  = r x F  = rFsin  ixi = jxj = kxk = 0 ixj = k (ijkijk) 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 x F = - F x r  F        r  +             -    CCW +

Figure 11.2  The vector product A  B is a third vector C having a magnitude AB sin  equal to the area of the parallelogram shown. The direction of C is perpendicular to the plane formed by A and B, and this direction is determined by the right-hand rule. Fig. 11.2, p.338

  = rFsin Active Figure 11.1  The torque vector  lies in a direction perpendicular to the plane formed by the position vector r and the applied force vector F. At the Active Figures link at http://www.pse6.com, you can move point P and change the force vector F to see the effect on the torque vector. Fig. 11.1, p.337

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.

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.

Ch 11 Angular Momentum 11.2 Angular Momentum: The Nonisolated System L = r x p Single Particle: net = dL/dt System: ext = dLtot/dt       

Active Figure 11.4  The angular momentum L of a particle of mass m and linear momentum p located at the vector position r is a vector given by L = r  p. The value of L depends on the origin about which it is measured and is a vector perpendicular to both r and p. At the Active Figures link at http://www.pse6.com, you can change the position vector r and the force vector F to see the effect on the angular momentum vector. Fig. 11.4, p.340

P11.12 (p.330) P11.18 (p.331)

Ch 11 Angular Momentum 11.3 Angular Momentum of a Rotating Rigid Object Lz = I ext = I

Ch 11 Angular Momentum 11.3 Angular Momentum of a Rotating Rigid Object Lz = I ext = I P11.27 (p.332)

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)   

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) (b) no difference The answer depends on the rotational inertia of the dumbbell.

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) (b) no difference The answer depends on the rotational inertia of the dumbbell.

Ch 12 Static Equilibrium and Elasticity 12.1 The Rigid Body in Equilibrium Fext = 0 ext = 0  

P12.40 (p.358)

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? No, the weight will be too large. No, the weight will be too small. Yes.