Answers to Chapter 8

1. In each revolution the wheel rolls a distance equal to its circumference, so we have

                L = N(pD);

                7.0 x 10³m = Np(0.68m), which gives N =  3.3 x 10³revolutions.

2. (a) We find the angular acceleration from

                q = w₀t + 1/2a t;

                (20 rev)(2p rad/rev) = 0 + 1/2a [(1min)(60s/min)]², which gives  a = 0.070 rad/s².

     (b) We find the final angular speed from

                 w = w₀ + a t = 0 + (0.070 rad/s²)(60s) = 4.2 rad/s =  40rpm.

3. (a) If there is no slipping, the linear tangential acceleration of the pottery wheel and the rubber wheel must be the same:

                 a_tan = R₁a₁ = R₂a₂;

                 (2.0cm)(7.2 rad/s²) = (25.0cm)a₂, which gives a₂ =  0.58 rad/s².

     (b) We find the time from

                 w = w₀ + a t;

                 (65 rev/min)(2p rad/rev)/(60s/min) = 0 + (0.58 rad/s²) t, which gives  t =  12s.

4. We assume clockwise motion, so the frictional torque is counterclockwise.         If we take the clockwise direction as positive, we have                  tnet = rF1 - RF2 + RF3 - tfr                    = (0.10m)(35N) - (0.20m)(30N) + (0.20m)(20N) - 0.40m.N                    =  1.1 m.N (clockwise)

5.  (a) Because we can ignore the mass of the rod, for the moment of inertia we have

                    I = m_ballR²

                      = (1.05kg)(0.900m)² = 0.851 kg.m².    

      (b) To produce constant angular velocity, the net torque must be zero:

                     t_net = t_applied - t_friction= 0 or

                     t_applied = FR = (0.0800 N)(0.900m) = 0.0720 m.N.

6. (a) Because we ignore the mass of the arm, for the moment of inertia we have                        I = mballd2 = (3.6kg)(0.30m)2 = 0.324kg.m2.              The angular acceleration of the ball-arm system is                        a = atan/d1 = (7.0m/s2)/(0.30m) = 23.3 rad/s2.               Thus we find the required torque from                         t = Ia                            = (0.324kg.m2)(23.3 rad/s2) =  7.5m.N.        (b) Because the force from the triceps muscle is perpendicular to the line from the axis,               we find the force from                         F = t /d2 = (7.5m.N)/(0.025m) =  3.0 x 102N.

7. The work done increases the kinetic energy of the merry-go-round:

                         W = DKE = 1/2Iw² - 0 = 1/2(1/2Mr²)(Dq /Dt)²

                              = 1/4(1640kg)(8.20m)²[(1rev)(2p rad/rev)/(8.00s)]² =  1.70 x 10⁴J

8. (a) We approximate the mass distribution as a solid cylinder. the angular momentum is

                          L = Iw = 1/2mR²w = 1/2(55kg)(0.15m)²[(3.5rev/s)(2p rad/rev)] =  14kg.m²/s.

    (b) If the arms do not move, the moment of inertia will not change. We find the torque from the change in angular momentum:

                          t = DL/Dt = (0 - 13.6kg.m/s)/(5.0s) =  -2.7m.N.