Question

Three objects having identical masses and radii are a disc, a ring and a solid sphere. Arrange the masses according to their increasing order of Moment of Inertia about their central axis.

Answer 1

When attempting questions like these , keep in mind the concepts of moment of inertia, and the concepts of ring, disc and sphere because many people get confused between the three circular masses. Keep in mind that there is a difference in solid sphere and hollow sphere as well.
Complete step-by-step solution:
Moment of inertia is basically the quantitative measure of the rotational inertia of the body, which is the opposition or resistance that the body exhibits to having its speed of rotation about an axis altered by the application of a torque, which is another name for turning force.
The axis for moment of inertia differs from one point to other, and the moment of inertia changes for central axis, for diameter et cetera. It is defined as the sum of the products obtained by multiplying the mass of each particle of matter of a given body by the square of its distance from the axis provided.
In this question, the axis we are provided with is the central axis.
Let’s find out the Moment of inertia of the Ring
Assuming the mass of ring to be $M$and radius to be $R$
\Rightarrow $$$$dm = (\dfrac{m}{{2\pi R}})dx
Next we calculate, $dl = (dm){R^2}$
\Rightarrow $$$$dl = \left[ {(\dfrac{m}{{2\pi R}})dx} \right]{R^2}
Substituting the values we get;
\Rightarrow $$$$dl = (\dfrac{m}{{2\pi R}}){R^2}dx
Using integration we get;

\Rightarrow I=(\dfrac{m}{2\pi })\left[ x \right]_{0}^{2\pi R} \\\ \Rightarrow I=(\dfrac{mR}{2\pi })\left[ 2\pi R-0 \right] \\\ \end{array}$$ And finally we get moment of inertia of Ring which is $$\Rightarrow $$$$I = M{R^2}$$ …… (Equation 1) Similarly, by establishing the relation for surface mass density$$\sigma$$which is defined as mass per unit surface area and assuming the disc to be uniform we can find out the moment of inertia of the disc as well. The moment of inertia of disc is; $$\Rightarrow $$$$I = \dfrac{1}{2}M{R^2}$$ …… (Equation 2) For solid sphere, obtaining moment of inertia can be done from two ways; 1\. Take the solid sphere and slice it into infinitesimally thin solid cylinders 2\. Then sum up the moments of extremely small thin discs in the required given axis from left to right. Like we did in cases of Ring and disc, we can derive the formula for moment of inertia of solid sphere in a similar way. The moment of inertia of solid sphere is; $$\Rightarrow $$$$I = \dfrac{2}{5}M{R^2}$$…… (Equation 3) Comparing (1), (2) and (3), assuming the masses and radius to be identical, we can conclude that moment of inertia of Solid sphere < disc < ring **Note:** Hollow sphere or a spherical shell has a moment of inertia which is determined by the formula, $$I = M{R^2}$$which is similar to the moment of inertia of the ring. Moment of inertia of rod whose axis passing its centre is given by $$I = \dfrac{1}{{12}}M{L^2}$$but if axis passes through end then moment of inertia changes into$$I = \dfrac{1}{3}M{L^2}$$.