Physics 1710 —Warm-up Quiz

1. Physics 1710—Warm-up Quiz Answer Now ! 0 40% 57 of 140 0 Two 2.0 kg disks, both of radius 0.10 m are sliding (without friction) and rolling, respectively, down an incline. Which will reach the bottom first? • Rolling disk wins. • Sliding disk wins. • Tie.

2. Physics 1710—Chapter 11 Rotating Bodies 0 Kinetic Energy of sliding disk: K1 = ½ mv2 = mg(h-z); v =√[2g(h-z)] Kinetic Energy of rolling disk: K2 = ½ mv2+ ½ I ω2= mg(h-z)= ½ mv2+ ½ ( ½ mr2) (v/r)2= 3/4 mv2; v =√[4/3 g(h-z)] Solution: Slider wins!

3. Physics 1710—Chapter 11 Rotating Bodies 0 a c Consider two spindles rolling down a ramp: b c Which one will win and why?

4. Physics 1710—Chapter 11 Rotating Bodies No Talking! Think! Confer! 0 Which one will win and why? Peer Instruction Time

5. Physics 1710—Chapter 11 Rotating Bodies 0 1′ Lecture The energy of rotation K = ½ I ⍵ 2 Torque (“twist”) is the vector product of the “moment” and a force. τ = r x F τ = I ⍺ = I d⍵/dt Angular momentum L is the vector product of the moment arm and the linear momentum. L= r x p. τ = d L/dt

6. Physics 1710—Chapter 11 Rotating Bodies 0 R r Moment of Inertia—sphere I = ∭ R 2 ρ d V N.B. : R 2 = r 2 – z 2 R 2 = r 2 – (r cos θ)2 =r 2 (1– cos 2 θ) I = ∫02π dφ∫0π (1– cos 2 θ) sin θdθ∫0a r 4 ρdr = [2π][2- 1/3(2)][1/5 a5] ρ = [4π][2/3][1/5 a5] ρ =[4π/3][2/5 a5] ρ = 2/5 M a2

7. Physics 1710—Chapter 11 Rotating Bodies 0 Kinetic Energy of Rotation: K = ½ Σi mi vi 2 K = ½ Σi mi (Ri ωi ) 2 K = ½ Σi mi R i2ωi2 For rigid body ωi = ω K = ½ [Σi mi R i2] ω 2 K = ½ I ω 2 With I =Σi mi R i2 = the moment of inertia.

8. Physics 1710—Chapter 11 Rotating Bodies θ Fg θ 0 F = m a = mr dω/dt rF = rFg sinθ= mr2dω/dt Torque = r x F = I α τ = r x F,|τ |= rFsinθ Fsinθ Why do round bodies roll down slopes? The torque is the “twist.”

9. Physics 1710—Chapter 11 Rotating Bodies 0 r X Torque and the Right Hand Rule: F r x F

10. Physics 1710—Chapter 11 Rotating Bodies 0 Vector Product: C = A x B Cx = Ay Bz – Az By Cyclically permute: (xyz), (yzx), (zxy) |C| =√[Cx2 + Cy2 + Cz2] = AB sin θ Directed by RH Rule.

11. Physics 1710—Chapter 11 Rotating Bodies 0 Vector Product: A x B = - B x A A x ( B + C ) = A x B + A x C d/dt ( A x B ) = d A /dt x B + A x d B/dt i x i = j x j = k x k = 0 i x j = - j x i = k j x k = - k x j = i k x i = - i x k = j

12. Physics 1710—Chapter 11 Rotating Bodies τ r F 0 τ = r x F Torque Bar:

13. Physics 1710—Chapter 11 Rotating Bodies A C B F2 F1 0 τ = r x F Teeter-totter: Where should the fulcrum be place to balance the teeter-totter?

14. Physics 1710—Chapter 11 Rotating Bodies 0 45% 63 of 140 Answer Now ! 0 Where should the fulcrum be place to balance the teeter-totter?

15. Physics 1710—Chapter 11 Rotating Bodies ? 0 Which way will the torque ladder move? Torque Ladder

16. Physics 1710—Chapter 11 Rotating Bodies 0 45% 63 of 140 Answer Now ! 0 Which way will the torque ladder move? • Clockwise • Counterclockwise • Will stay balanced

17. Physics 1710—Chapter 11 Rotating Bodies ? r sin θ r 0 Which way will the torque ladder move? Torque Ladder

18. Physics 1710—Chapter 11 Rotating Bodies 0 F = m a Or F = dp/dt Then: r x F = d (r xp)/dt Torque = τ = d L/dt Second Law of Motion L = r xp is the “angular momentum.”

19. Physics 1710—Chapter 11 Rotating Bodies 0 Angular Momentum: L = r x p The angular momentum is the vector product of the moment arm and the linear momentum. ∑ T = d L/dt The net torque is equal to the time rate of change in the angular momentum.

20. Physics 1710—Chapter 11 Rotating Bodies 0 Angular Momentum: Proof: ∑ T = r x ∑F = r x d p/dt And d L/dt = d( r x p) /dt = d r/dt x p + r x d p/dt. But p = m d r/dt , therefore d r/dt x p = 0 d L/dt = r x d p/dt And thus ∑ T = d L/dt.

21. Physics 1710—Chapter 11 Rotating Bodies 0 Torque = τ = d L/dt If τ = 0, then L is a constant. Second Law of Motion L = constant means angular momentum si conserved.

22. Physics 1710—Chapter 11 Rotating Bodies 0 Rotating Platform Demonstration

23. Physics 1710—Chapter 11 Rotating Bodies 0 Analysis: Why does an ice skater increase her angular velocity without the benefit of a torque? L = r x p = r x ( m v) = r x ( m r x ⍵) Li = mi ri 2 ⍵ Lz = (∑i mi ri 2 ) ⍵ Lz = I ⍵; & ⍵ = Lz / I Therefore, a decrease in I ( by reducing r) will result in an increase in ⍵.

24. Physics 1710—Chapter 11 Rotating Bodies 0 Summary: The total Kinetic energy of a rotating system is the sum of the rotational energy about the Center of Mass and the translational KE of the CM. K = ½ ICM⍵ 2 + ½ MR 2⍵ 2 τ = r x F

25. Physics 1710—Chapter 11 Rotating Bodies 0 • Angular momentum L is the vector product of the moment arm and the linear momentum. • L = r x p • The net externally applied torque is equal to the time rate of change in the angular momentum. • ∑ τz = d Lz /dt = Iz ⍺ Summary: