Phy 101 Week 4 Lecture Notes
Phy 101 Week 4 Lecture Notes PHY 101 - M001
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This 5 page Class Notes was uploaded by Bryce Caplan on Tuesday September 29, 2015. The Class Notes belongs to PHY 101 - M001 at Syracuse University taught by K. Foster in Fall 2015. Since its upload, it has received 39 views. For similar materials see Major Concepts of Physics I in Physics 2 at Syracuse University.
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Date Created: 09/29/15
Phy 101 Lecture 6 92115 Ball in a circle A ball is placed into a metal ring and rolled around the periphery If a section of the ring is removed and the ball is rolled around the periphery it will go out of the ring in the direction tangent to where it leaves the ring Two carts Two carts are situated on a floor with a person on each cart Each end of a rope is given to the people on the carts They pull the rope and both move closer to each other The two then come close and push off of each other and both move away from where they started In both cases both carts moved Pulley carts A cart is on a track and attached to a pulley mechanism The other end is attached to a weight When the weight drops the cart is pulled with a constant acceleration Spring carts A spring is connecting two carts whenever a force is exerted to move one cart the other cart experiences the same force Tension a force transmitted through an object usually a rope Tension of an object is equal to the force with which the object pulls the rope Two ropes are attached to a third rope that is attached to a mass The third rope suspends the mass vertically and the two ropes are attached at an angle The ropes all have the same amount of tension because the total tension in the rope that is attached to the mass is the force of gravity times the mass and the two ropes attached to that rope have a total tension of the vertical and horizontal directions When the tension in both directions is added up it comes out to the same value as that experienced by the rope that is directly connected to the mass Two masses are attached to a pulley one has a mass of 70 kilograms and is at rest on the ground while the other which has a mass of 56 kilograms is suspended in the air What is the force of tension on the pulley To approach this problem the first thing we notice is that the 70 kilogram mass is not suspended meaning that the ground underneath it is providing the force normal to gravity This causes there to be no tension on the rope from the 70 kg mass so it must all come from the suspended 56 kg mass Because mass is in Newtons all that has to be done to answer this question is multiply the 56 kg mass by 98ms2 effecting an answer of 5488N A pulley has two masses attached to it One mass is hanging off the edge of a table while the other is sitting on that table The mass hanging off of the table is falling How quickly is the mass accelerating toward the ground In order to answer this question a free body diagram would be very helpful First we have to figure out what s happening with the mass that s falling It is not moving side to side so we can ignore the x direction We know that the force of gravity is greater than any other force acting on it vertically because it is falling We also know that the tension from the string holding it up is there just less than the pull of gravity Ideally we would be given the masses of the two objects and the friction of the table Assuming that we have those we can figure out the tension of the rope We figure out the force of friction on the mass that s on the table Then because we know that s how much the mass on the table is pulling back we know that that is the tension in the string The tension in the string is what s counteracting the pull of gravity on the other object so we have to find the net force With the net force we can divide out the mass of the falling object and figure out the acceleration This question is also a good example of apparent weight Apparent weight is the force one object exerts on another due to gravity but if both are falling then the apparent weight is less The apparent weight of the mass that was falling is the amount of force of tension because that was how hard the box was pulling on the string Free body diagrams and free fall When an object is in free fall there is nothing preventing the object from continuing to fall so it will just keep accelerating with gravity An object in free fall therefore has no apparent weight because the only object in the system is the mass that is falling A free body diagram would only have one arrow coming off of the mass and that would be the force due to gravity acting on the mass Phy 101 Lecture 7 Sept 23 2015 The exam is pen and paper You will have to make a free body diagram on the test Air or other motion resistance We re going to stop ignoring air resistance and friction Assume that Fdforce of dragbv2 b is a parameter that depends on the size and shape of the object basically surface area that the force is acting on Since Fd is proportional to v2 can the object be in equilibrium Because Fd w ma we can set aO then substitute Fd and get bv2 mg 0 Then we can figure out v from that equation Dropping a coffee lter A coffee filter is dropped above a sonic ranger distance measuring deviceThe sonic ranger measures a steady drop in position and a slight increase in velocity from the time of the drop The velocity very quickly becomes constant This is due to the high amount of air drag on the shape of the coffee