# 5.2 Rotational Kinematics, Moment of Inertia

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1 5 ANGULAR MOTION 5.2 Rotational Kinematics, Moment of Inertia Name: 5.2 Rotational Kinematics, Moment of Inertia Rotational Kinematics In (translational) kinematics, we started out with the position x, then defined velocity v to be the rate of change of position, and acceleration to be the rate of change of velocity: v = dx dt, a = dv dt = d2 x dt 2 In rotational kinematics, things are quite analogous. To indicate how far something rotated, we use an angle φ (typically measured in radians). We can then define angular velocity ω (how fast something rotates) to indicated how fast that angle changes in time, and angular acceleration α to be the rate of change of angular velocity: ω = dφ dt, α = dω dt = d2 φ dt 2 A point rotating that is the distance r from the center of rotation rotates an angle of 2π in one revolution. It moves once around the circle, ie., a distance of 2πr. So we can see how the angular quantities are related to the corresponding quantities for linear motion: x = rφ, v = rω, a = rα where now x is the distance moved along the circle, v the tangential velocity, and a the tangential acceleration. Note: Angular velocity is very similar to something like rpm. If a wheel turns at a frequency f of one revolution per second, it goes through an angle of 2π in that one second, that is, its angular velocity is 2π/s. More generally: ω = 2π f 195

2 5.2 Rotational Kinematics, Moment of Inertia 5 ANGULAR MOTION 1. An ant is resting on a pottery wheel, 20 cm from center, and a scorpion is 40 cm from the center. Suddenly, somebody turns on the pottery wheel, and it begins to accelerate at a rate of 0.3 rad/s 2. (a) How many full revolutions has the ant gone through after 30 seconds (since the pottery wheel was turned on)? (b) What total distance has the ant traveled in meters in that time? (c) How many revolutions has the scorpion gone through in the same amount of time, and what total distance has it traveled in meters? (d) What is the angular velocity of the ant 30 seconds after the pottery wheel was turned on? (e) What is the linear velocity of the ant (the tangential velocity ) at that time? 196

3 5 ANGULAR MOTION 5.2 Rotational Kinematics, Moment of Inertia (f) What are the angular velocity and tangential velocity of the scorpion at that time? (g) If the coefficient of static friction between the ant s feet and the pottery wheel is 2 (yes, it is higher than 1), at what time would the ant start slipping on the wheel? (Note that the tangential acceleration is very very small compared to the centripetal acceleration by the time he slips, so it can be ignored in the final analysis. For extra credit, solve it including the tangential acceleration and show that you get essentially the same answer.) Moment of Inertia Mass is also referred to as inertia a measure of how difficult it is to change an object s velocity (i.e., make it accelerate). Newton s 2nd Law says that F = ma, so the larger the mass the more force we need to apply to get the same acceleration. For rotational dynamics, we now know about torque, which is like a rotational force, trying to make an object rotate. We also know about angular acceleration. What s missing is some quantity that corresponds to mass, describing how difficult it is to make something rotate given a certain amount of torque. We use the terms moment of inertia or rotational inertia to describe how difficult it is to make an object rotate or more specifically, how difficult it is to give it an angular acceleration. We symbolize this moment of inertia with the letter I. We can find the moment of inertia for a single point mass at some distance r from a pivot point by rewriting Newton s 2nd Law: F = ma Since τ = rf (we assume the force is at 90 degrees to the lever arm), and α = a/r, we can rewrite Newton s 2nd Law in terms of the rotational quantities: τ = mr 2 α 197

4 5.2 Rotational Kinematics, Moment of Inertia 5 ANGULAR MOTION or, putting it into the same form as Newton s 2nd Law, τ = Iα, where I = mr 2 So the moment of inertia for a point mass is I = mr 2, where r is the distance from the pivot point. But, what if we don t just have one point mass a distance r away from our pivot point? Say we have three masses (m 1 through m 3 ) connected on a rod at different distances from the pivot (r 1 through r 3 ). Each of them has a moment of inertia equal to its mass times lever arm squared. So, the total moment of inertia would be I = m 1 r m 2r m 3r 2 3 = m i ri 2 If we have an object that is a continuous shape (e.g., a rod or a disk), then to find its moment of inertia, we need to add up the effect of each piece of mass (dm) producing a moment of inertia equal to its own mass times its lever arm squared. This requires integrating that lever arm times the mass dm over the entire shape of the object, I = r 2 dm = r 2 ρ dv if we let the mass of each piece of the material equal the density of the object (assumed constant) times the volume of that piece. 2. Does a hammer have the same moment of inertia regardless of whether you swing it from its handle end or from the opposite end? Explain. i 198

