# Moment of inertia

1 vote
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The moment of inertia of hollow sphere (mass M) of inner radius R and outer radius 2R having material of uniform density about a diametric axis is---

I tried to divide the problem into two parts the moment of inertia of the sphere of radius 2R($I_1$) and the moment of inertia of smaller sphere of radius R($I_2$) Then moment of body=$I_1-I_2$

But I am not getting the correct answer

Here is the solution I found but couldn't getting how they used $I=1/2 M R^2$.it is not the moment of inertia of a sphere. edited Oct 12, 2017

1 vote

Your approach should give you the correct answer. The official solution is incorrect.

For a solid sphere the moment of inertia about the center is $I=\frac25 MR^2$, not $\frac12 MR^2$ as the solution implies. The latter is the MI for a disc or solid cylinder. There are formulae for each of these at the website https://en.wikipedia.org/wiki/List_of_moments_of_inertia.

The calculation of the masses of the two spheres is correct. More simply, if the mass of the solid sphere of radius $R$ is $m$ then by proportion the mass of a solid sphere of radius $2R$ is $2^3m=8m$ and the mass of the hollow sphere is $M=8m-m=7m$.

The moment of inertia of the hollow sphere is the difference for the two solid spheres :
$I=I_{2R}-I_R=\frac25(8m)(2R)^2-\frac25mR^2=\frac25(32-1)mR^2=\frac25 31 \frac{M}{7} R^2=\frac{62}{35}MR^2$.

Using the formula for a hollow sphere we get the same answer :
$I=\frac25 M (\frac{R_2^5-R_1^5}{R_2^3-R_1^3})=\frac25 M (\frac{32-1}{8-1})\frac{R^5}{R^3}=\frac{62}{35} MR^2$.

answered Oct 13, 2017 by (28,466 points)
selected Oct 15, 2017 by Navin
Thank you that means the solution is wrong.