Physics 2 Summer 2002

1. Physics 2Summer 2002 Scott Fraser email: scottf@physics.ucsb.edu office hours: M-F, 2-2:30 or by appt.

2. Course Information • Online at http://class.physics.ucsb.edu • Login: only needed to check grades • Physics 2 page: click “view list of classes” • Check the links regularly for updates!

3. Course Text • University Physics (10th edition),Young & Freedman • We will cover Chapters 10-18 • You should be familar with Chapters 1-9

4. Tips for Success • Always do the pre-lecture reading • Work examples in the text for yourself • Algebra first, numbers (if any!) last • Chapter 10 homework due this Friday!(4pm in locked PHYSICS 2 box in lobby)

5. Chapter 10 Dynamics of Rotational Motion

6. Pure Rotation

7. Show Fig. 9-7, 9-8 Rotation Revisited • See Fig. 9-7, 9-8 • Chapter 9: “Rotation happens.” • Chapter 10: “Why? What causes a?”

8. Torque (t) • Torque = (lever arm) x (force) • unit = N·m (not J) • coordinate-dependent:“torque of F about O”

9. Show Fig. 10-4 Torque is a vector • direction Fig.10-4: R-hand rule • magnitude

10. Newton’s 2nd Law for Rotation Fixed rotation axis: only F1,tan causes torque

11. Newton’s 2nd Law for Rotation • Single particle (m1) • Rigid body: sum over mi (each ai= a )

12. Demonstration:Newton’s 2nd Law for Rotation Perform demonstration

13. Solving Rotation Problems:Newton’s 2nd Law • Linear form • Rotational form

14. Exercise 10-13 • m1 and m2 move: d =1.20 m during t = 0.800 s • Values: m1 = 2.00 kg, m2 = 3.00 kg, R = 0.075 m • Find: T1, T2, I (of pulley about its rotation axis)

15. See Example 10-4 Now use these equations to do Exercise 10-13

16. Rotation with Translation

17. An Example:Rolling Without Slipping

18. Rolling without Slipping • ground sees: wheel translating with speed vcm • center of wheel sees: rim rotating with speed Rw • no slip: vcm= Rw (skid: Rw < vcm , slide: Rw > vcm)

19. Rolling without Slipping • two motions:translation of CM, rotation about CM • superposition: at each point on wheel, v = vcm+ v / • ground sees: point 1 of wheel momentarily at rest

20. Newton’s 2nd Lawfor Pure Rotation • We used the linear version (single particle) • To get the rotational version (rigid body)

21. Newton’s 2nd Lawfor Rotation and Translation • Recall linear relation for CM (total mass M) • Claim: an analog result (rigid body)

22. Newton’s 2nd Lawfor Rotation and Translation • Stext = Icma • Two conditions needed for this to hold: • Axis through CM must be a symmetry axis • Axis must not change direction

23. Newton’s 2nd Lawfor Rotation and Translation • Translation of CM (total mass M) • Rotation about axis through CM

24. Example: The Yo-Yo • Example 10-8:yo-yo = ‘axle’Icm= Icylinder • More generally:yo-yo = ‘spool’Icm= Ispool • But just draw the axle

25. Example: The Yo-Yo • Let’s draw the free body diagram for the yo-yo

26. Example: The Yo-Yo

27. Example: The Yo-Yo • String: no slipping • Newton’s 2nd Law Use these equations to find acm and T for arbitrary Icm

28. Exercise 10-15 • We found results for yo-yo with any Icm(not just Icm= Icylinder) • Exercise 10-15:yo-yo = hoopIcm= Ihoop (= MR2)

29. Exercise 10-15

30. Exercise 10-15 • Icm= Ihoop = MR2 • M = 0.18 kgR = 0.080 m • Find tension T • Find t, w when hoop has fallen a distance h= 0.75 m from rest Now use expressions for acm and T for arbitrary Icm (from the Yo-Yo Example)

31. Rotation and Translation:Energy • Pure rotation • Translation and rotation

32. Exercise 10-16 • Redo Exercise 10-15, but now use energy • Icm= Ihoop = MR2 • vcm= Rw • Find w when hoop has fallen a distance h= 0.75 m from rest

33. Exercise 10-16 • Redo Exercise 10-15, but now use energy • Icm= Ihoop = MR2 • vcm= Rw • Find w when hoop has fallen a distance h= 0.75 m from rest

34. Work and Power Done by Torque

35. Work and Power Done by Torque

36. Work Done by Net Torque (St)

37. Exercise 10-21 • R = 2.40 mI = 2100 kg·m2 • Initially at rest (w1=0) • The child applies:Ftan = 18.0 N during Dt = t2 – t1 = 15.0 s • Find w2 , W, P Do the calculation, using conservation of energy

38. Angular Momentum for a single particle for rigid bodies

39. Angular Momentum of a Particle • angular momentum: L • rotational analog of linear momentum, p • you can guess the definition of L?

40. Angular Momentum of a Particle • unit = kg·m2/s • coordinate-dependent:“angular momentum L about O”

41. Angular Momentum of a Particle • vector • magnitude Do Exercise 10-29 (a)

42. Angular Momentum of a Particle • For circular motion in the xy plane, f= 90°

43. Angular Momentum of a Rigid Body • consider a rigid body in xy plane(no extent in z) • sum over particles:L = Iw for whole body • what if the body extends in z direction?

44. Angular Momentum of a Rigid Body • consider special case: • extended rigid body has symmetry axis (here z) • and also rotates about the symmetry axis

45. Angular Momentum of a Rigid Body • so our special case is: • If rigid body rotates about a symmetry axis, then

46. Angular Momentum: Particle • particle (moving relative to some origin O) • ‘orbital’ angular momentum

47. Angular Momentum: Rigid Body • rigid body(rotating about a symmetry axis) • ‘spin’ angular momentum

48. Exercise 10-30 • Earth has two kinds of angular momentum: • orbital Lorb • (particle on circular orbit: r = 1.5×1011 m) • spin Lrot • (Earth: M = 6.0×1024 kg, R = 6.4×106 m) Find the values of the two angular momenta

49. special case: If rigid body’s rotation axis equals its symmetry axis... general case: If no symmetry axis, or rotation axis is not the symmetry axis... Angular Momentum of (Extended) Rigid Body

50. Torque and Angular Momentum • True for any system of particles!(rigid or not, symmetric or not) Prove this for case of single particle Do Exercise 10-29 (b)