Question

Four particles each of mass $m$ are lying symmetrically on the rim of a disc of mass $M$ and radius $R$. The moment of inertia of this system about an axis passing through one of the particles and perpendicular to the plane of the disc is :

• A. $16m{R}^2$
• B. $(3M+16m)\frac{R^2}{2}$
• C. $(3m+16M)\frac{R^2}{2}$
• D. $Zero$

Answer 1

The correct option is B $(3M+16m)\frac{R^2}{2}$

Given:

Mass of the disc $=M$

Mass of each particle $=m$

The radius of the disc $=R$

As the axis of rotation is passing through one of the particles (At the circumference of the disc) and perpendicular to the plane of the disc.

So, by using the parallel axis theorem, the moment of inertia of the disc can be given as:

$I_d=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$

Now the moment of inertia of the particle about which the body is revolving is zero.

Now, we can observe that two of the particles are at a distance of $\sqrt{2}R$ and one particle is at a distance of $2R$ from the rotational axis.

Thus the moment of inertia of the particles about the axis of rotation can be given as:

$I_p=2m\times (\sqrt{2}R{)}^2+m(2R{)}^2$

$I_p=4m{R}^2+4m{R}^2=8m{R}^2$

Now, the moment of inertia of the combined system can be given as $I$:

$I=I_p+I_d$

$\Rightarrow I=\frac{3}{2}M{R}^2+8m{R}^2$

On rearranging further:

$\Rightarrow I=(3M+16m)\frac{R^2}{2}$

Hence, the correct option is $\left(B\right)$