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

Step 1: Understand the formula for Moment of Inertia. For a uniform solid disc rotating about its central axis, the moment of inertia is $I = \frac{1}{2} M R^{2}$.

Step 2: Relate Mass to Radius. The discs are made of the same material (iron) and have negligible thickness. Let the thickness be $t$ and density be $\rho$. The mass $M$ is given by $M = \rho \times V = \rho \times (\pi R^{2} \times t)$. Since $\rho$ and $t$ are constant for both discs, $M \propto R^{2}$.

Step 3: Determine the proportionality of I. Substitute $M \propto R^{2}$ into $I = \frac{1}{2} M R^{2}$. This gives $I \propto (R^{2}) \times R^{2}$, which means $I \propto R^{4}$.

Step 4: Calculate the ratio. We are looking for the ratio $\frac{I_{1}}{I_{2}} = \left(\frac{R_{1}}{R_{2}}\right)^{4}$. Given that $R_{2} = 2R_{1}$, we substitute to find $\frac{I_{1}}{I_{2}} = \left(\frac{R_{1}}{2R_{1}}\right)^{4} = \left(\frac{1}{2}\right)^{4} = \frac{1}{16}$.

Step 5: Find the value of x. Comparing $\frac{I_{1}}{I_{2}} = \frac{1}{16}$ with the required format $\frac{1}{x}$, we find x = 16.