity vs time graph  ax= ∆V /xt o A larger slope means that the velocity is increasing at a faster rate o Velocity has magnitude and direction, therefore acceleration has both magnitude and direction (it’s a vector) o The average acceleration of an object during a time interval ∆t is:  a = ∆V /∆t  When is accelerative negative o Acceleration can be + or – o If an object is moving in the +X direction, and it is slowing down, then the slope of the V vs t graph is negative o An object can have negative acceleration and still speed up  Consider an object moving in the –X direction. Its velocity is always negative, but is increasing in magnitude  Determining the change in velocity from the acceleration o The slope of V vs t graph:  ax= ∆V /xt o Supposes the time starts at t =0, then:  ∆V (t) = V + a t x 0x x o This says that V xs a function of time, t, and has the initial value V0x  Displacement from a V vs t graph o Displacement X-X between t and t is the area between the V vs t 0 0 x graph line and the t-axis (x-axis)  Area = X-X 0  Position as a function of time o Area of a rectangle = V t 0x o Area of a triangle = ½ (V -V )t x 0x  Position of an object during linear motion with constant acceleration o Initial position at t=0, is0X o Initial velocity at t=0, V 0x o The acceleration, a xemains constant for all t o The position, as a function of time is:  X(t)=X +0 t +0x a t x2  Three equations of motion: o X(t)=X +0 t 0x½ a t x 2 o Vx(t)=V 0x t x  Use this equation for solving for t, then substitute into first equation for t o 2a (X-X )=V -V2 2 x 0 x 0x  Alternate equation for linear motion with constant acceleration Newton’s Mechanics  Describing interactions o We have two objects, but are usually interested in describing the motion of one of them at a time  This object is called the system  All objects that are not part of the system but can interact with it are in the system’s environment o Interactions between the system object and objects in it environment are called external interactions  External interactions can affect the motion of the system object  Force and force diagrams o Force is a vector quantity that characterizes how hard (magnitude) and in what direction an external object pushes or pulls on a system object o The SI unit for force is F o Arrows represent the forces  Length of the arrow indicates the strength of the force  Direction of the arrow point related to the direction in which the force is exerted on the object o Includes forces exerted on the system object o Shows the forces at a single instant in time  Normal forces o Forces perpendicular to a surface are called normal forces o We usually label normal forces with the letter N o Normal forces are contact forces due to object touching each other o Normal forces are always perpendicular to a surface, but not necessarily vertical  Adding forces graphically o Draw the force head to tail o The net force is the vector that goes from the tail of the first vector to the head of the second vector  Measuring force magnitudes o A spring is a simple device that can be used for measuring forces o The amount it is compressed or stretched can be calibrated using a known standard force o An unknown force that compresses or stretches the spring by the same amount must be equal to the force used for calibration o Springs are often linear—twice the force stretches the spring twice as much  Relating force and motion o A net force on an object changes its motion o If the net force is in the direction of its velocity, then the object speeds up o If the net force is in the direction opposite the velocity, then the object slows down  Inertial reference frames o An inertial reference frame is one in which an observer:  Sees that the velocity of the system object does not change if no other object exert forces on it or sees no change in the velocity if the net force exerted by all objects on the system object is zero  Non-inertial reference frames o In a non-inertial reference frame, the velocity of the system object can change even though the sum of forces exerted on it is zero  Unless you are driving at a constant velocity in a straight line, this is not an inertial reference frame  The velocity of the coffee cup might suddenly chance if the car accelerates  Newton’s first law of motion o For an observer in an inertial reference frame, an object will continue to move at constant velocity (or remain at rest)  When no other object exerts force on the system object or when the net force exerted on the system object is zero o Inertia refers to the characteristic that objects have for continuing to move at constant velocity when the net force exerted by other objects is zero  Force and acceleration o We observe a linear relation between force and acceleration  As acceleration increases, the force also increases  Newton’s second law of motion o The resulting acceleration of a system object is the proportional to the net force acting on the system object o The resulting acceleration is in the same direction as the net force acting on the system object o They are proportional: if the net forces doubles, the resulting acceleration will also double  Mass o Mass is a measure of the amount of matter  SI unit for mass is m, which is measured in kg  Newton’s second law of motion o We observe that the acceleration of system object, when subjected to the same force, is