In Fig.$\mathbf{10}\mathbf{-}\mathbf{45}\mathbf{\text{a}}$ , an irregularly shaped plastic plate with uniform thickness and density (mass per unit volume) is to be rotated around an axle that is perpendicular to the plate face and through point $\mathbf{\text{O}}$. The rotational inertia of the plate about that axle is measured with the following method. A circular disk of mass$\mathbf{0}\mathbf{.500}\mathbf{\text{\hspace{0.17em}kg}}$ and radius$\mathbf{2.00}\mathbf{\text{\hspace{0.17em}cm}}\mathbf{}$ is glued to the plate, with its center aligned with point $\mathit{O}$(Fig.$\mathbf{10}\mathbf{-}\mathbf{45}\mathbf{\text{b}}$ ). A string is wrapped around the edge of the disk the way a string is wrapped around a top. Then the string is pulled for $\mathbf{5.00}\mathbf{\text{\hspace{0.17em}s}}$. As a result, the disk and plate are rotated by a constant force of $\mathbf{0.400}\mathbf{\text{\hspace{0.17em}N}}$that is applied by the string tangentially to the edge of the disk. The resulting angular speed is $\mathbf{114}\mathbf{\text{\hspace{0.17em}rad}}\mathbf{/}\mathbf{\text{s}}$. What is the rotational inertia of the plate about the axle?

1. The radius is,$\mathrm{R}=0.02\text{\hspace{0.17em}m}$.
2. The force is,$\mathrm{F}=0.400\text{\hspace{0.17em}N}$.
3. The time is,$\mathrm{t}=5\text{\hspace{0.17em}sec}$.
4. The angular speed is,$\mathrm{\omega }=114\text{\hspace{0.17em}rad}/\text{s}$.
5. The mass of disc is, $\mathrm{M}=0.5\text{\hspace{0.17em}kg}$.

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.

$\mathbf{RF}\mathbf{=}\mathbf{I\alpha }$

Where,
$\mathbf{\alpha }\mathbf{=}\frac{\mathbf{\omega }}{\mathbf{t}}$

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,

$\mathrm{RF}=\mathrm{\alpha t}$

Where,
$\mathrm{\alpha }=\frac{\mathrm{\omega }}{\mathrm{t}}$

Therefore,

$\begin{array}{c}\mathrm{RF}=\frac{\mathrm{I\omega }}{\mathrm{t}}\\ \mathrm{I}=\frac{\mathrm{RFt}}{\mathrm{\omega }}\end{array}$

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

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

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{I}=\frac{\left(0.02\text{\hspace{0.17em}m}\right)\left(0.4\text{\hspace{0.17em}N}\right)\left(5\text{\hspace{0.17em}s}\right)}{114\text{\hspace{0.17em}rad}/\text{s}}-\frac{1}{2}\left(0.5\text{\hspace{0.17em}kg}\right)\times {\left(0.02\mathrm{m}\right)}^2\\ =2.51\times {10}^{-4}{\text{\hspace{0.17em}kg.m}}^{\text{2}}\end{array}$

Hence the rotational inertia of the plate about axle is $2.51\times {10}^{-4}{\text{\hspace{0.17em}kg.m}}^2$.