Question

Four point masses, each of mass $m$ are present at the four corners of a square of side length equal to $a$. Axis of rotation is passing through through one particle as shown in the figure below. The moment of inertia about the axis of rotation is

• A. $4m{a}^2$
• B. $3m{a}^2$
• C. $(2+\sqrt{2})m{a}^2$
• D. $2\sqrt{2}m{a}^2$

Answer 1

The correct option is B $3m{a}^2$

As we can see from the figure, axis of rotation passes through the point mass $4$. Hence $MOI$ for point mass $4$ will be zero.

Distance of point mass $1$ and $3$ from the axis of rotation
$=a\cos {45}^{\circ }=\frac{a}{\sqrt{2}}$

Distance of point mass $2$ from the axis of rotation is $=\sqrt{a^2+a^2}=\sqrt{2}a$

Hence, moment of inertia
$I=I_1+I_2+I_3+I_4$
$I=m{\left(\frac{a}{\sqrt{2}}\right)}^2+m(\sqrt{2}a{)}^2+m{\left(\frac{a}{\sqrt{2}}\right)}^2+0$
$I=\frac{m{a}^2}{2}+2m{a}^2+\frac{m{a}^2}{2}+0$
$I=3m{a}^2$