inversely proportional to their mass o a system(ΣF on the systemm system  a  is the acceleration of the system object in the x-direction sstem  ΣF on the systemhe next force in the x-direction acting on the system object  m is the mass of the system object o Forces that point in the negative direction have a negative component o Forces that point in the positive directions have a positive component o Units of Newton’s are: 2  1 N = 1kg*m/s  Gravitational force law o In the absence of air resistance all objects fall straight down with the same acceleration  This acceleration has a magnitude of approximately 9.8 m/s 2 near Earth’s surface o The earth (E) exerts the only force on falling objects (O)  F = m *a = m (9.8m/s ) 2 E on O 0 0 0 o We define g such that  F E on Om *0  g = 9.8N/kg  Weight o The weight of the object on a planet is the force that the planet exerts on the object o In everyday language, the normal force that a scale exerts on you (which balances the downward force you exert on it) is your weight  Newton’s third law of motion o When two objects interact, object 1 exerts a force on object 2, but object 2 exerts a force on object 1  These forces are equal in magnitude but opposite in direction  F object 1 on object 2object 2 on object 1 o These forces are exerted on different objects and cannot be added to find the sum of the forces acting on one object o Tips:  The forces in Newton’s third law are exerted on two different objects  This means that the two forces will never appear on the same force diagram  They should not be added together to find the net force on one of the objects  You have to choose the system object and consider only the forces exerted on it o Example:  A book is lying on a table  The book is the system object  The book interacts with the table (normal force pushing upwards) and the earth (gravity force pushing downwards) o The table exerts sufficient normal force to keep the book from sinking into the surface o The book mist remain in contact with the table to be influence by this force  If the mass of the book was 2kg, then the Earth exerts a force FE on B ,z*g =z(2kg)(-9.8N/kg) = -19.6N  Therefore, F T on B ,z19.6N o If the book slides with constant velocity along the surface of the table, the only interactions are still just due to the table and the Earth (normal force and gravity) o But when it slides off the table, the only force is gravity o The book accelerations (in the downward direction) 2  az= F zm = -19.6N/2kg = -19.6kg*m/s /2kg = -9.8 m/s Apply Newton’s Laws   Example: o You are approaching a red light. You slowdown from 50km/h (13.89m/s) to rest over a distance of 100m. If your mass is 80kg, what force do you feel?  F = (m*a ) and a = (V 2 – V2 )/2(X-X ) x 2 2x x 0x 0 2  F = (m* (V x– V 0x)/2(X-X0) = (80kg*(0-13.89m/s) )/2(100m-0) = 77N  Forces in two dimensions o A small box is being pulled by two spring scales o Each scales is at angle θ with respect to the x-axis o How can we replace one scale with two scales that lie along the x- and y-axes?  Pythagoras’s theorem:  F = F + F 2  F = (F + F ) 2 1/2 x y x y  Fy= F sinθ  F = F cosθ  x  F = 5N (given) and suppose θ = 37˚  Fx= 5N cos(37) = 4N  Fy= 5N sin(37) = 3N  F = (4N + 3N )2 1/2= 5N o Net force is the same (zero)  Newton’s second law in two dimensions o The sum of scalar components of the forces along the x-axis equals mass time acceleration along the x-axis o The sum of scalar components of the forces along the y-axis equals mass time acceleration along the y-axis o If possible, we might pick the direction of the x-axis to be in the direction of the acceleration o Then the component of the acceleration along the other axis will be zero o This would simplify the problem, but might not always be possible o This carefully about which choice of geometry makes the problem easier o Example:  Princess Anna Karenina pushes a 2kg book onto a slippery wall, exerting a force of 21N at an angle of 45˚ above the floor  What is the normal force of the wall on the book? o Forces acting on book:   F E on Bforce due to Earth gravity  FA on Bforce due to Anna pushing  NW on Bnormal force of the wall  The acceleration in the x-direction is zero  aB ,x ΣF xm B 0 (book can only slide up/down wall)  FA on B ,x1N cos(45) = 14.8N  NW on B ,x-14.8N  Gravity exerts a force in the –y direction  FE on B ,y(2kg)(9.8N/kg) = -19.6N  Vertical component due to Anna  F = 21N sin(45) = 14.8N A on B ,y  Acceleration  AB ,y (-19.6N + 14.8N)/2kg = 2 -2.4m/s  Does the book slide up or down the wall? o The book slides down the wall  Incline planes o Gravity is always in the downward vertical direction o The normal force is always perpendicular to a surface o The motion is along the surface of the plane o Sometimes it is easier to align the x-axis with the direction of motion rather than the usual horizontal direction:  a x (ΣF )xm = ((m*g)sinθ)/m = g*sinθ  a x g*sinθ  Tension in string o We assume that strings will not stretch o Strings can only pull o The tension, T, in a string is the same everywhere o Example:  m 1 boxs on a surface attached to m 2 boxhat is attached to a pulley string rig hanging off the surface  m 1ccelerates in the +x direction  m 2ccelerates in the –y direction  The magnitude of the acceleration are equal because the string doesn’t stretch  a1= T/m 1  a1= -a 2 (T-F E on m2/m2= (T-m *g2/m = -2/m 2  Solve for tension: o T(1/m 2 1/m ) =1g o T = g(m m1/m2+m 1 2  Substitute it in to solve for a: o a1= T/m = 1(m /m +m 2 = 1a 2 2  Friction o There are two kinds of friction  Static friction: the maximum horizontal force that can be applied before an object starts to move  Kinetic friction: the horizontal force that acts in the direction opposite the motion of an object o How can we observe these forces?  