5 5 ANGULAR MOTION 5.2 Rotational Kinematics, Moment of Inertia 3. Suppose you have a solid sphere and a hula hoop of the same radius and same total mass. Which should have a higher moment of inertia? Discuss with your group (don t use the book!), and justify your answer qualitatively. 4. A merry-go-round is essentially a large disk and the moment of inertia of a flat disk is I = 1 2 mr2. A merry-go-round with a mass of 200 kg and diameter of 6 meters is initially stationary, until Kenny starts pushing it. Kenny, who has a mass of 30 kg, pushes with a constant force of 30 N as he runs around with the spinning merry-go-round. (a) How fast will he have to run to keep up with the merry-go-round after 20 seconds, assuming the merry-go-round was initially stationary? (b) How many revolutions will the merry-go-round go through in those 20 seconds? (c) If Cartman had been sitting on the edge of the merry-go-round while Kenny was pushing it (pushing with the same force), would the merry-go-round still have the same angular velocity after 20 seconds as you found with Cartman not sitting on it? Explain in terms of the moment of inertia, and how it would affect things. 199

6 5.2 Rotational Kinematics, Moment of Inertia 5 ANGULAR MOTION Up to this point, we treated pulleys as if they had no mass, and thus no moment of inertia. That meant that no net torque was required to give a pulley an angular acceleration. That is what allowed us to say that the tension in a rope was the same on both sides of a pulley. However, now we can treat pulleys as having mass (as real pulleys do) thus having a moment of inertia. That means that if a rope going over a pulley is accelerating, and therefore the pulley is undergoing angular acceleration, there must be a net torque on the pulley. 5. If a pulley that does have mass (and therefore moment of inertia) is undergoing an angular acceleration, can the tension in the rope going over the pulley be the same on both sides? Explain. (Consider the diagram of a pulley with a rope going over it, with the rope on the left having tension T 1 and the right having T 2. If T 1 = T 2, is there a net torque on the pulley?) 6. A pulley of radius r = 10 cm and mass M = 2 kg has a massless string going over it, attached to masses A and B, as shown to the right. If B has a mass of 100 grams and A has a mass of 50 grams, how fast will B accelerate? The pulley is basically a solid disk, with moment of inertia 1 2 Mr2. (a) Draw Free Body Diagrams of each block and of the pulley, giving each unique force a unique name. 200

7 5 ANGULAR MOTION 5.2 Rotational Kinematics, Moment of Inertia (b) Apply Newton s 2nd Law linearly to the blocks and angularly to the pulley. Consider how the accelerations of the blocks relate to each other, and how they relate to the angular acceleration of the pulley. Here and in the following (and really, always): Do not put in numbers until you have a symbolic! (c) Use your equations to solve for the acceleration of block B. (d) If we solved the same problem, but made the assumption that the pulley has no mass (so we would say the tension is the same on both sides, as we did earlier in the semester), we would have found an acceleration of a = g m 2 m 1 m 2 +m 1. If you let the mass of the pulley M be zero, does your solution simplify to this? Should it? (e) If the blocks started out at the same elevation and stationary, how long will it take until block A is 2 meters above block B? (f) How fast is each block going when they are 2 meters apart (vertically)? (g) Let s consider energy now. Is there any work done by non-conservative forces on the 201

8 5.2 Rotational Kinematics, Moment of Inertia 5 ANGULAR MOTION system as a whole? Explain. (h) Let s say the initial state is when the blocks are stationary and at the same elevation so let s call that starting elevation h i = 0. What is the initial energy of the entire system at that time? (You don t need to count the gravitational potential energy of the pulley itself, since the pulley only rotates it doesn t change elevation, though it wouldn t hurt to include it.) (i) When the blocks are 2 meters apart, what is the total energy of the blocks combined? Is energy conserved? How about if we consider the pulley massless? 202

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