Answer
(a) $I = 0.502~kg~m^2$
(b) The motor must provide a torque of $16.8~N \cdot m$
Work Step by Step
(a) We can find the rotational inertia:
$I = \frac{1}{2}MR^2$
$I = \frac{1}{2}(20.0~kg)(0.224~m)^2$
$I = 0.502~kg~m^2$
(b) We can express the angular speed in units of rad/s:
$1200~rpm \times \frac{2\pi~rad}{1~rev} \times \frac{1~min}{60~s} = 40\pi~rad/s$
We can find the angular deceleration as the wheel slows to a stop:
$\omega_f = \omega_0+\alpha~t$
$\alpha = \frac{\omega_f - \omega_0}{t}$
$\alpha = \frac{0-40\pi~rad/s}{60.0~s}$
$\alpha = -2.09~rad/s^2$
We can find the frictional torque:
$\tau_f = I~\alpha$
$\tau_f = (0.502~kg~m^2)(-2.09~rad/s^2)$
$\tau_f = -1.05~N \cdot m$
We can find the angular acceleration as the wheel speeds up:
$\omega_f = \omega_0+\alpha~t$
$\alpha = \frac{\omega_f - \omega_0}{t}$
$\alpha = \frac{40\pi~rad/s-0}{4.00~s}$
$\alpha = 31.4~rad/s^2$
We can find the required torque that the motor must provide:
$\sum \tau = I~\alpha$
$\tau_m+\tau_f = I~\alpha$
$\tau_m = I~\alpha-\tau_f$
$\tau_m = (0.502~kg~m^2)(31.4~rad/s^2)-(-1.05~N \cdot m)$
$\tau_m = 16.8~N \cdot m$
The motor must provide a torque of $16.8~N \cdot m$.
