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 $1/x$, where $x =$ _____.

Hint:

Since the thickness and material are the same, mass is proportional to the square of the radius. Combined with the moment of inertia formula, I is proportional to $R^4$.

Answer 1

Correct Answer: 16

The moment of inertia of a solid disc about an axis passing through its center and perpendicular to its plane is given by $I = \frac{1}{2}MR^2$, where $M$ is the mass of the disc and $R$ is its radius.

Since both discs are made of iron and have negligible thickness, their mass depends on their area. For a disc, $\text{Area} = \pi R^2$. Thus the mass is proportional to $R^2$: $M \propto R^2$.

Substituting this proportionality into the moment of inertia formula: $I = \frac{1}{2}MR^2 \implies I \propto R^2 \times R^2 \implies I \propto R^4$.

Therefore, $\frac{I_2}{I_1} = \left(\frac{R_2}{R_1}\right)^4$. Given $R_2 = 2R_1$, we have $\frac{I_2}{I_1} = \left(\frac{2R_1}{R_1}\right)^4 = 2^4 = 16$.

Hence, $\frac{I_1}{I_2} = \frac{1}{16}$. Comparing with the given form $\frac{I_1}{I_2} = \frac{1}{x}$, we obtain $x = 16$.