Perform an experiment using spring scales to measure horizontal and vertical forces o Static friction force is parallel to the surfaces of two objects that are not moving in relation to each other and opposes the tendency of one object to move across another o Static friction force changes magnitude to prevent motion, up to a maximum value called the maximum static friction force o Sometimes friction is a necessary phenomenon for movement—for example when walking on a flat horizontal sidewalk o Static friction prevents your shoes from slipping at the start and end of a step  Static friction and the normal force o A surface really exerts one force, but we can analyze each component separately  Magnitude of maximum static friction o Maximum static friction is independent of surface area it’s in contact with o Maximum static friction is proportional to the normal force  Measuring the static friction force o The hypothesis that the maximal static friction force depends on mass is not correct  There is a constant ratio between the maximum static friction force and the normal force  Relationship between normal force and friction force o The normal force and friction force are two perpendicular components of the same force—the force a surface exerts on an object o The ratio between the maximum static friction force and the normal force is constant in all trials o The proportional constant is different for different surfaces; the proportionality depends on the types of contact surfaces o The proportionality constant is greater for two rough surfaces contracting each other and smaller for 2 smoother surfaces contacting o This ratio is the coefficient of static friction:  μ s f s maxN  Coefficient of static friction o The coefficient of static friction is a measure of the relative difficulty of sliding two surfaces across each other o The easier it is to slide on surface on the other ,the smaller the coefficient o This coefficient is unit-less and typically has a value between 0 and 1 o μ s f s maxN is reasonable only in situations in which the following conditions hold:  Relatively light objects are resting on relatively firm surfaces  The objects never cause the surfaces to deform significantly (for example, they do not involve a tire sinking into mud)  Kinetic friction o Kinetic indicates that the surfaces in contact are moving relative to each other o A similar relationship exists as between the friction force and the normal force, but with two important differences:  Under the same conditions, the magnitude of the kinetic friction force is always lower than the maximum static friction force  The resistive force exerted by the surface on the moving object has a constant value  Assumptions for our kinetic friction model o Our equation f kμ *k is reasonable only in situations in which the following conditions hold:  It cannot be used for rolling objects  It makes the same assumption about the rigidity of the surfaces as the model for static friction  The objects cannot be moving at high speed o This equation does not have general applicability, but it is useful for rigid surfaces and objects moving at everyday speeds  What causes friction? o Even the slickest surfaces have bumps that can hook onto the bumps on another surface o If two surfaces are too smooth, friction increases again:  This is due to attraction between particles at the surface that are too close together without typical surfaces bumpiness  Determining friction experimentally o A shoe is being pulled horizontally with a spring scale until the shoe begins to slide o Just before the shoe starts to slide, its acceleration is zero, and the scale indicates the maximum force of static friction that the tile exerts on the shoe  0 = T Scale on S ms T on S max o Tip:  The magnitude of the normal force that a surface exerts on an object does not necessarily equal the magnitude of the gravitational force that Earth exerts on the object, especially when the object is on an inclined surface  Example: o If a driver slams on the brakes, the tires can lock, causing the car to skid. Police officers use the length of the skid marks to estimate the initial speed of the vehicle. A car involved in a minor accident left 18m of skid marks on a horizontal road. After inspecting the car and road surface, the police officer decided the coefficient of kinetic friction was 0.80. The speed limit was 15.6m/s on that street. Was the car speeding?  