Question

What is the moment of inertia of a solid sphere of density \(\rho\) and radius \(\mathrm{R}\) about its diameter? \(\\{\mathrm{A}\\}(105 / 176) \mathrm{R}^{5} \rho\) \(\\{\mathrm{B}\\}(176 / 105) \mathrm{R}^{5} \rho\) \(\\{C\\}(105 / 176) R^{2} \rho\) \(\\{\mathrm{D}\\}(176 / 105) \mathrm{R}^{2} \rho\)

Short answer

The correct answer for the moment of inertia of a solid sphere of density \(\rho\) and radius R about its diameter is \(\frac{8\pi}{5} R^5 \rho\), which is not present in the given options.

Step-by-step solution

Moment of Inertia Formula

Moment of inertia (I) is found by integrating the square of the distance from the axis of rotation (r) multiplied by the mass (dm) over the entire volume of the object. For a sphere, we can use spherical coordinates to find the following formula for the moment of inertia about the z-axis: \[I = \int_{V} r^2 dm\]

Express dm in terms of the density and volume element

We know the density \(\rho\) and we can express mass (dm) in terms of density and volume element (dV): \[dm = \rho dV\]

Convert volume element from Cartesian Coordinates to Spherical Coordinates

We have to change the volume element from Cartesian coordinates to spherical coordinates. In spherical coordinates, the volume element is given by: \[dV = r^2 \sin\theta dr d\theta d\phi\]

Substitute dm and dV in the Moment of Inertia Formula

Now we can substitute for dm and dV in the moment of inertia formula: \[I = \int_{V} r^2 (\rho r^2 \sin\theta dr d\theta d\phi)\]

Integrate over the volume of the sphere

Performing the integration over the volume of the sphere with limits for r: 0 to R, for θ: 0 to π, and for φ: 0 to 2π: \[I = \rho \int_{0}^{\mathrm{R}} \int_{0}^{\pi} \int_{0}^{2 \pi} r^4 \sin\theta dr d\theta d\phi\] \[I = \rho \left[ \int_{0}^{\mathrm{R}} r^4 dr \right] \left[ \int_{0}^{\pi} \sin\theta d\theta \right] \left[ \int_{0}^{2 \pi} d\phi \right]\] After performing these integrations, we get: \[I = \rho \left[ \frac{1}{5} R^5 \right] \left[ -\cos\theta \Big|_{0}^{\pi} \right] \left[ \phi \Big|_{0}^{2\pi} \right] \] \[I = \rho \left[ \frac{1}{5} R^5 \right] [2] [2\pi] \] \[I = \frac{8\pi}{5} R^5 \rho\]

Compare the result with the given options

Comparing the result we obtained with the given options, none of the options MATCHES with our moment of inertia calculation result. If this is an exercise, it's possible that there is a minor mistake in the options or the provided statement. However, it's important to note that our calculated correct answer is: \[I = \frac{8\pi}{5} R^5 \rho\] One might argue that the closest option to the answer is option A with the calculated answer having the same power in the radius (\(R^5\)) and the appearance of scalar factors (\(\frac{105}{176}\) in A) resembling the \(\frac{8\pi}{5}\), but there is a mismatch due to the missing \(\pi\) factor.