Question

Two iron solid discs of negligible thickness have radii \( R_1 \) and \( R_2 \) and moment of inertia \( I_1 \) and \( I_2 \), respectively. For \( R_2 = 2R_1 \), the ratio of \( I_1 \) and \( I_2 \) would be \( \frac{1}{x} \), where \( x \) is:

Hint:

Remember that the moment of inertia scales with the square of the radius for simple geometric bodies like discs and spheres.

Answer 1

Correct Answer: 16

Step 1: Formula for Moment of Inertia of a Solid Disc

The moment of inertia of a solid disc about its central axis is: \[ I = \frac{1}{2}MR^2 \]

Step 2: Moments of Inertia of Discs \( I_1 \) and \( I_2 \)

For the first disc: \[ I_1 = \frac{1}{2} M_1 R_1^2 \] For the second disc: \[ I_2 = \frac{1}{2} M_2 R_2^2 \]

Step 3: Given Condition

Given: \[ R_2 = 2R_1 \]

Assuming the discs are of the same material and negligible thickness, mass is proportional to area: \[ \frac{M_2}{M_1} = \left( \frac{R_2}{R_1} \right)^2 = \left( \frac{2R_1}{R_1} \right)^2 = 4 \] Thus: \[ M_2 = 4M_1 \]

Step 4: Ratio of \( I_1 \) to \( I_2 \)

Substitute values: \[ I_1 = \frac{1}{2} M_1 R_1^2 \] \[ I_2 = \frac{1}{2} \times 4M_1 \times (2R_1)^2 = \frac{1}{2} \times 4M_1 \times 4R_1^2 = 8M_1R_1^2 \]

The ratio \( \frac{I_1}{I_2} \) is: \[ \frac{I_1}{I_2} = \frac{\frac{1}{2} M_1 R_1^2}{8M_1R_1^2} = \frac{1}{2 \times 8} = \frac{1}{16} \]

Therefore: \[ \boldsymbol{x = 16} \]