Question

A sphere of radius $R$ is cut from a larger solid sphere or radius $2R$ as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is:

• A. $\frac{7}{8}$
• B. $\frac{7}{40}$
• C. $\frac{7}{57}$
• D. $\frac{7}{64}$

Answer 1

Answer: (C)

$\frac{7}{57}$

Answer 2

Let $R_{large} = 2R$ be the radius of the larger sphere and $R_{small} = R$ be the radius of the smaller sphere. Let $\rho$ be the density of the solid sphere.

The mass of the larger sphere is $M_{large} = \rho \cdot \frac{4}{3}\pi R_{large}^3 = \rho \cdot \frac{4}{3}\pi (2R)^3 = \rho \cdot \frac{32}{3}\pi R^3$. The mass of the smaller sphere is $m_{small} = \rho \cdot \frac{4}{3}\pi R_{small}^3 = \rho \cdot \frac{4}{3}\pi R^3$.

The larger sphere is centered at the origin. The Y-axis passes through the center of the larger sphere. The moment of inertia of the larger solid sphere about the Y-axis is $I_{large} = \frac{2}{5}M_{large}R_{large}^2 = \frac{2}{5}(\rho \frac{32}{3}\pi R^3)(2R)^2 = \frac{2}{5}(\rho \frac{32}{3}\pi R^3)(4R^2) = \frac{256}{15}\pi \rho R^5$.

From the figure, we assume the smaller sphere is centered at a distance of $R$ from the origin along the X-axis. So, the center of the smaller sphere is at $(R, 0, 0)$. The Y-axis is the axis of rotation. The distance from the center of the smaller sphere to the Y-axis is $d = R$.

The moment of inertia of the smaller sphere about an axis passing through its center and parallel to the Y-axis is $I_{small, center} = \frac{2}{5}m_{small}R_{small}^2 = \frac{2}{5}(\rho \frac{4}{3}\pi R^3)R^2 = \frac{8}{15}\pi \rho R^5$. Using the parallel axis theorem, the moment of inertia of the smaller sphere about the Y-axis is $I_{small} = I_{small, center} + m_{small}d^2 = \frac{2}{5}m_{small}R^2 + m_{small}(R)^2 = \frac{7}{5}m_{small}R^2$. Substituting the value of $m_{small}$: $I_{small} = \frac{7}{5}(\rho \frac{4}{3}\pi R^3)R^2 = \frac{28}{15}\pi \rho R^5$.

The rest part of the sphere is the larger sphere with the smaller sphere removed. By the principle of superposition, the moment of inertia of the rest part about the Y-axis is the moment of inertia of the larger sphere minus the moment of inertia of the smaller sphere about the Y-axis. $I_{rest} = I_{large} - I_{small} = \frac{256}{15}\pi \rho R^5 - \frac{28}{15}\pi \rho R^5 = \frac{228}{15}\pi \rho R^5$.

The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is: Ratio = $\frac{I_{small}}{I_{rest}} = \frac{\frac{28}{15}\pi \rho R^5}{\frac{228}{15}\pi \rho R^5} = \frac{28}{228} = \frac{7}{57}$.

Therefore, the ratio is $\frac{7}{57}$.