Question

The moment of inertia of a disc of uniform thickness of mass M having internal and external radii r and R respectively about an axis through its centre and perpendicular to the plane of the disc is
A. $\dfrac{{M\left( {{R^2} - {r^2}} \right)}}{4}$
B. $\dfrac{{M\left( {{R^2} - {r^2}} \right)}}{2}$
C. $\dfrac{{M\left( {{R^2} + {r^2}} \right)}}{2}$
D. $\dfrac{{M\left( {{R^2} + {r^2}} \right)}}{4}$

Answer 1

- Hint: In this question, we will use the concept of moment of inertia of a body. We should know that moment of inertia of a rigid body about a fixed axis is defined as the sum of products of the masses of the particles constituting the body and the squares of their respective distances from the axis of rotation, i.e. $I = {m_1}{r_1}^2 + {m_2}{r_2}^2 + {m_3}{r_3}^2 + ........$
$I = \sum\limits_{i = 1}^n {{m_i}{r_i}^2}$.

Complete step-by-step solution -

Given that, M is the mass of the disc of outer radius R and inner radius r.
Then surface mass density = $\sigma = \dfrac{{mass}}{{area}}$.
$\sigma = \dfrac{M}{{\pi \left( {{R^2} - {r^2}} \right)}}$.
Mass of elementary ring of radius $x$ and thickness $dx$
$= \dfrac{M}{{\pi \left( {{R^2} - {r^2}} \right)}} \times 2\pi x{\text{ }}dx \\\ = \dfrac{{2Mx{\text{ }}dx}}{{\left( {{R^2} - {r^2}} \right)}} \\\$
Moment of inertia of ring about an axis passing through the centre and perpendicular to the plane is
$\Rightarrow dI = \dfrac{{2Mx{\text{ }}dx}}{{\left( {{R^2} - {r^2}} \right)}}{x^2} \\\ \Rightarrow dI = \dfrac{{2M{x^3}{\text{ }}dx}}{{\left( {{R^2} - {r^2}} \right)}} \\\$
Integrating both sides from R to r, we get
$\Rightarrow I = \int\limits_r^R {\dfrac{{2M{x^3}{\text{ }}dx}}{{\left( {{R^2} - {r^2}} \right)}}}$
$\Rightarrow I = \dfrac{{2M}}{{\left( {{R^2} - {r^2}} \right)}}\left[ {\dfrac{{{R^4} - {r^4}}}{4}} \right]$
Using identity [ $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)(a + b)$ , $\left( {{R^4} - {r^4}} \right)$ can also be written as $\left( {{R^2} - {r^2}} \right)\left( {{R^2} + {r^2}} \right)$ ], we get
$\Rightarrow I = \dfrac{M}{{\left( {{R^2} - {r^2}} \right)}}\left[ {\dfrac{{\left( {{R^2} - {r^2}} \right)\left( {{R^2} + {r^2}} \right)}}{2}} \right] \\\ \Rightarrow I = \dfrac{{M\left( {{R^2} + {r^2}} \right)}}{2} \\\$
Hence, the correct answer is option(C).

Note: In this type of questions, we should have some basic knowledge of the moment of inertia of a body. Then we will find out the surface mass density and then we will find the mass of the elementary ring. After that we will put this value in the formula of inertia and then we will integrate it with limit R to r. solving it step by step, we will get the required answer.