Physics 2211: Lecture 37

1. Physics 2211: Lecture 37 • Work and Kinetic Energy • Rotational Dynamics Examples • Atwood’s machine with massive pulley • Falling weight and pulley • Translational and Rotational Motion Combined • Rotation around a moving axis • Important kinetic energy theorem

2. Consider the work done by a force acting on an object constrained to move around a fixed axis. For an infinitesimal angular displacement d:   d axis • Analog of Work • For constant  : W =  D • W will be negativeif  and Dq have opposite signs!

3. Work & Kinetic Energy • Recall the Work/Kinetic Energy Theorem: WNET =K • This is true in general, and hence applies to rotational motion as well as linear motion. • So for an object that rotates about a fixed axis:

4. M R F Example: Disk & String • A massless string is wrapped 10 timesaround 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 Nuntil it has unwound. (Assume the string does not slip, and that the disk is initially not spinning). • How fast is the disk spinning after the string has unwound?

5. M R • So W = (.1 m)(10 N)(20rad) = 62.8 J F  Dq Disk & String • The work done is W =  Dq • The torque is = RF (since = 90o) • The angular displacement Dq is2 rad/rev x 10 rev.

6. Recall that I for a disk about its central axis is given by: M R  So Disk & String

7. ExampleWork & Energy • 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. • Which disk has the biggest angular velocity after the pull ? w2 w1 (1)disk 1 (2)disk 2 (3)same F F

8. But we know So since I1 = I2 w1=w2 d ExampleSolution • The work done on both disks is the same! • W = Fd • The change in kinetic energy of each will therefore also be the same since W = DK. w2 w1 F F

9. Review: Torque and Angular Acceleration   • This is the rotational analog of FNET = ma • Torque is the rotational analog of force: • The amount of “twist” provided by a force. • 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. F2 F1 ExampleRotations • 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. • Forces F1 and F2 are applied as shown. What is F2 / F1 if the angular acceleration of the wheels is the same? (a) 1 (b) 2 (c) 4

11. We know and but so Since R2 = 2R1 F2 F1 ExampleSolution

12. Work & Power • The work done by a torque  acting through a displacement D is given by: • The power provided by a constant torque is therefore given by:

13. y x M  R T2 T1 T1 T2 a • For the pulley use • T1R - T2R m2 m2 m1 m1 a m2g (Since for a disk) m1g Atwood’s Machine with Massive Pulley • A pair of masses are hung over a massive disk-shaped pulley as shown. • Find the acceleration of the blocks. • For the hanging masses use SF = ma • T1- m1g = m1(-a) • T2- m2g = m2a

14. y x M  R T2 T1 T1 T2 a m2 m2 m1 m1 a m2g m1g Atwood’s Machine with Massive Pulley • We have three equations and three unknowns (T1, T2, a). Solve for a. T1- m1g = - m1a (1) T2- m2g = m2a (2) T1 - T2 = 1/2 Ma (3)

15. I  R T T m mg a L 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. • Starting at rest, what is the speed of the mass after it has fallen a distance L.

16. I  R T T m mg a L Falling weight & pulley • For the hanging mass use SF = ma • mg - T = ma • For the pulley + flywheel use =I • = TR = I • Realize that a = R • Now solve for a using the above equations.

17. I  R T T m where mg a L Falling weight & pulley • Using 1-D kinematics we can solve for the speed of the mass after it has fallen a distance L:

18. I  R T where T m mg a L Falling weight & pulley • Conservation of Energy Solution: y = 0

19. M R F Top view Rotation around a moving axis • 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. • What length of string L has unwound after the puck has moved a distance D?

20. The distance moved by the CM is thus • So the angular displacement is M A  R F Rotation around a moving axis • The CM moves according to F= MA • The disk will rotate about its CM according to =I

21. (a) (b) Divide (b) by (a):  F F D L 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: The length of string pulled out is L = R :

22. M A  R F Comments on CM acceleration • We just used =I for rotation about an axis through the CM even though the CM was accelerating! • The CM is not an inertial reference frame! Is this OK??(After all, we can only use F = ma in an inertial reference frame). • YES!We can always write =Ifor an axis through the CM. • This is true even if the CM is accelerating. • We will prove this when we discuss angular momentum!

23. 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 = 0 * = KCM = KREL * Important kinetic energy theorem • Consider the total kinetic energy of a system of two masses:

24. So is the kinetic energy of the center of mass (M is total mass). = KCM = KREL Important kinetic energy theorem • Thus KRELis the kinetic energy due to motion relative to the center of mass.

25. KCM KREL where but Connection with rotational motion • For a solid object rotating about its center of mass:

26. VCM  Translational & rotational motion combined • For a solid object which rotates about its center or mass and whose CM is moving: