Physics 101: Lecture 15, Pg 1 Physics 101: Lecture 15 Rolling Objects l Today’s lecture will cover Textbook Chapter Exam III

Physics 101: Lecture 15, Pg 2Overview l Review  K rotation = ½ I  2 è Torque = Force that causes rotation  = F r sin  è Equilibrium  F = 0  = 0 l Today   = I    nergy conservation revisited 05

Physics 101: Lecture 15, Pg 3 Linear and Angular LinearAngular Displacement x  Velocity v  Acceleration a  Inertia mI KE ½ m v2 ½ I  2 N2L F=ma  = I  Momentum p = mv L = I  Today 08

Physics 101: Lecture 15, Pg 4 Rotational Form Newton’s 2 nd Law  = I  è Torque is amount of twist provide by a force »Signs: positive = CCW è Moment of Inertial like mass. Large I means hard to start or stop from spinning. l Problems Solved Like N2L è Draw FBD è Write N2L 09

Physics 101: Lecture 15, Pg 5 The Hammer! You want to balance a hammer on the tip of your finger, which way is easier A) Head up B) Head down C) Same 15 30% 64% 6% by putting it [the head] towards you finger there is less torque due to gravity If the most mass is farther from the axis of rotation (my finger), the moment of inertia will be greater and less torque will be needed to balance it. This is mainly from personal experience. I don't have a great physics answer. But I believe that it has something to do with the distance of the mass from the finger. For some reason, a greater distance makes it easier to balance. WHY is this person balancing a hammer...shouldn't they be doing work!?!

Physics 101: Lecture 15, Pg 6 The Hammer! You want to balance a hammer on the tip of your finger, which way is easier A) Head up B) Head down C) Same  = I  m g R sin(  ) = mR 2  mg R Torque increases with R Inertia increases as R 2 g sin(  ) / R =  Angular acceleration decreases with R! So large R is easier to balance % 64% 6%

Physics 101: Lecture 15, Pg 7 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 m R T mg  a L 18 What method should we use to solve this problem A)Conservation of Energy (including rotational)  and then use kinematics Either would work, but since it asks for time, we will use B.

Physics 101: Lecture 15, Pg 8 Falling weight & pulley... For the hanging mass use  F = ma è mg - T = ma For the flywheel use  = I   TR sin(90) = I  Realize that a =  R l Now solve for a using the above equations. I m R T mg  a L 21

Physics 101: Lecture 15, Pg 9 Falling weight & pulley... l Using 1-D kinematics we can solve for the time required for the weight to fall a distance L: I m R T mg  a L where 23

Physics 101: Lecture 15, Pg 10 Tension….. m2 m1 T1T1 T2T2 T 1 < T 2 since T 2 – T 1 = m 2 a. It takes force to accelerate block 2. m3 F m3m3 m1m1 m2m2 T1T1 T2T2 T 2 > T 1 since RT 2 – RT 1 = I 2 . It takes force (torque) to accelerate the pulley. 09 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 Compare the tensions T 1 and T 2 as block 3 falls A) T 1 T 2

Physics 101: Lecture 15, Pg 11Rolling 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? What is the velocity of the bottom of the wheel relative to the ground? 28 x y 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) +2m/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) +2m/s E) +4 m/s + 2 m/s -2 m/s 2 m/s

Physics 101: Lecture 15, Pg 12Rolling 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  R I M 29

Physics 101: Lecture 15, Pg 13Rolling... l Static friction f causes rolling. It is an unknown, so we must solve for it. First consider the free body diagram of the object and use  F NET = Ma cm : In the x direction Mg sin  - f = Ma cm Now consider rotation about the CM and use  = I  realizing that  = Rf and a =  R R M  f y x 33 Mg 

Physics 101: Lecture 15, Pg 14 Rolling... We have two equations: Mg sin  - f = Ma l We can combine these to eliminate f:  A R I M For a sphere: 36

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!  W = F d cos  d is zero for point in contact l No dissipated work, energy is conserved l Need to include both translation and rotation kinetic energy.  K = ½ m v 2 + ½ I  2 38

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. H useand 43 Translational: K T = ½ M v 2 Rotational: K R = ½ I  2 Rotational: K R = ½ (½ M R 2 )  V/R) 2 = ¼ M v 2 = ½ K T

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. è If 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 48 K i + U i = K f + U f

Physics 101: Lecture 15, Pg 18Summary  = I  l Energy is Conserved è Need to include translational and rotational

Physics 101: Lecture 15, Pg 19 Torque ACT l Which pulley will make it drop fastest? 1) Small pulley 2) Large pulley 3) Same 25 Larger R, gives larger acceleration.