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Show that the moment of inertia of a uniform solid sphere
Chapter 3, Problem 3.32(choose chapter or problem)
Show that the moment of inertia of a uniform solid sphere rotating about a diameter is \(\frac{2}{5} M R^{2}\). The sum (3.31) must be replaced by an integral, which is easiest in spherical polar coordinates, with the axis of rotation taken to be the \(z\) axis. The element of volume is \(dV=r^2dr\sin\theta\ d\theta\ d\phi\). (Spherical polar coordinates are defined in Section 4.8. If you are not already familiar with these coordinates, you should probably not try this problem yet.)
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QUESTION:
Show that the moment of inertia of a uniform solid sphere rotating about a diameter is \(\frac{2}{5} M R^{2}\). The sum (3.31) must be replaced by an integral, which is easiest in spherical polar coordinates, with the axis of rotation taken to be the \(z\) axis. The element of volume is \(dV=r^2dr\sin\theta\ d\theta\ d\phi\). (Spherical polar coordinates are defined in Section 4.8. If you are not already familiar with these coordinates, you should probably not try this problem yet.)
ANSWER:Step 1 of 3
The expression for the density of the sphere is given by,
\(\rho=\frac{M}{V} \dots \dots (1)\)
Here, M is the mass of the sphere, and V is volume of the sphere.
The volume of sphere is given by.
\(V=\frac{4}{3} \pi R^{3} \dots \dots (2)\)
Here, R is radius of sphere.
Substitute all the values in equation (1).
\(\begin{aligned}
\rho & =\frac{M}{\frac{4}{3} \pi R^{3}} \\
& =\frac{3 M}{4 \pi R^{3}}
\end{aligned} \dots \dots (3)\)
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