Question

The moment of inertia of a hollow cylinder of mass $M$ and inner radius $R_1$ and outer radius $R_2$ about its central axis is
(A). ${\displaystyle \frac{1}{2}}M(R_2^2-R_1^2)$
(B). ${\displaystyle \frac{1}{2}}M(R_1^2+R_2^2)$
(C). ${\displaystyle \frac{1}{2}}M(R_1^2-R_2^2)$
(D). ${\displaystyle \frac{1}{2}}M(R_2-R_1{)}^2$

Answer 1

Hint: You can start the solution by calculating the mass per unit cross section area. Then divide the cylinder into an inner and outer cylinder. Then find the mass of the inner and outer cylinder by using the equation ${\displaystyle \frac{M}{\pi (R_2^2-R_1^2)}}\times \pi R^2$ . Then use the equation $I={\displaystyle \frac{M{R}_{}^2}{2}}$ to find the moment of inertia of the inner and the outer cylinder. Then calculate the difference between the moment of the inertia of the outer and inner cylinder to reach the solution.

Complete step-by-step answer:
Here we are given a hollow cylinder with a mass $M$ and inner radius $R_1$ and outer radius $R_2$ .
So the total area of cross section of the cylinder is
Area $=\pi (R_2^2-R_1^2)$
The mass $M$ of the hollow cylinder is distributed over a cross section area of $\pi (R_2^2-R_1^2)$ .
So the mass per unit cross section area is ${\displaystyle \frac{M}{\pi (R_2^2-R_1^2)}}$
In this problem we have a hollow cylinder, let’s divide it into two parts: a bigger cylinder with a radius $R_2$ and a smaller cylinder with a radius $R_1$ .
The mass of the outer cylinder is
 $M_{outer}={\displaystyle \frac{M}{\pi (R_2^2-R_1^2)}}\times \pi R_2^2$
 $M_{outer}={\displaystyle \frac{M}{(R_2^2-R_1^2)}}\times R_2^2$
Similarly the mass of inner cylinder is
 $M_{inner}={\displaystyle \frac{M}{\pi (R_2^2-R_1^2)}}\times \pi R_1^2$
 $M_{inner}={\displaystyle \frac{M}{(R_2^2-R_1^2)}}\times R_1^2$
The moment of inertia of the outer cylinder is
 $I_{outer}={\displaystyle \frac{M_{outer}{R}_2^2}{2}}$
 $I_{outer}={\displaystyle \frac{\left({\displaystyle \frac{M}{R_2^2-R_1^2}}\times R_2^2\right)R_2^2}{2}}$
 $I_{outer}={\displaystyle \frac{M{R}_2^4}{2(R_2^2-R_1^2)}}$
The moment of inertia of the inner cylinder is
 $I_{inner}={\displaystyle \frac{M_{inner}{R}_1^2}{2}}$
 $I_{inner}={\displaystyle \frac{\left({\displaystyle \frac{M}{R_2^2-R_1^2}}\times R_1^2\right)R_1^2}{2}}$
 $I_{inner}={\displaystyle \frac{M{R}_1^4}{2(R_2^2-R_1^2)}}$
The net moment of inertia is the difference in the moment of inertia of the outer cylinder and the movement of inertia of the inner cylinder
 $I_{net}=I_{outer}-I_{inner}$
$$I_{net}={\displaystyle \frac{M{R}_2^4}{2(R_2^2-R_1^2)}}-{\displaystyle \frac{M{R}_1^4}{2(R_2^2-R_1^2)}}$$
$$I_{net}={\displaystyle \frac{M(R_2^4-R_1^4)}{2(R_2^2-R_1^2)}}$$
$$I_{net}={\displaystyle \frac{M(R_2^2-R_1^2)(R_2^2+R_1^2)}{2(R_2^2-R_1^2)}}$$$$[\because A^2-B^2=(A-B)(A+B)]$$
$$I_{net}={\displaystyle \frac{M(R_2^2+R_1^2)}{2}}$$
Hence, option B is the correct choice.

Note: In this problem we divided the hollow cylinder into an outer bigger cylinder and smaller cylinder, found out the moment of inertia of outer cylinder and inner cylinder individually. In this question we will not use the value of moment of inertia of a cylinder around its central diameter i.e. $${\displaystyle \frac{1}{4}}M{R}^2+{\displaystyle \frac{1}{12}}M{L}^2$$.