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Physics 1710 —Warm-up Quiz

Physics 1710 —Warm-up Quiz. Answer Now !. 10. 0% 0 of 1.

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Physics 1710 —Warm-up Quiz

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  1. Physics 1710—Warm-up Quiz Answer Now ! 10 0% 0 of 1 What torque must be applied to a discus (assume a short cylindrical shape) that has a diameter of 20. cm and a mass of 3.0 kg, if it leaves the hand of the discus thrower spinning at a rate of 180 rpm if he takes 0.5 sec to throw the disc? • 120. Nm • 5.61 x10-01 Nm • 2.41 x10-02 Nm • 1.60 x10-03 Nm • 4.01 x10-04 Nm

  2. Physics 1710—Chapter 11 Rotating Bodies ⍵ = 2π‧(180)/60 = 18.7 rad/s I = ½ MR2 = ½ (3.0) (0.10)2 = 0.015 kg m 2 ⍵ = αt; α = ⍵ /t = (18.7 rad/s)/(0.5 sec)= 37.4 /s2 τ = I α = (0.015 kg m 2)(37.4 /s2) τ =5.61 x10 – 1Nm Solution:

  3. Physics 1710—Chapter 11 Rotating Bodies No Talking! Think! Confer! How much force must you exert on a bucket crank handle that is 30.0 cm long to lift a bucket of water of mass 25.0 kg if the rope is 15.0 m long and the cylinder is 10.0 cm in diameter? Peer Instruction Time

  4. Physics 1710—Chapter 11 Rotating Bodies 10 0% 0 of 1 Answer Now ! How much force must you exert on a bucket crank handle that is 30.0 cm long to lift a bucket of water of mass 25.0 kg if the rope is 15.0 m long and the cylinder is 10.0 cm in diameter? • 16. N • 24. N • 41. N • 82. N • 735. N

  5. Physics 1710—Chapter 11 Rotating Bodies τ = r1 F1 sin θ = r1 mg =(0.05 m)(9.8*25.0 N) = 12.2Nm τ = r2 F2 F2 = τ /r2 = (12.2 Nm)/(0.3 m) = 41. N Solution:

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

  7. Physics 1710—Chapter 11 Rotating Bodies 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 Review

  8. Physics 1710—Chapter 11 Rotating Bodies 1′ Lecture Angular momentum about an axis z is equal to the product of the moment of inertia of the body about that axis and the angular velocity about z. In the absence of torques, the angular momentum is conserved.

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

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

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

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

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

  16. Physics 1710—Chapter 11 Rotating Bodies 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.”

  17. Physics 1710—Chapter 11 Rotating Bodies 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.

  18. Physics 1710—Chapter 11 Rotating Bodies 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.

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

  20. Physics 1710—Chapter 11 Rotating Bodies Rotating Platform Demonstration

  21. Physics 1710—Chapter 11 Rotating Bodies No Talking! Think! Confer! Why does the platform spin faster when he brings his arms in? Peer Instruction Time

  22. Physics 1710—Chapter 11 Rotating Bodies 10 0% 0 of 1 Answer Now ! Why does the platform spin faster when he brings his arms in? • He increases his angular momentum. • He increase his moment of inertia. • He decrease his moment of inertia. • He pushes against the inertia of the weights. • None of the above

  23. Physics 1710—Chapter 11 Rotating Bodies Analysis: Like an ice skater. 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 ⍵ z Lz = (∑i mi ri 2 ) ⍵z Lz = Iz ⍵z ; & ⍵z = Lz / Iz Therefore, a decrease in I ( by reducing r) will result in an increase in ⍵ even if dL/dt = 0!

  24. Physics 1710—Chapter 11 Rotating Bodies Rotating Platform Demonstration

  25. Physics 1710—Chapter 11 Rotating Bodies 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 ⍵.

  26. Physics 1710—Chapter 11 Rotating Bodies Iω τ r F = mg L = Iω Torque = τ = d L/dt = d( Iω)/dt Second Law of Motion & Gyroscopic Precession τ = r x F = d L/dt = I(dω/dt); dω/dt = I-1τ

  27. Physics 1710—Chapter 11 Rotating Bodies Iω τ Torque = τ = d L/dt = d( Iω)/dt Gyroscopic Precession (Top view)

  28. Physics 1710—Chapter 11 Rotating Bodies Fundamental Angular Momentum Fundamental unit of angular momentum = ℏ ℏ= 1.054 x 10 -34 kg‧m/s2 ICM⍵ ≈ ℏ ⍵≈ ℏ / ICM = 1.054 x 10 -34 kg‧m/s2 / (1.95 x 10 -46 kg‧m) = 5.41 x 10 11 rad/s

  29. Physics 1710—Chapter 11 Rotating Bodies 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

  30. Physics 1710—Chapter 11 Rotating Bodies • 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:

  31. Physics 1710—Chapter 11 Rotating Bodies Summary: Angular momentum about an axis z is equal to the product of the moment of inertia of the body about that axis and the angular velocity about z. L = I ⍵ Lz = Iz⍵ In the absence of torques, the angular momentum is conserved. In the presence of torques the angular moment will change with time.

  32. Physics 1710—Chapter 11 Rotating Bodies Good luck on the exam tonight at 4:30 p.m.

  33. Physics 1710—Chapter 11 Rotating Bodies 10 0% 0 of 1 Answer Now ! Where should the fulcrum be place to balance the teeter-totter?

  34. Physics 1710—Chapter 11 Rotating Bodies 10 0% 0 of 1 Answer Now ! Which way will the torque ladder move? • Clockwise • Counterclockwise • Will stay balanced

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