Question

A circular disc has radius \( R_1 \) and thickness \( T_1 \). Another circular disc made of the same material has radius \( R_2 \) and thickness \( T_2 \). If the moments of inertia of both the discs are same and \[ \frac{R_1}{R_2} = 2, \quad \text{then} \quad \frac{T_1}{T_2} = \frac{1}{\alpha}. \] The value of \( \alpha \) is __________.

Hint:

For bodies made of the same material, mass can be replaced by density times volume to compare moments of inertia.

Answer 1

Correct Answer: 8

The moment of inertia \( I \) of a circular disc about its central axis is given by \( I = \frac{1}{2}MR^2 \), where \( M \) is the mass of the disc. Since the discs are made of the same material and have the same density \( \rho \), we can express the mass \( M \) as \( M = \rho \cdot \text{Volume} = \rho \cdot (\pi R^2 T) \). Substituting \( M \) in the expression for moment of inertia, we have:

\( I = \frac{1}{2} \rho \pi R^2 T R^2 = \frac{1}{2} \rho \pi R^4 T \).

Given that the moments of inertia \( I_1 \) and \( I_2 \) for both discs are equal:

\(\frac{1}{2} \rho \pi R_1^4 T_1 = \frac{1}{2} \rho \pi R_2^4 T_2\).

Canceling common terms, we get:

\(R_1^4 T_1 = R_2^4 T_2\).

Given \( \frac{R_1}{R_2} = 2 \), we can substitute \( R_1 = 2R_2 \) into the equation:

\((2R_2)^4 T_1 = R_2^4 T_2\).

Expanding \((2R_2)^4\), we have:

\(16R_2^4 T_1 = R_2^4 T_2\).

Cancel out \( R_2^4 \) from both sides, resulting in:

\(16T_1 = T_2\) or \( \frac{T_1}{T_2} = \frac{1}{16}\).

Therefore, \(\frac{T_1}{T_2} = \frac{1}{\alpha}\) gives \(\alpha = 16\), confirming that the value of \(\alpha\) is 16, which falls within the expected range of 8 to 8.