Question

Four identical solid spheres each of mass $m$ and radius $a$ are placed with their centres on the four corners of a square of side $b$. The moment of inertia of the system about one side of square where the axis of rotation is parallel to the plane of the square is:

• A. $\frac{4}{3}m{a}^2$
• B. $\frac{8}{5}m{a}^2+m{b}^2$
• C. $\frac{4}{5}m{a}^2+2m{b}^2$
• D. $\frac{8}{5}m{a}^2+2m{b}^2$

Answer 1

The correct option is D

$\frac{8}{5}m{a}^2+2m{b}^2$

Step 1. Given Data,

Mass of four identical sphere $=m$

Radius of four identical sphere $=a$

Side of square $=b$

Let the Moment of inertia is $I$.

Axis of rotation is a straight line with the reference of which, other points of the body gets rotated around it.

Step 2. We have to find the moment of inertia $I$ of the system about one side of the square where the axis of rotation is parallel to the plane of the square,

From the figure, the moment of inertia $I$ could be calculated as,

Since, we have to find the moment of inertia $I$ of the system about one side of the square where the axis of rotation is parallel to the plane of the square,

We know that,

Parallel axis theorem: “The moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its center of mass $I_Z$ and the product of its mass and the square of the distance between the two parallel axes $M{a}^2$.”

We know that, Moment of inertia $I$ of a sphere (radius $a$ and mass $m$)$=\frac{2}{5}m{a}^2$

Therefore, Moment of inertia $I$ $=$moment of inertia of two spheres $+$moment of inertia of two spheres along the axis of rotation $b$

$$\begin{gathered} \Rightarrow I=\frac{2}{5}m{a}^2+\frac{2}{5}m{a}^2+\left[\frac{2}{5}m{a}^2+m{b}^2\right]+\left[\frac{2}{5}m{a}^2+m{b}^2\right] \\ \Rightarrow I=4\times \frac{2}{5}m{a}^2+2m{b}^2 \\ \Rightarrow I=\frac{8}{5}m{a}^2+2m{b}^2 \end{gathered}$$

Hence, correct option is $\text{'}D\text{'}$,