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Ch 11 Angular Momentum 11.1 The Vector Product and Torque = r x F = rFsin Direction: Right hand...

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Page 1: Ch 11 Angular Momentum 11.1 The Vector Product and Torque  = r x F  = rFsin Direction: Right hand rule ixi = jxj = kxk = 0 ixj = k (ijkijk) Note: r.
Page 2: Ch 11 Angular Momentum 11.1 The Vector Product and Torque  = r x F  = rFsin Direction: Right hand rule ixi = jxj = kxk = 0 ixj = k (ijkijk) Note: r.

Ch 11 Angular Momentum

11.1 The Vector Product and Torque

= r x F = rFsinDirection: Right hand rule

ixi = jxj = kxk = 0 ixj = k (ijkijk)

Note: r x F = - F x r

r

F

+

-

CCW +

Page 3: Ch 11 Angular Momentum 11.1 The Vector Product and Torque  = r x F  = rFsin Direction: Right hand rule ixi = jxj = kxk = 0 ixj = k (ijkijk) Note: r.

Fig. 11.2, p.338

Page 4: Ch 11 Angular Momentum 11.1 The Vector Product and Torque  = r x F  = rFsin Direction: Right hand rule ixi = jxj = kxk = 0 ixj = k (ijkijk) Note: r.

Fig. 11.1, p.337

= rFsin

Page 5: Ch 11 Angular Momentum 11.1 The Vector Product and Torque  = r x F  = rFsin Direction: Right hand rule ixi = jxj = kxk = 0 ixj = k (ijkijk) Note: r.

CT1: A ladybug sits at the outer edge of a merry-go-round that is turning and is slowing down. The vector expressing her angular velocity is

A. in the +x direction.B. in the –x direction.C. in the +y direction.D. in the –y direction.E. in the +z direction.F. in the –z direction.G. zero.

Page 6: Ch 11 Angular Momentum 11.1 The Vector Product and Torque  = r x F  = rFsin Direction: Right hand rule ixi = jxj = kxk = 0 ixj = k (ijkijk) Note: r.

CT2: A ladybug sits at the outer edge of a merry-go-round that is turning and is slowing down. The vector expressing her angular acceleration is

A. in the +x direction.B. in the –x direction.C. in the +y direction.D. in the –y direction.E. in the +z direction.F. in the –z direction.G. zero.

Page 7: Ch 11 Angular Momentum 11.1 The Vector Product and Torque  = r x F  = rFsin Direction: Right hand rule ixi = jxj = kxk = 0 ixj = k (ijkijk) Note: r.

Ch 11 Angular Momentum

11.2 Angular Momentum: The Nonisolated System

L = r x p

Single Particle: net = dL/dt

System: ext = dLtot/dt

Page 8: Ch 11 Angular Momentum 11.1 The Vector Product and Torque  = r x F  = rFsin Direction: Right hand rule ixi = jxj = kxk = 0 ixj = k (ijkijk) Note: r.

Fig. 11.4, p.340

Page 9: Ch 11 Angular Momentum 11.1 The Vector Product and Torque  = r x F  = rFsin Direction: Right hand rule ixi = jxj = kxk = 0 ixj = k (ijkijk) Note: r.

P11.12 (p.330)

P11.18 (p.331)

Page 10: Ch 11 Angular Momentum 11.1 The Vector Product and Torque  = r x F  = rFsin Direction: Right hand rule ixi = jxj = kxk = 0 ixj = k (ijkijk) Note: r.

Ch 11 Angular Momentum11.3 Angular Momentum of aRotating Rigid Object

Lz = I ext = I

Page 11: Ch 11 Angular Momentum 11.1 The Vector Product and Torque  = r x F  = rFsin Direction: Right hand rule ixi = jxj = kxk = 0 ixj = k (ijkijk) Note: r.

Ch 11 Angular Momentum11.3 Angular Momentum of aRotating Rigid Object

Lz = I ext = I P11.27 (p.332)

Page 12: Ch 11 Angular Momentum 11.1 The Vector Product and Torque  = r x F  = rFsin Direction: Right hand rule ixi = jxj = kxk = 0 ixj = k (ijkijk) Note: r.

Ch 11 Angular Momentum

11.4 The Isolated System: Conservation of Angular Momentum

If ext = dLtot/dt = 0, then Ltot is conserved.

P11.35 (p.333)

Page 13: Ch 11 Angular Momentum 11.1 The Vector Product and Torque  = r x F  = rFsin Direction: Right hand rule ixi = jxj = kxk = 0 ixj = k (ijkijk) Note: r.

CT3: A force F is applied to a dumbbell for a time interval t, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed?

A. (a)B. (b)C. no differenceD. The answer depends on the rotational inertia of the dumbbell.

Page 14: Ch 11 Angular Momentum 11.1 The Vector Product and Torque  = r x F  = rFsin Direction: Right hand rule ixi = jxj = kxk = 0 ixj = k (ijkijk) Note: r.

CT4: A force F is applied to a dumbbell for a time interval t, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater energy?

A. (a)B. (b)C. no differenceD. The answer depends on the rotational

inertia of the dumbbell.

Page 15: Ch 11 Angular Momentum 11.1 The Vector Product and Torque  = r x F  = rFsin Direction: Right hand rule ixi = jxj = kxk = 0 ixj = k (ijkijk) Note: r.

Ch 12 Static Equilibrium and Elasticity

12.1 The Rigid Body in Equilibrium

Fext = 0 ext = 0

Page 16: Ch 11 Angular Momentum 11.1 The Vector Product and Torque  = r x F  = rFsin Direction: Right hand rule ixi = jxj = kxk = 0 ixj = k (ijkijk) Note: r.

P12.40 (p.358)

Page 17: Ch 11 Angular Momentum 11.1 The Vector Product and Torque  = r x F  = rFsin Direction: Right hand rule ixi = jxj = kxk = 0 ixj = k (ijkijk) Note: r.

CT5: A girl has a large, docile dog she wishes to weigh on a small bathroom scale. She reasons that she can determine her dog's weight by the following method. First she puts the dog's two front feet on the scale and records the scale reading. Then she places the dog's two back feet on the scale and records the scale reading. Assume the dog is standing. She thinks that the sum of the readings will be the dog's weight. Is she correct?

A. No, the weight will be too large.

B. No, the weight will be too small.

C. Yes.


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