Answer
$\sum\tau=8.03\times10^{-4}N.m$
Work Step by Step
1) Find the angular acceleration of the CD
We have $\omega_0=0$, $\omega=21rad/s$ and $t=0.8s$. Therefore, $$\alpha=\frac{\omega-\omega_0}{t}=26.25rad/s^2$$
2) The net torque acting on the CD is $$\sum\tau=I\alpha$$
From Table 9.1, the moment of inertia of the disk $T=\frac{1}{2}mr^2$
$$\sum\tau=\frac{1}{2}mr^2\alpha$$
We have $m=17g=1.7\times10^{-2}kg$ and $r=6cm=6\times10^{-2}m$
$$\sum\tau=8.03\times10^{-4}N.m$$
