Question

The moment of inertia of a hollow cylinder of mass M and inner radius R1 and outer radius R2 about its central axis is ___

Answer 1

by the principle of superposition we can say inertia of hollow cylinder is subtraction of moment of inertia of full solid cylinder and inner solid cylinder

$I = I_{out} - I_{in}$

now for the mass of the two parts we can use

$M_{out} = \frac{M}{\pi(R_2^2 - R_1^2)}*\pi R_2^2$

$M_{out} = \frac{M}{(R_2^2 - R_1^2)}*R_2^2$

similarly mass of inner solid cylinder

$M_{in} = \frac{M}{(R_2^2 - R_1^2)}*R_1^2$

Now moment of inertia is given as

$I = \frac{M_{out} R_2^2}{2} - \frac{M_{in} R_1^2}{2}$

$I = \frac{\frac{M}{(R_2^2 - R_1^2)}*R_2^4}{2} - \frac{ \frac{M}{(R_2^2 - R_1^2)}*R_1^4}{2}$

$I = \frac{\frac{M}{(R_2^2 - R_1^2)}}{2}(R_2^4 - R_1^4)$

$I = \frac{M(R_2^2 + R_1^2)}{2}$

so above is the moment of inertia of the hollow cylinder