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Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium CCW +

Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium CCW +

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Page 1: Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium  CCW +

Ch11-1 Torque

Chapter 11: Rotational Dynamics and Static Equilibrium

CCW +

Page 2: Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium  CCW +

Applying a Torque

Page 3: Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium  CCW +

Ct1: You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is least effective in loosening the nut?

A B

C D

Page 4: Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium  CCW +

Ch11-1 Torque = rF = rFsin

Chapter 11: Rotational Dynamics and Static Equilibrium

Page 5: Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium  CCW +

Torque = moment arm x force = rF = rsinF

Chapter 11: Rotational Dynamics and Static Equilibrium

line of action of the force

moment arm

The moment arm is the perpendicular distance from rotation axis to the line of action of the force.

Page 6: Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium  CCW +

P11.2 (p.350)

Page 7: Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium  CCW +

Ch11-2 Torque and Angular Acceleration

= I Torques summed about fixed axis O.

P11.15 (p.351)

Chapter 11: Rotational Dynamics and Static Equilibrium

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Page 8: Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium  CCW +

CT2: Two wheels with fixed hubs, each having a mass of 1 kg, start from rest, and forces are applied as shown. Assume the hubs and spokes are massless, so that the rotational inertia is I = mR2. In order to impart identical angular accelerations, how large must F2 be?

A. 0.25 NB. 0.5 NC. 1 ND. 2 NE. 4 N

Page 9: Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium  CCW +

Ch11-3 Zero Torque and Static Equilibrium

F = 0 and = 0 P11.22 (p.351)

Chapter 11: Rotational Dynamics and Static Equilibrium

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Page 10: Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium  CCW +

CT3: P11.22c If the mass of the teeter-totter were doubled, the answers to parts a and b for the position of the applied force would

A. double.

B. halve.

C. stay the same.

Page 11: Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium  CCW +

P11.31 (p.352)

Page 12: Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium  CCW +

Ch11-4 Center of Mass and Balance

Near the Earth, the center of mass is the balance point of an object.

Chapter 11: Rotational Dynamics and Static Equilibrium

Page 13: Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium  CCW +

Ch11-5 Dynamic Applications of Torque

We can finally relax the simplification that a pulley is massless. P11.44 (p.353)

Chapter 11: Rotational Dynamics and Static Equilibrium

Page 14: Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium  CCW +

CT4: P11.44a The 25N tension is

A. greater than the tension on the other side.

B. less than the tension on the other side.

C. the same as the tension on the other side.

Page 15: Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium  CCW +

Ch11-6 Angular Momentum

p = mv (kgm/s)

L = I (fixed axis) (kgm2/s)

F = p/t = L/t P11.55 (p.353)

Chapter 11: Rotational Dynamics and Static Equilibrium

Page 16: Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium  CCW +

Ch11-7 Conservation of Angular Momentum

If Fext = 0, then ptot is conserved

If = 0, then Ltot is conserved P11.63 (p.354)

Chapter 11: Rotational Dynamics and Static Equilibrium

Page 17: Ch11-1 Torque Chapter 11: Rotational Dynamics and Static Equilibrium  CCW +

Ch11-8 Rotational Work and Power

W = Fx = (one dimension or fixed axis)

W = K Power = W/t = /t =

P11.68 (p.354)

Chapter 11: Rotational Dynamics and Static Equilibrium