Question

The moment of inertia of a circular disc of mass M and radius R about an axis passing through the centre of mass is \[{I_o}\]. The moment of inertia of another circular disc of the same mass and thickness but half the density about the same axis is :
A. \[\dfrac{{{I_o}}}{8}\]
B. \[\dfrac{{{I_o}}}{4}\]
C. \[2{I_o}\]
D. \[4{I_o}\]

Answer 1

Hint:The moment of inertia of a circular disc of mass M due to rotation about an axis passing through its centre of mass is \[M{R^2}/2\]. The density of any solid body depends upon its mass and volume. From that, it is found that density and moment of inertia are inversely proportional to each other while other physical parameters are kept constant.

Formula Used:
The moment of inertia of a circular disc of mass M due to rotation about an axis passing through its centre of mass is,
\[{I_C} = M{R^2}/2\]. ---- (1)
Where \[{I_C}\] = moment of inertia of the circular disc about an axis passing through its centre of mass
M = mass of the circular disc
R = Radius of the disc
Density, \[\rho = \dfrac{{mass}}{{volume}}\]---- (2)

Complete step by step solution:
Given: A circular disc (disc 1) of mass M and radius R with moment of inertia of the circular disc about an axis passing through its centre of mass = \[{I_o}\].

Let the thickness of this disc be t and its density be \[\rho \]. A second circular disc 2 of the same mass M and thickness t but the half density of the previous disc. Let its density be \[\rho '\] and radius \[R'\] and its moment of inertia about an axis passing through its centre of mass be \[{I_o}'\].

For disc 1
Using equation (1), we get,
\[{I_o} = M{R^2}/2\]---- (3)
And \[\text{volume} = \text{Area} \times \text{Thickness}\]\[ = \pi {R^2}t\]
Therefore, using equation (2),
\[\rho = \dfrac{\text{mass}}{\text{volume}}\]= \[\dfrac{M}{{\pi {R^2}t}}\]
\[\Rightarrow M = \pi {R^2}t.\rho \]----(4)
Substituting (4) in (3), we get,
\[{I_o} = \pi {R^4}t.\rho /2\]---(5)

For disc 2, according to the question,
\[\rho = 2\rho '\]
Therefore, its density is
\[\rho ' = \dfrac{M}{{\pi {R^{'2}}t}} = \dfrac{\rho }{2}\]
So, \[M = \dfrac{{\pi tR{'^2}t\rho }}{2}\]---- (6)
Comparing equations (4) and (6),
\[{(R')^2} = {(2R)^2}\]---(7)
\[\Rightarrow I{'_o} = \pi {\left( {R'} \right)^4}t.\rho /2 = \pi {\left( {2R} \right)^4}t.\rho /2\]----(8)
Dividing equations (5) and (8), we get,
\[\therefore {I_o}' = 2{I_{{o_{}}}}\]

Hence option C is the correct answer.

Note: Moment of inertia of a particular solid depends on the dimensions of the solid as well as the axis of rotation. If the density is halved, the moment of inertia is also impacted. To determine the moment of inertia in different reference axes of rotation, we use parallel axis theorem and perpendicular axis theorem.