Relation between a, V , 0nd X:  Ax= (V 2x– V20x)/2(X-X0)  Newton’s second law:  Fx= ma = x N =kμ mg k  Initial position, 0 =0, Final position, Vx=0 2  V x = 2a x = 1/2gk 2 1/2  Vx= (2μ gk) = (2*0.80*9.8m/s *18m)  Vx= 16.8m/s  16.8m/s > 15.6m/s, so the driver was speeding  Using Newton’s third law to explain how static friction helps a car start and stop o Increasing or decreasing a car’s speed involves static friction between the tire’s region of contact and the pavement  To move faster, turn the tire faster. The tire then pushes back harder on the pavement, and the pavement pulls forward more on the tire  To slow down, turn the tire more slowly. The tire pulls forward on the pavement, which in turn pushed back on the tire to accelerate the car backward (pavement pushes back on the tire) o The force responsible for this backward and forward push is the static friction force that the road exerts on the tires  Other types of friction o Rolling friction is caused by surfaces of rolling objects indenting slightly as they turn  This friction is decreased in tired that have been inflated to a higher pressure o The friction that air or water exerts on a solid object moving through the air or water is called drag force Projectile Motion   Projectile motion o Projectiles are objects launched at an angle with respect to some horizontal surface o Throw a ball straight up while moving on roller blades  As long as you do not change your speed or direction while the ball is in flight, it will land back in your hands  Qualitative analysis of the projectile motion in the y-axis  A ball thrown straight up in the air by a person moving horizontally on rollerblades will land in the person/s hand  Earth exerts a gravitational force on the ball, so its upward speed decreased until it stops at the highest point, and then its downward speed increases until it returns to your hands  With respect to you on the rollerblades, the ball simply moves up and down  Qualitative analysis of the projectile motion in the x-axis  A ball thrown straight up in the air by a person moving horizontally on rollerblades will land in the person/s hand  The ball is also moving horizontally  No object exerts a horizontal force on the ball. Thus, according to Newton’s first law, the ball’s horizontal velocity does not change once it is released and is the same as the person’s horizontal component of velocity  Quantitative analysis of projectile motion: acceleration o The equations of motion for velocity and constant acceleration are used to analyze projectile motion quantitatively o The x-component (in the horizontal direction) of a projectile’s acceleration is zero o The y-component (in the vertical direction) of a projectiles acceleration is –g  The force is mg—the force of gravity that Earth exerts on the projectile  Quantitative analysis of projectile motion: velocity o For a projectile launched at speed V at0an angle θ with respect to the horizontal axis:  The x-component of the velocity remains constant during the flight because the acceleration in the x-direction is zero  Quantitative analysis of projectile motion: using kinematic equations o Projectile motion in the x-direction:  X(t) = X +0V cos0)t  This can be used to determine how far the projectile travels in the horizontal direction during that time interval o Projectile motion in the y-direction: 2  Y(t)= Y +0(V sin0)t – ½ gt  This can be used to determine the time interval for the projectiles flight  Best angle for farthest flight o What is the angle at which you should throw the rock so that it travels the longest horizontal distance, assuming it is always thrown with the same initial speed?  Y(t)= Y +0(V sin0)t – ½ gt 2  Initial height: 0 =0  Final height: y(t)=0. 2  (V0sinθ)t – ½ gt = 0  t = (2V 0inθ)/2 (substitute into the X(t) equation)  X = X 0 (V c0sθ)t  Initial position: 0 =0  Maximum distance is when θ = 45˚  Example: o Dave is to be shot at a cannon that is oriented at 45˚ above the horizontal o What is the muzzle velocity needed for him to land in a net located 40m away at the same height as the muzzle of the cannon?  X = (V 0osθ)(2V s0nθ/g)  2cosθsinθ=sin2θ  X = (V 0in2θ/g)  When θ=45˚, sin2θ=1, so 1/2 2 1/2  V0=(xg) ((40m)(9.8m/s )) = 20m/s Uniform Circular Motion   The qualitative changes in velocity o At any instance, the instantaneous velocity is tangent to the path along which the object moves o In circular motion, the system object travels in a circle and the velocity is always tangent to the circle o Even if an object is moving with constant speed around a circle, its velocity changes direction o A change in velocity means there is acceleration  Estimating the direction of acceleration o The method is used to estimate the direction of acceleration of any object during a small time interval ∆t=t -tf i  Tips for estimating the direction of the acceleration o Make sure that you choose initial and final points at the same distance before and after the points at which you are estimating the acceleration direction o Draw a long velocity arrow so that when you put them tail to tail, you can clearly see the direction of the velocity change arrow o Make sure that the velocity change arrow points from the head of the initial velocity to the head of the final velocity so that vI+ ∆v  v f  Newton’s second law and circular motion o The sum of the forces on an object moving at constant speed along a circular path points toward the center as the object accelerates o When the object moves at constant speed along a circular path, the net force has no tangential component  Factors that might affect acceleration o Imagine your experience in a car driving around a traffic circle  The faster the car moves around the circle, the greater the risk that the car will skid off the road  For the same speed, there is a greater risk of skidding on the inner lane (smaller radius)  We guess that that acceleration depends