filter Description of uniform circular motion A spinning DVD has different velocities and accelerations at different points Because of that we use a simpler lingo and say how many times an object rotates per second This requires us to use angles and angular displacement instead of distance Angular velocity D Spinning disk demo A disk spins on a surface with two blocks on it One block is on the edge and one is near the center The disk is spun and the block on the outside is seen to have a much higher velocity than that on the inside Keep perspective in mind counterclockwise from one perspective is clockwise from the opposite A CD makes one complete revolution every tenth of a second The angular velocity of a point on the outside is the same as that of a point on the inside because the angle is the same If the point on the inside moves 90 degrees so does the point on the outsideThe velocity no adjective is greater on the outside however Radians A radian is a measure of a circle using the distance around the circle However because circles can have different radii that would make everything super complicated to remember So what we do is we use the formula Circumference2 7V and say that r doesn t really matter Thus a full circle is 2nradians a half is 7 a quarter is 752 etc If you want a simple formula ifx is the number of degrees lx go radians Relation between Linear and Angular speed Arc length is the path of a point moving in a circular path during an angular displacement ofAG Thus srAG or in spoken word displacement is equal to the radius of the circle times the absolute value of the change of the angle If you have a graphing calculator or scientific calculator it is very helpful to set them to radians Period and frequency iff is frequency and T is period can be put into this equation f lT 2atr Speed T Period The amount of time it takes to go in a complete circle Frequency The number of times in a given amount of time an object will rotate uniform circular motion A centrifuge is spinning at 5400 rpm Using this information figure out how fast it goes in Hz First we need to know that Hz is a measurement that only means number of times something repetitive happens in a second something like swinging back and forth or going in a circle Because it s unspecified in what it does every second the units of Hz the unit of frequency is just 1s Something per second So we need to figure out what s happening The centrifuge is spinning so then it must be going in a circle a number of times per second so how many times Well it s spinning at 5400 rpm or rotations per minute Okay so that s almost what we need but it isn t in the right units We want rotations per second That will require changing the units from minutes to seconds Because there is one minute every 60 seconds we need to multiply it so that the minutes cancel out and we re left with rotations per second That gives us 5400rm160ms which equals 90 rs The next part of the question wants to know how fast the edge is going if it has a 14 cm long radius Well we need to translate 90rs into something that can be read as ms This means we have to find the rotational velocity which is in radians per second Because one rotation is 27 radians we can just multiply 90rs by 27 radr That gives us 18075 rads Because of the formula V oo r we can multiply 1807 rads by 014m or 14 cm in order to get the answer 79ms Translational v Rotational motion Translational motion is motion that moved along an axis If a box is pushed down a slope that s translational motion Rotational motion is motion around a center A wheel spinning around an axle is an example of rotational motion These two motions are related easily using the equation V or or in words translational velocity is equal to rotational velocity times the radius of rotation Also if a wheel is spinning on the ground without slipping the distance it will travel in one full rotation is equal to the circumference of the circle So if a wheel on a surface makes a full rotation without slipping it moves 27 r units of distance If it makes half of a rotation without slipping it will move 7 r units of distance etc Radial Acceleration Change in velocity around a circle is called radial acceleration The acceleration vector always points inward to the center of the circle All radial acceleration really is is the change in velocity of something moving around in a circle Conservation of momentum This is a conservation law meaning that all values in the initial calculations will be preserved after the interaction has happened In in equation if p represents momentum 219 pr Momentum is a vector equal to mass times velocity or pmv Medicine Ball Demo Two people face each other One person is sitting on a cart and the other is standing They toss a heavy ball back and forth Each time the person on the cart catches or throws the ball that person moves further away from the person who is just standing there In this demo whenever the person on the cart catches the ball they move back because the momentum of the ball is conserved and pushes the person and the cart in the direction that the ball was going before it was caught When the person on the cart threw the ball back they exerted a force on the ball that was met with an equal and opposite force from the ball pushing the person on the cart back even further The person standing didn t move because that person exerted a force on the ground to counteract the force experienced by acquiring and losing the ball
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