Answer
(a) $I = 1.35\times 10^{-5}~kg~m^2$
(b) The required muscular force is $526.5~N$
Work Step by Step
(a) We can find the rotational inertia:
$I = \frac{1}{3}MR^2$
$I = \frac{1}{3}(0.028~kg)(0.038~m)^2$
$I = 1.35\times 10^{-5}~kg~m^2$
(b) We can find the angular acceleration:
$\omega_f = \omega_0+\alpha~t$
$\alpha = \frac{\omega_f - \omega_0}{t}$
$\alpha = \frac{175~rad/s - 0}{1.50\times 10^{-3}~s}$
$\alpha = 1.17\times 10^5~rad/s^2$
We can find the required force:
$\tau = I~\alpha$
$r~F = I~\alpha$
$F = \frac{I~\alpha}{r}$
$F = \frac{(1.35\times 10^{-5}~kg~m^2)(1.17\times 10^5~rad/s^2)}{3.00\times 10^{-3}~m}$
$F = 526.5~N$
The required muscular force is $526.5~N$.
