1. 1 7/26/04 Midterm 2 – Next Friday (7/30/04)  Material from Chapters 7-12 I will post a practice exam on Monday Announcements

2. 2 7/26/04 Last Time … Rotation Position  x Velocity  = d  /dt v = dx/dt Acceleration  = d  /dt a = dv/dt Translation Kinetic Energy K = ½I  2 K=½mv 2 Mass I =  m i r i 2 m F = ma  = I  Newton’s 2 nd Law

3. 3 7/26/04 Example: (Problem 11.49) During a launch from the board, a diver’s angular speed about her center of mass changes from zero to 6.20 rad/s in 220 ms. Her rotational inertia about her center of mass is 12.0 kg m 2. During the launch, what are the magnitudes of a) her average angular acceleration and b) the average external torque on her from the board?

4. 4 7/26/04 Example: a) Average angular acceleration b) Average external torque

5. 5 7/26/04 Example: (Problem 11.53) The world’s heaviest hinged door has a mass of 44,000 kg, a rotational inertia about a vertical axis through its hinges of 8.7x10 4 kg m 2, and a width of 2.4 m. Neglecting friction, what steady force applied to the door’s outer edge and perpendicular to the plane of the door can move it from rest through an angle of 90° in 30 s?

6. 6 7/26/04 Solution: Initial and final positions will be: Using a standard kinematics equation, Finally,

7. 7 7/26/04 Solution: Plugging in our value for  :

8. 7/26/04 8 Chapter 12 Rolling, Torque, and Angular Momentum

9. 9 7/26/04 Rolling Without Slipping How do we describe an object that is rolling? Consider the case of pure translation: v cm

10. 10 7/26/04 Rolling Without Slipping Consider the case of pure rotation: v cm

11. 11 7/26/04 Rolling Without Slipping Combining the two: v cm + = 2 v cm v=0

12. 12 7/26/04 Rolling: Another View At any instant the wheel rotates about the point of contact  vcvc  vcvc  s = R  P

13. 13 7/26/04 I about point of contact Parallel Axis Theorem K = ½I P  2 I P = I cm + MR 2 P Rolling: Another View

14. 14 7/26/04 Rolling Down a Hill h  vcvc Conservation of Energy: U i = K f For a point mass: For an extended object:

15. 15 7/26/04 Rolling Down a Hill For a disc: For a ring:  vcvc h Notice there is no mass or radial dependence!

16. 16 7/26/04 How to Solve Torque Problems 1) Draw a picture 2) Pick an origin (axis of rotation usually a good choice) 3) Sum torques about the origin  net = I  4) Sum the forces in every direction (if necessary) F net = ma 5) Solve for unknowns

17. 17 7/26/04 The Forces of Rolling Constraint of Rolling without Slipping: Mg Mg sin  Mg cos  FNFNF Before when we drew a free body diagram, we drew all the forces from the center of mass. Cannot do that with torques! Note: Friction causes rolling (otherwise it would just slide)

18. 18 7/26/04 The Forces of Rolling Pick the center of mass as the origin. Sum the forces and torques. Mg Mg sin  Mg cos  FNFNF Find the linear acceleration of the center of mass.

19. 19 7/26/04 The Forces of Rolling Mg Mg sin  Mg cos  FNFNF Recall: Linear acceleration doesn’t depend on mass or radius!!

20. 20 7/26/04 Another Way to Solve the Problem What if we pick a different origin? Pick point of contact. No torque from normal force or friction! Same result! (Don’t need to sum forces) Mg Mg sin  Mg cos  FNFNF

21. 21 7/26/04 Angular Momentum Recall: linear motion Using the correspondence with linear motion, angular momentum should have the form: If no torque, then L is a constant. We have conservation of angular momentum!

22. 22 7/26/04 Conservation of Angular Momentum

23. 23 7/26/04 Conservation of Angular Momentum If I changes, then  must change to keep L constant (no external torques) Example: Ice skaters pulling arms in L i =I i  i L f =I f  f

24. 24 7/26/04 Angular Momentum of a Particle Can also define angular momentum for a particle with a linear velocity v r is vector from origin to particle r v r  v  L is big r || v  L is 0 (Example: something circling the origin)

25. 25 7/26/04 Angular Momentum of a Particle: Is this new definition of L consistent with the old one? Same as before!

26. 26 7/26/04 Example: Constant velocity particle: Is L really constant? Direction is constant as well! r v θ d

27. 27 7/26/04 Torque and Angular Momentum A) No torque: L is constant L = Iω if you change I, ω changes to keep L constant This allows skaters and divers to spin really really fast (they studied their physics!) L=Iω  =dL/dt B) If I is constant and ω changes, there must have been a torque 

28. 28 7/26/04 Angular Momentum of a Group of Particles: For a group of particles: The net torque on the system will be: No need to worry about internal torques

29. 29 7/26/04 A merry-go-round with radius 3 m and I mgr =90 kg m 2 is spinning at  0 =5 rad/s. A 100 kg pig is then dropped on it at a radius of 2 meters. His landing takes  t=1s. 3 m 2 m Example: Merry Go Round

30. 30 7/26/04 Example: Merry Go Round What is the final angular velocity of the system?

31. 31 7/26/04 What is the average torque on the merry-go-round during the landing? Example: Merry Go Round Note: This is an internal torque

32. 32 7/26/04 What is the average linear tangential acceleration of the rim of the merry-go-round? Example: Merry Go Round

33. 33 7/26/04 Example: FOR SALE m = 5 g M = 2.2 kg v = 300 m/s ℓ = 0.2 m v ℓ /2 ℓ θ A bullet hits a sign: how high does it go?

34. 34 7/26/04 Example: m = 5 g M = 2.2 kg v = 300 m/s ℓ = 0.2 m Before bullet hits: After bullet hits: v ℓ/2 ℓ θ

35. 35 7/26/04 Example: K i +U i = K f +U f ℓ /2 ℓ θ h

36. 36 7/26/04 Precession Right hand rule  dL/dt out of the page! mg h L Consider a spinning gyroscope:

37. 37 7/26/04 Precession L=I   = mgh ΔL =  Δt ΔL = mgh Δt ΔΔ LfLf LiLi ΔLΔL mg h L

38. 38 7/26/04 The Bicycle Why is it hard to fall over? L gets larger as the bike goes faster A torque is needed to change L If we lean to one side (i.e. fall over)  = dmg x y z mg d  in the x direction Turns the bike! d