Question

For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through $O$ (the centre of mass) and $O\text{'}$ (corner point) is:

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{8}$

Answer 1

The correct option is C

$\frac{1}{4}$

Step 1: Given data

Length of the rectangle ,$a$ $=80cm$

The breadth of the rectangle ,$b$$=60cm$

Moment of inertia $=I$

Moment of inertia about the axis perpendicular to the sheet and passing through $O$ $=I_O$

Moment of inertia at the point $O^{\text{'}}$ $=I_{O^{\text{'}}}$

Total mass $=M$

Step 2: To find the ratio of moment of inertia

We know

$I_O=\frac{M}{12}\left(a^2+b^2\right)$

$I_O=\frac{M}{12}\left({\left(80\right)}^2+{\left(60\right)}^2\right)$

Using the Parallel axis theorem

$I_{O^{\text{'}}}=I_O+M\left({\left(\frac{b}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2\right)$

$=\frac{M{a}^2}{12}+\frac{M{b}^2}{12}+\frac{M{b}^2}{4}+\frac{M{a}^2}{4}$

$$\begin{gathered} {I}_{O\text{'}}=\frac{M}{12}\left(a^2+b^2\right)+M\left(\frac{a^2}{4}+\frac{b^2}{4}\right) \\ {I}_{O\text{'}}=I_O+M{\left(50\right)}^2 \end{gathered}$$

Then the ratio of moment of inertia

$$\begin{gathered} \frac{I_O}{I_{O\text{'}}}=\frac{{\displaystyle \frac{M}{12}}\left(10000\right)}{{\displaystyle \frac{M}{12}}\left(10000\right)+M\left(2500\right)} \\ =\frac{1}{4} \end{gathered}$$

Hence the ratio of moment of inertia $\frac{I_O}{I_{O\text{'}}}=\frac{1}{4}$

Therefore the correct answer is Option C