Question

The figure shows two solid discs with radius R and r respectively. If mass per unit area is same for both, what is the ratio of MI of bigger disc around axis AB (which is $\perp$ to the plane of the disc and passing through its centre) to MI of smaller disc around one of its diameters lying on its plane ? Given 'M' is the mass of the larger disc. (MI stands for moment of inertia) 

 

• A. $2R^2 : r^2$
• B. $R^2 : r^2$
• C. $2R^4 : r^4$
• D. $2r^4 : R^4$

Hint:

Know the standard formulas for Moment of Inertia. For a solid disc of mass M and radius R: about a perpendicular axis through the center is $\frac{1}{2}MR^2$, and about a diameter is $\frac{1}{4}MR^2$. The latter can be quickly derived using the perpendicular axis theorem.

Answer 1

The Correct Option is C

Let $\sigma$ be the mass per unit area, which is the same for both discs.
Mass of the bigger disc, $M = \sigma \times (\text{Area}) = \sigma(\pi R^2)$.
Mass of the smaller disc, $m = \sigma \times (\text{Area}) = \sigma(\pi r^2)$.
Moment of inertia of the bigger disc ($I_{big}$) about the axis AB, which is perpendicular to its plane and passes through its center, is:
$I_{big} = \frac{1}{2} M R^2 = \frac{1}{2} (\sigma \pi R^2) R^2 = \frac{1}{2} \sigma \pi R^4$.
Moment of inertia of the smaller disc ($I_{small}$) about one of its diameters is:
From the perpendicular axis theorem, $I_z = I_x + I_y$. For a disc, $I_{diameter} = I_x = I_y$.
The MI about the perpendicular axis through the center is $I_z = \frac{1}{2}mr^2$.
So, $2 \times I_{diameter} = \frac{1}{2}mr^2 \implies I_{diameter} = \frac{1}{4}mr^2$.
$I_{small} = \frac{1}{4}mr^2 = \frac{1}{4}(\sigma \pi r^2)r^2 = \frac{1}{4}\sigma \pi r^4$.
Now, we find the ratio of $I_{big}$ to $I_{small}$:
$\frac{I_{big}}{I_{small}} = \frac{\frac{1}{2} \sigma \pi R^4}{\frac{1}{4} \sigma \pi r^4}$
Cancel the common term $\sigma \pi$:
$\frac{I_{big}}{I_{small}} = \frac{1/2}{1/4} \times \frac{R^4}{r^4} = 2 \frac{R^4}{r^4}$.
So, the ratio $I_{big} : I_{small}$ is $2R^4 : r^4$.