Physics Midterm Announcements and Practice Exam

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

2. Translation Rotation Position x  Velocity v = dx/dt  = d/dt Acceleration a = dv/dt  = d/dt Mass m I = Smiri2 Newton’s 2nd Law F = ma  = I Kinetic Energy K=½mv2 K = ½Iw2 Last Time…

3. 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 m2. 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. Example: a) Average angular acceleration b) Average external torque

5. 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 8.7x104 kg m2, 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. Solution: Initial and final positions will be: Using a standard kinematics equation, Finally,

7. Solution: Plugging in our value for a:

8. Chapter 12 Rolling, Torque, and Angular Momentum

9. Rolling Without Slipping How do we describe an object that is rolling? Consider the case of pure translation: vcm vcm vcm

10. Rolling Without Slipping Consider the case of pure rotation: vcm vcm

11. vcm vcm vcm Rolling Without Slipping Combining the two: vcm 2 vcm vcm = + vcm v=0

12.   vc vc q s = Rq P Rolling: Another View At any instant the wheel rotates about the point of contact

13. P Rolling: Another View K = ½IP2 I about point of contact Parallel Axis Theorem IP = Icm + MR2

14. Rolling Down a Hill Conservation of Energy: For a point mass: Ui = Kf For an extended object: h vc q

15. Rolling Down a Hill h vc q For a disc: Notice there is no mass or radial dependence! For a ring:

16. How to Solve Torque Problems • Draw a picture • Pick an origin (axis of rotation usually a good choice) • Sum torques about the origin tnet= Ia • Sum the forces in every direction (if necessary) Fnet = ma • Solve for unknowns

17. The Forces of Rolling FN FF Mg sin q Mg cos q Mg 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) Constraint of Rolling without Slipping:

18. The Forces of Rolling FN FF Mg sin q Mg cos q Mg Find the linear acceleration of the center of mass. Pick the center of mass as the origin. Sum the forces and torques.

19. The Forces of Rolling FN FF Mg sin q Mg cos q Mg Recall: Linear acceleration doesn’t depend on mass or radius!!

20. FN FF Mg sin q Mg cos q Mg 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)

21. 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. Conservation of Angular Momentum

23. Conservation of Angular Momentum If I changes, then w must change to keep L constant (no external torques) Example: Ice skaters pulling arms in Li=Iiwi Lf=Ifwf

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

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

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

27. Torque and Angular Momentum A) No torque: L is constant L=Iωt=dL/dt 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!) B) If I is constant and ω changes, there must have been a torque 

28. 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. 2 m 3 m Example: Merry Go Round A merry-go-round with radius 3 m and Imgr=90 kg m2 is spinning at w0=5 rad/s. A 100 kg pig is then dropped on it at a radius of 2 meters. His landing takes Dt=1s.

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

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

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

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

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

35. ℓ/2 h θ ℓ Example: Ki+Ui = Kf+Uf

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

37. h L Li ΔL mg Δf Lf Precession L=I ΔL = Δt t = mgh ΔL = mgh Δt

38. z y d x d t mg 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) t = dmg in the x direction Turns the bike!