Answer
See answers.
Work Step by Step
a. Calculate the angular momentum using equation 8–18.
$$L_i = I \omega = \frac{1}{2}MR^2 \omega = 0.5(2.8 kg)(0.28m)^2(\frac{1300 \times 2\pi rad}{60s})\approx 15 \frac{kg \cdot m^2}{s}$$
b. The required torque is the change in angular momentum per second. The final L is zero.
$$\tau=\frac{0-L_i}{6.0s}=-2.5 N \cdot m$$
The negative sign verifies that the torque reduces the initial angular momentum.