on both v and r  Dependence of acceleration on speed o ∆v ’ = 2∆v  o ∆t’ = ∆t/2  Time interval is only half as long, because the object is moving twice as fast o The magnitude of the acceleration is proportional to the square on the veloci2y  a rv  Dependence of acceleration on radius o If v remains constant, how long does it take the object to move in a circle of radius r? o The circumference of a circle is C = 2Πr  If the radius increases by a factor of 2, then the circumference increases by a factor of 2  The net change in velocity is the same  Since velocity is constant, it will take twice as long to travel around the circle o The magnitude of the acceleration is inversely proportional to the radius  a r 1/r o In fact, the radial acceleration is:  a r v /r  Radial acceleration o For motion in a circle of radius r with constant speed v, the radial acceleration is: 2  a r v /r o The acceleration points towards the center of the circle  Period of circular motion o The period (T) is the time interval it takes for an object to travel once around an entire circular path o For constant speed, circular motion, we divide the circumference by the velocity to get:  T = C/v = 2Πr/v  Example: o What is the radial acceleration when you sleep in your bed in Ecuador (0˚)?  24 hours in a full circular motion for Earth (86400s)  Radius of earth is r=6400km (6.4x10 m) 6 2  a r v /r  We know r, so we need to find v:  V = c/T = 2Πr/T 2 2 2 2  a = v /r = 1/r (2Πr/T) = 4Π r/T  4Π (6.4x10 m)/(86400s ) = 0.034m/s 2 o What is the radial acceleration when you sleep in your bed in West Lafayette (40˚ N)?  Radius of circular path:  RWL = r cosθ 2 2 2 2 2 6  a = 4Π r/T  a WL = 4Π r cosθ/T = (4Π (6.4x10 m)cos(40))/ ( 86400s ) = 0.026m/s 2 o Highway curves are banked to prevent cars from skidding off the road. The angle of the bank depends on the expected speed and the radius of the curvature. 2  Radial acceleration: a r v /r  Radial force: F r ma = rv /r 2  Fris zero when curve is banked correctly  N cosθ = mg 2  N sinθ = mv /r  tanθ = sinθ/cosθ = v /gr  Suppose r=250m and v=100km/hr (27.78m/s)  tanθ = v /gr = (27.78m/s) /(9.8m/s )(250m) = 0.31  θ = tan (0.31) = 17˚ o A certain road has a bump that is approximately circular with a radius of a curvature r = 30m. What speed do you have to drive over the bump to feel weightless (normal forces vanish)?  Ns on y, y, v = gr 1/2 2 1/2  V = (gr) = ((9.8m/s )(30m)) = 17.1m/s  Tips for circular motion o There is no special force that causes the radial acceleration of an object moving at constant speed along a circular path o This acceleration is caused by all of the forces exerted on the system object by other objects o Add the radial components of these regular forces o This sum is what causes the radial acceleration of the system object  Conceptual difficulties with circular motion o When sitting in a car that makes a sharp left turn, you feel throw outward, inconsistent with the idea that the net force points toward the center of the circle (inward) o Because the car is accelerating as it rounds the curve, passengers in the car are not in an inertial reference frame o As you turn left, the car itself goes left but the you continue to travel straight because the net force exerted on you is zero Planetary Motion  Planetary motion o Newton was among the first to hypothesize that the Moon moves in a circular orbit around the Earth because Earth pulls on it, continuously changing the direction of the Moon’s velocity o He wondered if the force exerted by Earth on the Moon was the same type of force that Earth exerted on falling objects, such as an apple falling from a tree o Newton compared the acceleration of the Moon if it could be modeled as a point particle near Earth’s surface to the acceleration of the moon observed in its orbit  Motion of the Moon 2 o a = v r 8 o The orbit of the moon is approximately circular with r = 3.8x10 m = 60R E o The period of rotation is T=27.3 days (2358720s) o The speed of the moon around the earth is:  v = 2Πr/T o The accelerati2n i2 there2ore: 8 2 -3 2  a = 4Π r/T = 4Π (3.8x10 m)/(2358720s) = 2.69x10 m/s o You might wonder why if Earth pulls on the Moon, the Moon does not come closer to the Earth in the same way that an apple falls from a tree o The difference in these two cases is the speed of the objects. The apple is at rest with respect to Earth before it leaves the tree, and the Moon is moving tangentially o If Earth stopped pulling on the Moon, it would fly away along a straight line  Dependence of gravitational force on mass o Newton deduced F E on Mm moonby recognizing that acceleration would equal g only if the gravitational force was directly proportional to the system object’s mass o Newton deduced F E on M m earth applying Newton’s third law of motion  The law of universal gravitation o Newton deduced:  FE on MF M on E m earthmoonr2 o Gravitational constant:  G = 6.67x10 -1N*m /kg 2  This indicates that this law works anywhere in the universe for any two masses  Kepler’s first law of planetary motion o The orbits of all planets are ellipses, with the Sun located at one of the ellipse’s foci o The shape of the planetary orbits is close to circular for most planets. In those cases, the foci of the ellipse are very close to the center of the circular orbit that most closely approximates the ellipse  Kepler’s second law of planetary motion o