Exam II Physics 101 Lecture 15 Rolling Objects

Exam II Physics 101: Lecture 15 Rolling Objects Today’s lecture will cover Textbook Chapter 8. 5 -8. 7 Physics 101: Lecture 15, Pg 1

Overview l Review èKrotation = ½ I w 2 èTorque = Force that causes rotation t = F r sin èEquilibrium SF=0 St=0 l Today èS t = I a (rotational F = ma) èEnergy conservation revisited Physics 101: Lecture 15, Pg 2

Linear and Angular Displacement Velocity Acceleration Inertia KE N 2 L Momentum Linear Angular x v w a a m I ½ m v 2 ½ I w 2 F=ma t = Ia p = mv L = Iw Today Physics 101: Lecture 15, Pg 3

Rotational Form Newton’s l. S nd 2 Law t=Ia èTorque is amount of twist provided by a force » Signs: positive = CCW èMoment of Inertia like mass. Large I means hard to start or stop spinning. l Problems Solved Like N 2 L èDraw FBD èWrite N 2 L Physics 101: Lecture 15, Pg 4

The Hammer! You want to balance a hammer on the tip of your finger, which way is easier 29% A) Head up Why am I balancing a hammer on my finger? It sounds dangerous. 63% B) Head down 8% C) Same I just tried it in my home and I guess it is easier to balance the hammer with the head up. Angular acceleration is smaller the larger the radius the larger the moment of inertia. Physics 101: Lecture 15, Pg 5

The Hammer! You want to balance a hammer on the tip of your finger, which way is easier 29% A) Head up 63% B) Head down R 8% C) Same t=Ia Key idea: higher angular m g R sin( ) = m. R 2 a means more acceleration Angular acceleration Torque Inertia difficult to balance. increases decreases with R! increases with R as R 2 So large R is easier to balance. g sin( ) /What R = ais angular acceleration? mg Physics 101: Lecture 15, Pg 6

Falling weight & pulley l A mass m is hung by a string that is wrapped around a pulley of radius R attached to a heavy flywheel. The moment of inertia of the pulley + flywheel is I. The string does not slip on the pulley. Starting at rest, how long does it take for the mass to fall a distance L. I R T m a What method should we use to solve this problem? A) Conservation of Energy (including rotational) mg L B) St = Ia and then use kinematics Either would work, but since it asks for time, we will use (B). Physics 101: Lecture 15, Pg 7

Falling weight & pulley. . . l For the hanging mass use SF = ma èmg - T = ma l For the flywheel use S = I I R èTR sin(90) = I l Realize that a = R T m l Now solve for a, eliminate T: a mg L Physics 101: Lecture 15, Pg 8

Falling weight & pulley. . . l Using 1 -D kinematics we can solve for the time required for the weight to fall a distance L: I R T m a mg L where Physics 101: Lecture 15, Pg 9

Torque ACT l Which pulley will make it drop fastest? 1) Small pulley 2) Large pulley 3) Same Larger R, gives larger acceleration. Physics 101: Lecture 15, Pg 10 25

Tension… m 1 T 1 m 2 T 2 m 3 F Compare the tensions T 1 and T 2 as the blocks are accelerated to the right by the force F. A) T 1 < T 2 B) T 1 = T 2 C) T 1 > T 2 T 1 < T 2 since T 2 – T 1 = m 2 a. It takes force to accelerate block 2. T 1 m 2 Compare the tensions T 1 and T 2 as block 3 falls A) T 1 < T 2 B) T 1 = T 2 C) T 1 > T 2 m 1 T 2 m 3 T 2 > T 1 since RT 2 – RT 1 = I 2 a. It takes force (torque) to accelerate the pulley. Physics 101: Lecture 15, Pg 11

Rolling y A wheel is spinning clockwise such that the speed of the outer rim is 2 m/s. What is the velocity of the top of the wheel relative to the ground? + 2 m/s What is the velocity of the bottom of the wheel relative to the ground? -2 m/s x 2 m/s You now carry the spinning wheel to the right at 2 m/s. What is the velocity of the top of the wheel relative to the ground? A) -4 m/s B) -2 m/s C) 0 m/s D) +2 m/s E) +4 m/s What is the velocity of the bottom of the wheel relative to the ground? A) -4 m/s B) -2 m/s C) 0 m/s D) +2 m/s E) +4 m/s Physics 101: Lecture 15, Pg 12

Rolling l An object with mass M, radius R, and moment of inertia I rolls without slipping down a plane inclined at an angle with respect to horizontal. What is its acceleration? l Consider CM motion and rotation about the CM separately when solving this problem I M R Physics 101: Lecture 15, Pg 13

Rolling. . . Static friction f causes rolling. It is an unknown, so we must solve for it. l First consider the free body diagram of the object and use SFNET = Macm : In the x direction Mg sin - f = Macm l l Now consider rotation about the CM and use S = I realizing that = Rf and a = R M y R x Mg Physics 101: Lecture 15, Pg 14 f

Rolling. . . Mg sin - f = Ma l We have two equations: l We can combine these to eliminate f: I A M R For a sphere: Physics 101: Lecture 15, Pg 15

Energy Conservation! l Friction causes object to roll, but if it rolls w/o slipping friction does NO work! èW = F d cos l No d is zero for point in contact dissipated work, energy is conserved l Need to include both translational and rotational kinetic energy. èK = ½ m v 2 + ½ I w 2 Physics 101: Lecture 15, Pg 16

Translational + Rotational KE l Consider a cylinder with radius R and mass M, rolling w/o slipping down a ramp. Determine the ratio of the translational to rotational KE. Translational: KT = ½ M v 2 Rotational: K R = ½ I w 2 use and Rotational: KR = ½ (½ M R 2) (V/R)2 = ¼ M v 2 H = ½ KT Physics 101: Lecture 15, Pg 17

Rolling Act l Two uniform cylinders are machined out of solid aluminum. One has twice the radius of the other. èIf both are placed at the top of the same ramp and released, which is moving faster at the bottom? (a) bigger one (b) smaller one (c) same Ki + U i = K f + U f For a hollow cylinder, For a sphere, Physics 101: Lecture 15, Pg 18

Summary lt =Ia l Energy is Conserved èNeed to include translational and rotational Physics 101: Lecture 15, Pg 19