91Ó°ÊÓ

1. The radius is,$\mathrm{R}=0.02\text{″¾}$.
2. The force is,$\mathrm{F}=0.400\text{ N}$.
3. The time is,$\mathrm{t}=5\text{ s±ð³¦}$.
4. The angular speed is,$\mathrm{Ó\neg }=114\text{ r²¹»å}/\text{s}$.
5. The mass of disc is, $\mathrm{M}=0.5\text{ k²µ}$.

You can find the angular acceleration of the plate using the formula for torque in terms of force and distance as well as in terms of angular acceleration and rotational inertia. Using the formula for the rotational inertia of the disk and total rotational inertia of the system, you can find the rotational inertia of the plate. The formulas used are given below.

$\mathbf{RF}\mathbf{=}\mathbf{\pm õÎ\pm }$

Where,
$\mathbf{Î\pm }\mathbf{=}\frac{\mathbf{Ó\neg }}{\mathbf{t}}$

Moment of inertia,
$\mathbf{I}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{MR}}^{\mathbf{2}}$

Total moment of inertia,

role="math" localid="1660915909375" $\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{p}\mathbf{l}\mathbf{a}\mathbf{t}\mathbf{e}}\mathbf{+}{\mathit{I}}_{\mathbf{d}\mathbf{i}\mathbf{s}\mathbf{c}}$

You have,

$\mathrm{RF}=\mathrm{Î\pm ³Ù}$

Where,
$\mathrm{Î\pm }=\frac{\mathrm{Ó\neg }}{\mathrm{t}}$

Therefore,

$\begin{array}{c}\mathrm{RF}=\frac{\mathrm{IÓ\neg }}{\mathrm{t}}\\ \mathrm{I}=\frac{\mathrm{RFt}}{\mathrm{Ó\neg }}\end{array}$

Now, let the moment of inertia of a disc be,

$\mathrm{I}=\frac{1}{2}{\mathrm{MR}}^2$

Therefore, total moment of inertia of plastic plate and disc can be given bytheformula

$\begin{array}{c}\mathrm{I}={\mathrm{I}}_{\mathrm{plate}}+{\mathrm{I}}_{\mathrm{disc}}\\ {\mathrm{I}}_{\mathrm{plate}}=\mathrm{I}−{\mathrm{I}}_{\mathrm{disc}}\\ \mathrm{I}=\frac{\mathrm{RFt}}{\mathrm{Ó\neg }}−\frac{1}{2}{\mathrm{MR}}^2\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{I}=\frac{\left(0.02\text{″¾}\right)\left(0.4\text{ N}\right)\left(5\text{ s}\right)}{114\text{ r²¹»å}/\text{s}}−\frac{1}{2}\left(0.5\text{ k²µ}\right)×{\left(0.02\mathrm{m}\right)}^2\\ =2.51×{10}^{−4}{\text{ k²µ.m}}^{\text{2}}\end{array}$

Hence the rotational inertia of the plate about axle is $2.51×{10}^{−4}{\text{ k²µ.m}}^2$.