When a planet travels in an orbit, an imaginary line connecting the planet and the Sun continually sweeps out the same area during the same time interval, independent of the planet’s distance from the Sun  Kepler’s third law of planetary motion o The square of the period T of the planet’s motion (time interval to complete one orbit) divided by the cube of the semi-major axis of the orbit (which is half the maximum diameter of the elliptical orbit and the radius r of a circle) equals the same constant for all known planets:  T /r = K  Gravitational and inertial mass o ΣFx= ma i x (inertial mass) 2 o FE on m G( Mearthg/r ) (gravitational mass)  Geostationary satellites o Geostationary satellites stay at the same location in the sky. This is why a satellite dish always points in the same direction o These satellites must be placed at a specific altitude that allows the satellite to travel once around the Earth in exactly 24 hours while remaining above the equator o An array of such satellites can provide communications to all parts of Earth o Example:  At which altitude above the equator must the satellite orbit be to provide continuous communication to a stationary dish antenna on Earth?  Mass of Earth = 5.98x10 kg 24  The only force acting on the satellite is gravity  F r ma = r(M earthr ) = m(v /r)  V = 2Πr/T  G(M earth ) = 4Πr /T2 3 2 2  r = G(M earth)/ 4Π  r=(G(M earth )/4Π ) 1/3 -11 3 2 24 2 2 1/3  r=((6.67x10 m /s /kg)(5.98x10 kg)(86400s) /4Π ) = 4.23x10 m = 42300km  Are astronauts weightless in the ISS?—No o Earth exerts a gravitational force on them  This force causes the astronaut and space station to fall toward the earth at the same rate while they fly forward, staying on the same circular path o The astronaut is in free fall (as in the station)  If the astronaut stood on a scale in the space station, the scale would read zero even though the gravitational force is nonzero o Weight is a way of referring to the gravitational force, not the reading of the scale Impulse and Linear Momentum   First: accounting for mass o The mass of a log in a campfire decreases as the log burns. What happened to the lost mass? o If we choose only the log as the system, the mass of the system decreases as it burns o Air is needed for burning. What happens to the mass if we choose the surrounding air and the log as the system? o If we burn steel wool in a closed container the total mass doesn’t change, but if we burnt it in an open container the total mass would decrease  Law of constancy of mass o When a system of objects is isolated, its mass equals the sum of the masses of components and remains constant in time o When the system is not isolated, any change in mass is equal to the amount of mass leaving or entering the system  Accounting for changing mass o The mass is constant if there is no flow of mass in or out of the system o The mass changes in a predictable way if there is some flow of mass between the system and the environment  Observing collision of two carts o One quantity remains the same before and after the collision in each experiment: the sum of the products of the mass and x-velocity component of the system objects o Hypothesis to test: the sum of masses times velocity is the quantity characterizing motion that is constant in an isolated system  Important points about linear momentum o Linear momentum is a vector quantity; it is important to consider the direction in which the colliding objects are moving before and after the collision o Momentum depends on the choice of the reference frame. Different observers will measure different momenta for the same object o To establish that momentum is a conserved quantity, we need to ensure that the momentum of a system changes in a predictable way for the systems that are not isolated  Momentum of an isolated system is constant o To describe a system with more than two objects, we simply include a term on each side of the equation for each object in the system  Example: o Jen (50kg) and David (75kg), both on rollerblades push off each other abruptly. Each person coasts backward approximately constant speed. During a certain time interval, Jen travels 3m. How far does David travel during the same time interval?  What is Jens final velocity?  VJ ,x X/jf= -3m/t f  What is Jen’s final momentum?  P J ,x mv J J,x  What is David’s final momentum?  P D,x= m vD D,x= m (XD/t D =f-p J,x= -m (XJ/tJ) f  m DX /D )f= -m(X/tJ) J f  X D -(m/m )J = D(50Jg/75kg)(-3m) = 2m  Impulse due to a force exerted on a single object o A relationship can be derived from Newton’s laws and kinematics  a = (v –f )/(i -t)f=iΣF /m        mv -fmv = p i p f i = ΣF (t -tf) iimpulse  Both force and time interval affect the momentum: a small force exerted over a long time interval can change the momentum of an object by the same amount as a large scale force exerted over a short time interval  Impulse o The impulse J of a force is the product of the average force F  avg exerted on an object during a time interval (t -t) and that time f i interval:  J = F avg(tf– t i o Impulse is a vector quantity that points in the direction of the force. The impulse has a plus or minus sign depending on the orientation or the force relative to a coordinate axis. The SI unit for impulse is N*s = 2 (kg*m/s )*s = kg*m/s  Impulse-momentum equation for a single object o If the magnitude of the force changes during the time interval  considered, then we just use the average force F avg  Example: o A 60kg person is traveling in a car that is moving at 16m/s with respect to the ground when the car hits a barrier. The person is not wearing a seat belt, but is stopped by an air bag in a time interval of 20s. Determine the average force that the air bag exerts on the person while stopping him.  