Question

The moment of inertia of a square lamina about the perpendicular axis through its center of mass is 20 kg per meter square then its moment of inertia about an axis touching its side and in the plane of the lamina will be

Answer 1

Step 1: Given Data

Moment of inertia of a square lamina (${\mathrm{I}}_{\mathrm{z}}$) = $20kg{m}^2$

Let us consider,

$\mathrm{L}$ = Length of square

$\mathrm{M}$ = Mass of the square lamina

Step 2: Calculating moment of inertia about an axis touching its side and in the plane of the lamina

The moment of inertia must be specified with respect to a chosen axis of rotation. For a point mass, the moment of inertia is just the mass times the square of perpendicular distance to the rotation axis, $I=m{r}^2$.

$I_z=\frac{M{L}^2}{6}$

$$\begin{gathered} 20=\frac{{\mathrm{ML}}^2}{6} \\ \Rightarrow {\mathrm{ML}}^2=120{\mathrm{kgm}}^2 \end{gathered}$$

According to the parallel axis theorem,

$$\begin{gathered} {\mathrm{I}}_{\mathrm{x}}={\mathrm{I}}_{\mathrm{y}}=\frac{{\mathrm{I}}_{\mathrm{z}}}{2}=10{\mathrm{kgm}}^2 \\ {\mathrm{I}}_{\mathrm{L}}={\mathrm{I}}_{\mathrm{x}}+\mathrm{M}{\left(\frac{\mathrm{L}}{2}\right)}^2 \\ \Rightarrow {\mathrm{I}}_{\mathrm{L}}={\mathrm{I}}_{\mathrm{x}}+\frac{{\mathrm{ML}}^2}{4} \\ \Rightarrow {\mathrm{I}}_1=10+\frac{120}{4} \\ \Rightarrow {\mathrm{I}}_1=40{\mathrm{kgm}}^2 \end{gathered}$$

Hence, the moment of inertia about an axis touching its side and in the plane of the lamina is $40{\mathrm{kgm}}^2$.