1. Lecture 37, Page 1 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Physics 2211: Lecture 37 l Work and Kinetic Energy l Rotational Dynamics Examples çAtwood’s machine with massive pulley çFalling weight and pulley l Translational and Rotational Motion Combined çRotation around a moving axis çImportant kinetic energy theorem

2. Lecture 37, Page 2 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Work dd axis   Consider the work done by a force acting on an object constrained to move around a fixed axis. For an infinitesimal angular displacement d  : W will be negative if  and  have opposite signs! l Analog of For constant  : W = 

3. Lecture 37, Page 3 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Work & Kinetic Energy Recall the Work/Kinetic Energy Theorem: W NET =  K l This is true in general, and hence applies to rotational motion as well as linear motion. l So for an object that rotates about a fixed axis:

4. Lecture 37, Page 4 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Example: Disk & String l A massless string is wrapped 10 times around a disk of mass M = 40 g and radius R = 10 cm. The disk is constrained to rotate without friction about a fixed axis though its center. The string is pulled with a force F = 10 N until it has unwound. (Assume the string does not slip, and that the disk is initially not spinning). çHow fast is the disk spinning after the string has unwound? F R M

5. Lecture 37, Page 5 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Disk & String The work done is W =   The torque is  = RF (since  = 90 o )  The angular displacement  is 2  rad/rev x 10 rev. F R M So W = (.1 m)(10 N)(20  rad) = 62.8 J  

6. Lecture 37, Page 6 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Disk & String  R M Recall that  I  for a disk about its central axis is given by: So

7. Lecture 37, Page 7 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Example Work & Energy l Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same distance. Disk 1 has a bigger radius, but both have the same moment of inertia. Both disks rotate freely around axes though their centers, and start at rest. çWhich disk has the biggest angular velocity after the pull ? (1) (1) disk 1 (2) (2) disk 2 (3) (3) same FF 11 22

8. Lecture 37, Page 8 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Example Solution FF 11 22 d l The work done on both disks is the same! çW = Fd The change in kinetic energy of each will therefore also be the same since W =  K. So since I 1 = I 2  1 =  2 But we know

9. Lecture 37, Page 9 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Review: Torque and Angular Acceleration   l This is the rotational analog of F NET = ma l Torque is the rotational analog of force: çThe amount of “twist” provided by a force. Moment of inertia I is the rotational analog of mass Moment of inertia I is the rotational analog of mass  If I is big, more torque is required to achieve a given angular acceleration.

10. Lecture 37, Page 10 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Example Rotations l Two wheels can rotate freely about fixed axles through their centers. The wheels have the same mass, but one has twice the radius of the other. l Forces F 1 and F 2 are applied as shown. What is F 2 / F 1 if the angular acceleration of the wheels is the same? (a) 1 (b) 2 (c) 4 F1F1 F2F2

11. Lecture 37, Page 11 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Example Solution We know so F1F1 F2F2 Since R 2 = 2 R 1 but and

12. Lecture 37, Page 12 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Work & Power The work done by a torque  acting through a displacement  is given by: l The power provided by a constant torque is therefore given by:

13. Lecture 37, Page 13 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Atwood’s Machine with Massive Pulley l A pair of masses are hung over a massive disk-shaped pulley as shown. çFind the acceleration of the blocks. For the hanging masses use  F = ma  T 1  m 1 g = m 1 (-a)  T 2  m 2 g = m 2 a (Since for a disk) For the pulley use ç T 1 R - T 2 R m2m2 m1m1 R M y x  m2m2 m1m1 m2gm2g a T1T1 m1gm1g a T2T2 T1T1 T2T2

14. Lecture 37, Page 14 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Atwood’s Machine with Massive Pulley m2m2 m1m1 R M y x  m2m2 m1m1 m2gm2g a T1T1 m1gm1g a T2T2 T1T1 T2T2 l We have three equations and three unknowns (T 1, T 2, a). Solve for a. T 1  m 1 g =  m 1 a (1) T 2  m 2 g = m 2 a (2) T 1 - T 2 = 1/2 Ma (3)

15. Lecture 37, Page 15 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Falling weight & pulley 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. çStarting at rest, what is the speed of the mass after it has fallen a distance L. I m R T mg  a L T

16. Lecture 37, Page 16 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Falling weight & pulley For the hanging mass use  F = ma çmg - T = ma For the pulley + flywheel use  = I    = TR = I  Realize that a = R  l Now solve for a using the above equations. I m R T mg  a L T

17. Lecture 37, Page 17 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Falling weight & pulley l Using 1-D kinematics we can solve for the speed of the mass after it has fallen a distance L: where I m R T mg  a L T

18. Lecture 37, Page 18 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Falling weight & pulley l Conservation of Energy Solution: where y = 0 I m R T mg  a L T

19. Lecture 37, Page 19 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Rotation around a moving axis l A string is wound around a puck (disk) of mass M and radius R. The puck is initially lying at rest on a frictionless horizontal surface. The string is pulled with a force F and does not slip as it unwinds. çWhat length of string L has unwound after the puck has moved a distance D? F R M Top view

20. Lecture 37, Page 20 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Rotation around a moving axis l The CM moves according to F = MA l The distance moved by the CM is thus F MA  R The disk will rotate about its CM according to  = I  l So the angular displacement is

21. Lecture 37, Page 21 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Rotation around a moving axis So we know both the distance moved by the CM and the angle of rotation about the CM as a function of time: D F F  L The length of string pulled out is L = R  : (b) Divide (b) by (a): (a)

22. Lecture 37, Page 22 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Comments on CM acceleration We just used  = I  for rotation about an axis through the CM even though the CM was accelerating! çThe CM is not an inertial reference frame! Is this OK?? (After all, we can only use F = ma in an inertial reference frame). YES! YES! We can always write  = I  for an axis through the CM. çThis is true even if the CM is accelerating. çWe will prove this when we discuss angular momentum! F R MA 

23. Lecture 37, Page 23 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Important kinetic energy theorem l Consider the total kinetic energy of a system of two masses: Now write the velocities as a sum of the velocity of the center of mass and a velocity relative to the center of mass so = K CM = 0 * = K REL *

24. Lecture 37, Page 24 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm l Thus K REL is the kinetic energy due to motion relative to the center of mass. So is the kinetic energy of the center of mass (M is total mass). = K CM = K REL Important kinetic energy theorem

25. Lecture 37, Page 25 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Connection with rotational motion l For a solid object rotating about its center of mass: K CM K REL where but

26. Lecture 37, Page 26 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm l For a solid object which rotates about its center or mass and whose CM is moving:  V CM Translational & rotational motion combined