Initial momentum:  p=iv i  Final momentum:  P f0  Change in momentum:  p f p =i-mv=J=F i avg(tf– t i  Average force:  F =(-mv/ t – t) = -(60kg)(16m/s)/0.2s = -4800N avg i f i  Using Newton’s laws to understand the constancy of momentum o Newton’s third law provides a connection between our analysis of two colliding carts o Interacting objects at each instant exert equal-magnitude but oppositely directed forces on each other  Initial momentum = final momentum  m 1 1i+ m v 2 2i = m v 1 1f + m v2 2f  With an additional, external impulse:  m v1 1i + m v2 2i + J = m v 1 1f + m v2 2f  Determining the speed of a bullet o It might be easier to measure the distance it takes for an object to stope during a collision o The stopping distance can be measured after a collision, such as how far a car’s front end crumples or the depth of a hole in the left by a meteorite o Impulse-momentum tells us information about the stopping time; we must use kinematics to relate this to distance  Determining the stopping time interval from the stopping distance o Assume that the acceleration of the object while stopping is constant o The average velocity is the average of the initial and final velocities v avg= ½ (v -vf) i o The stopping displacement and the stopping time interval are related by:  Xf-X i v avg= (t fti = ½ (v +v fxt -ix) f i o Solving for the stopping time:  tf-ti= 2(X -Xf)/(iv +v fx ix  Example: o The record for the highest movie stunt fall without a parachute is 71m, help by an 80kg man. His fall stopped by a large air cushion, into which he sank about 4m. His speed was approximately 30m/s when he reached the top of the air cushion. Estimate the average force that the cushion exerted on this stunt diver’s body while stopping him.  Impulse-momentum relation:   p fp i -mv =Ji= F avg(tft)i  F avg = (p fp)i(t -f)i  We know the initial momentum:  p i -mv i  Time interval:  tf-i = 2(X -f)/i v +vfx ix  We know the initial and final velocity is zero  2  Favg = (p fp)i(t ft)i= -(pv)i2ix -x) f -imv ) 2(x -i ) f i  Favg= -(80kg)(36m/s) /2(-4m) = 12960N  Jet propulsion o Cars change velocity because of an interaction with the road; a ship’s propellers push water backward o A rocket in empty space has nothing to push against  If the rocket and fuel are at rest before the rocket fires its engines, the momentum is zero. Because there are no external impulses, after the rocket fires its engines, the momentum should still be zero  Burning fuel is ejected backward at high velocity, so the rocket must have a nonzero forward velocity  Thrust o Thrust is the force exerted by the fuel on a rocket during jet propulsion 6 o Typical rocket thrusts measure approximately 10 N, and exhaust speeds are more than 10 x the speed of sound o Thrust provides the impulse necessary to change a rocket’s momentum  The same principle is at work when you blow up a balloon, but the open valve and release it, and when you stand on a skateboard with a heavy ball and throw the ball away from you Work and Energy   Energy o Another conserved quantity is energy o Unlike momentum, energy is a scaler quantity o “Energy” is kind of like the propensity for an object to break another object  The more “energy” the system has, the easier it is to break the other object  Gravitational potential energy o The energy of an object-Earth system associated with the elevation of the object above Earth is called gravitational potential energy (symbol U ) g o The higher above the Earth the object is, the greater the gravitational potential energy  Kinetic energy o The energy due to an object’s motion is called kinetic energy o The faster the object is moving, the greater the kinetic energy  Elastic potential energy o The energy associated with an elastic object’s degree of stretch is called elastic potential energy (symbol U s o The greater the stretch (or compression), the greater the object’s elastic potential energy  Internal energy o If an object slides on a surface, the surfaces in contact can become warmer o Structural changes in an object can occur when an external force is applies o The energy associated with both temperature and structure is called internal energy (symbol U )m  Positive, negative, and zero work o When the direction of the external force exerted on the system is in the same direction (θ=0˚ and cos0˚=+1.0) as the object’s displacement, it does positive work, causing the system to gain energy o When the direction of the external force exerted on the system is opposite ((θ=180˚ and cos0˚=-1.0) the object’s displacement it does negative work, causing the energy of the system to decrease o When the direction of an external force is perpendicular (θ=90˚ and cos90˚=0) to an object’s displacement, it does zero work on the system, causing no change to its energy  Defining work as a physical property  o Work done by a constant external force F exerted on a system object while the system object undergoes a displacement d is:   W = F d cosθ o Where F is the magnitude of the force