Answer
The rotational inertia of the plate is $~~2.51\times 10^{-4}~kg~m^2$
Work Step by Step
We can find the angular acceleration:
$\alpha = \frac{\omega}{t}$
$\alpha = \frac{114~rad/s}{5.00~s}$
$\alpha = 22.8~rad/s^2$
Let $I_p$ be the rotational inertia of the plate.
Let $I_d$ be the rotational inertia of the disk.
We can find the rotational inertia of the plate:
$\tau = (I_p+I_d)~\alpha$
$\frac{r~F}{\alpha} = I_p+I_d$
$I_p = \frac{r~F}{\alpha} -I_d$
$I_p = \frac{r~F}{\alpha} -\frac{1}{2}mr^2$
$I_p = \frac{(0.0200~m)(0.400~N)}{22.8~rad/s^2} -\frac{1}{2}(0.500~kg)(0.0200~m)^2$
$I_p = 2.51\times 10^{-4}~kg~m^2$
The rotational inertia of the plate is $~~2.51\times 10^{-4}~kg~m^2$
