Question

If a solid sphere and solid cylinder of same mass and density rotate about their own axis the moment of inertia will be greater for
A. Solid sphere
B. Solid Cylinder
C. Both a and b
D. Equal both

Answer 1

Hint: We studied the moment of inertia in the chapter of rotational motion of the body. The moment of inertia is the quantity which decides the amount of torque needed for angular acceleration around the rotational axis. Simply it is the sum of the product of mass of all the particles with the square of distance from the axis of rotation. It is expressed as :$I = M{R^2}$ where $M$ is the mass of the body and $R$ is the distance from the axis of rotation.

Complete step by step answer:
The unit of moment of inertia is $kg{m^2}$ and its dimensional formula is $\left[ {M{L^2}{T^0}} \right]$.
In the rotational motion, the role of moment of inertia is the same as the mass in the linear motion. It depends on the distribution of mass relative to the axis of rotation, the density of the body, shapes and size of body. For different shapes of body it is different. Because its depends on the axis of rotation also then, for the same mass and same density of the bodies it varies

For the solid cylinder of mass $M$ and radius $R$, moment of inertia about the axis passing through its centre and parallel to its height is : ${I_C} = \dfrac{1}{2}M{R^2} = 0.5M{R^2}$.

For the solid sphere of same mass and same radius as of solid cylinder, the moment of inertia about the axis passing through its diameter and its centre is as follows; ${I_S} = \dfrac{2}{5}M{R^2}$$ = 0.4M{R^2}$

So from above discussion, we know that If a solid sphere and solid cylinder of same mass and density rotate about their own axis the moment of inertia will be greater for the solid cylinder.

So, the correct answer is “Option B”.

Note:
For calculation of moment of inertia about any given axis there are two theorem:
Parallel axis theorem and perpendicular theorem. Moment of inertia for some object around their geometrical axis: for disc, $I = \dfrac{1}{2}M{R^2}$, for rod about its centre, $I = \dfrac{1}{{12}}M{L^2}$ where $L$ is the length of rod, for solid sphere, $I = \dfrac{2}{5}M{R^2}$ where $R$ is the radius of sphere etc.