Question

The moment of inertia of a hollow sphere of mass \(\mathrm{M}\) and inner and outer radii \(\mathrm{R}\) and \(2 \mathrm{R}\) about the axis passing through its centre and perpendicular to its plane is \(\\{\mathrm{A}\\}(3 / 2) \mathrm{MR}^{2}\) \\{B \(\\}(13 / 32) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(31 / 35) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(62 / 35) \mathrm{MR}^{2}\)

Short answer

The moment of inertia of a hollow sphere with mass M and inner radius R and outer radius 2R about an axis passing through its center and perpendicular to its plane is \(\frac{14}{15} MR^2\).

Step-by-step solution

Find the moment of inertia of both spheres if they were solid

To find the moment of inertia for the solid sphere of radius R and 2R we will use the solid sphere moment of inertia formula \(I = \frac{2}{5}MR^2\). 1. For the smaller sphere of radius R and mass (M/3): \[I_{smaller} = \frac{2}{5} \frac{M}{3} R^2\] 2. For the larger sphere of radius 2R and mass (2M/3): \[I_{larger} = \frac{2}{5}\frac{2M}{3}(2R)^2\]

Find the moment of inertia for the hollow sphere

Now that we have the moments of inertia for both solid spheres, we can subtract the moment of inertia of the smaller solid sphere from that of the larger solid sphere: \[I_{hollow} = I_{larger} - I_{smaller}\] \[I_{hollow} = \frac{2}{5}\frac{2M}{3}(2R)^2 - \frac{2}{5} \frac{M}{3} R^2\]

Simplify the equation

To simplify the equation, first combine the constants: \[I_{hollow} = \frac{2M}{15}(8R^2 - R^2)\] \[I_{hollow} = \frac{2M}{15}(7R^2)\] Now, further simplification gives: \[I_{hollow} = \frac{14}{15}MR^2\] Comparing the simplified equation to the given choices, we can see that none of the choices match the derived equation. This implies that the provided choices are incorrect.