| Quiz 8 |
Ch. 8: rotational dynamics: 1*, 5, 7, 15*, 19, 24, 25, 27*, 29, 30, 33, 40, 45, 48, 52*, 59, 60( 60 is ec), 77, 83 (83 ec) TURN IN 19, 24, 25, 33, 45, 48, 52, 59 |
| 1. See class notes: the conversion factor is 3.14/180. where 3.14 = pi. |
| 5. Arc length s = R*Ø, where Ø = the angle
swept out about the center as the ball rolls. R = radius of
ball. The angle is (15)*(6.28), where 6.28 = 2·pi. The arc length s = x = 3.5 m = linear distance the center moves along the floor. Solve for R, then compute the diameter. |
| 7. (a) Angular velocity = omega = w = (2500 rev/min)*(6.28 rad/rev)*(1/60 min/sec) (b) Linear velocity v = R*w. Centripetal acceleration = Rw2. Read section 8.1 |
| 15. Average angular velocity = (w1 + w2)/2. Angle = (average angular velocity)*(220 seconds) w1 = 0. w2 = 15,000 rev/min. Please convert to rad/s as shown in the previous problem to get the angle in radians (rad). To get the the number of revolutions, divide the computed angle by 6.28. |
| 19. wf2 = wi2 +2*(alpha)*Ø, where Ø = (1500)*(6.28) rad and wi = (850)*(6.28 )*(1/60) rad/sec. The final angular velocity is zero. Find alpha. |
| 24. First try the
problem assuming the friction torque is zero just to
get practice with torques. "pos" is designated for the torques that cause counterclockwise (CCW) motion. net torque = pos - neg net torque = (0.24)*(28) - (0.12)*(35) - (0.24)*(18). The result is negative. That means the wheel rotates
clockwise (CW) under the influence of these torques. Now pretend
CW is the positive direction. In that case, net torque is: |
| 25. It should be clear that the rotation will be
CW, which can be taken as the pos direction. net torque = pos - neg net torque = L2*mg - L1*mg. |
| 27. See Table in fig. 8-21 for I for the solid uniform sphere. |
| 29. (a) I = mr2. See class notes. (b) net torque = 0 = (applied torque) - (1.2 )*(0.020). Solve for the applied torque. Note that I was not needed. |
| 33. I*alpha = net torque I*alpha = 4*R*F You need alpha. wf - wi = alpha*(change in time) The initial angular velocity is zero; wf = 32*(6.28 rad/rev)(1/60 min/sec). Solve for F given R and I = (1/2)*MR2. |
| 45. Use equation (8-16). VCM is given as is ICM from the Table in Fig 8-21. Use VCM = R*w to make the correct substitutions to compute the total kinetic energy. |
| 48. (a) See conceptual example 8-14 page 212. H = Dsin30 and D = 10.0 m (b) ratio = [(1/2)*MVCM2]/ [(1/2)*ICM*w2]; Get ICM from the Table in Fig 8-21. Use VCM = R*w to make the correct substitutions to compute the ratio. The mass M and radius R cancel. . |
| 52. (a) L = ICM*w. Get ICM from the Table in Fig 8-21. w = (1500 rev/min)* (6.28 rad/rev)*(1/60 min/sec) (b) I*alpha = net torque. wf - wi = alpha*(change in time). The final angular velocity is zero; w i =is given above. Find alpha and the net torque. Note that alpha and the net torque are both negative. The corresponds to the wheel slowing down as it moves in the counter-clockwise direction. |
| 59. Read examples 8-15 and 8-16.
Li = Lf . |
| more hints later. |