in N (always +), d is the magnitude of the displacement in meters (always +), and θ is the angle between the direction of F and the direction of d. The sign of cosθ determines the sign of work. Work is a scalar physical property. (unit of work is in J, J = N*m)  Important tip o It is tempting to equate the work done on a system with the force that is exerted on it o In physics, there must be a displacement of a system object for an external force to do work o Force and work are not the same thing  Example: o Two friends are cycling up a hill inclined at 8˚. The stronger cyclists helps his friends up the hill by exerting a 50N pushing force on his friend’s bicycle and parallel to the hill, while the friend moves a distance of 100m up the hill. The force exerted on the weaker cyclists and the displacement is in the same direction. Determine the work done by the stronger cyclist on the weaker cyclist.  W = F d cosθ = (50N)(100m)cos(8) = 4950N*m  Total energy o The total energy U of a system is the sum of all these energies in the system  Total energy = U = K + U + U g U s int o Hypothesis: if no work is done on the system, the energy of the system should not change; it should be constant  Is the energy of an isolated system constant? o When released from the same vertical height with respect to the table, the car lands the same distance from the table  Conservation of energy o Energy of an isolated system is constant and the different process inside the system convert energy from one form to another o We reason that work is a mechanism through which the energy of a non-isolated system changes  Work-energy bar charts o A work-energy bar chart indicates the relative amounts of a system's different types of energy in the initial state of a process, the work done on the system by external forces during the process, and the relative amounts of different types of energy in the system at the end of the process  Generalized work-energy principle o Generalized work-energy principle  The sum of the initial energies of a system plus the work dose on the system by external forces equals the sum of the final energies of the system  U i W = U f  (K + U + U + W = (K + U + U + ∆U ) i gi si f gf sf int  ∆U int int f Uint i  Work and energy o Total energy = U = K + U + U + Ug s int  K is the kinetic energy  Ugis the gravitational potential energy  Usis the elastic potential energy  Uintis the internal energy o An external force does work on a system and can change its total energy:  U i W = U f  Gravitational potential energy o A rope lifts a heavy box upward at a constant negligible velocity. The box is the system.  The box moves at constant velocity; the upward tension force of the rope on the box is equal in magnitude to the downward gravitational force Earth exerts on the box: m boxg  The rope does work on the box, changing its vertical position from y io y f  W R on B T R cosθ = T (yR– f)cosi˚  =mg(y – f) i  U gig(y – yf = i gf o Changes in gravitational potential energy only depend on changes in height of an object o It is convenient to define U togbe zero when the object is at the origin of the vertical axis o Then, at a time, the object has gravitational potential axis:  Ug=mgy o Units of gravitational potential energy are:  Kg(N/kg)m = Nm = J (joule)  Kinetic energy o The equation we found can be put in terms of properties of the cart using dynamic and kinematics:  m c c = F H on C 2 2  FH on C = K –fK = i m v – ½cm f c i o The kinetic energy of an object is:  K = ½ mv 2 o Kinetic energy also has units of joules:  Kgm /s = (kgm/s)m = Nm = J  Example: o You sit on the deck behind tour house. Several 5g acorns fall from the trees high above, just missing your chair and head. Use work-energy equation to estimate how fast one of these acorns is moving just before it reaches the level of your head.  Initial potential energy is:  U gmgy i  Initial kinetic energy is:  K i 0  Final potential energy:  U gf0  Final kinetic energy: 2  K f ½ mv = U = mggi i  Solve for v: 1/2  V = (2gy) i  Elastic potential energy o When you stretch or compress an elastic spring-like object, you have to pull or push harder as the object is stretched or compressed more  This force is not constant o This factor makes it more difficult to find a mathematical expression for the elastic potential energy stored by an elastic object when it has been stretched or compressed  Hooke’s law o Tow springs of the same length—one less stiff and the other more stiff —are pulled, and the magnitude and the distance that each spring stretches from its unstretched position are measured  More force is applied to the more stiff spring to pull them at equal distance o The magnitude of the force exerted by the scale of each spring is proportional to the distance that each spring stretches:  FScale on Springx o From Newton’s third law:  FSpring on Scalekx  Elastic force (Hooke’s law) o If any object causes a spring to stretch or compress, the spring exerts an elastic force on that object o If the object stretches the spring along the x-direction, the x- component of the force that the spring exerts on the object if:  FSpring on Objectkx o The spring constant, k, is a property of the spring and is measured in N/m  Elastic potential energy o To calculate the work done on the spring by such a variable force, we


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