Question

Moment of inertia of a uniform circular disc about a diameter is $I$. Its moment of inertia about an axis perpendicular to its plane and passing through a plane on its rim will be:
A. $5I$
B. $6I$
C. $3I$
D. $4I$

Answer 1

To solve this problem, we will first see the moment of inertia of a uniform circular disc having mass M and radius R about an axis passing through the center. Then we will use the perpendicular axis theorem to determine the moment of inertia about the diameter of the disc and compare it with the given moment of inertia. After that by using parallel axis theorem, we will finally determine the moment of inertia about an axis perpendicular to its plane and passing through a plane on its rim.

Formula used:
${I_C} = \dfrac{1}{2}M{R^2}$,
where, ${I_C}$ is the moment of inertia of a uniform circular disc about an axis passing through the center, $M$ is mass of the disc and $R$ is the radius of the disc.

Complete step by step answer:
We know that see the moment of inertia of a uniform circular disc having mass M and radius R about an axis passing through the center is:
${I_C} = \dfrac{1}{2}M{R^2}$
Let the moment of inertia about the diameter of the disc be ${I_D}$.Now, according to perpendicular axis theorem, the moment of inertia about the diameter of the disc can be given by:
${I_D} = \dfrac{{{I_C}}}{2} \\\ \Rightarrow {I_D} = \dfrac{{\dfrac{1}{2}M{R^2}}}{2} \\\ \Rightarrow {I_D} = \dfrac{1}{4}M{R^2}$
But, in the question, we are given that the moment of inertia of a uniform circular disc about a diameter is $I$.
I = \dfrac{1}{4}M{R^2} \\\ \Rightarrow M{R^2} = 4I \\\
Now, to find the moment of inertia about an axis perpendicular to its plane and passing through a plane on its rim, we will use the parallel axis theorem.
Let the required moment of inertia be ${I_R}$.
By using parallel axis theorem, we can say that
${I_R} = {I_C} + M{R^2}$
Here, we will put ${I_C} = \dfrac{1}{2}M{R^2}$
\Rightarrow {I_R} = \dfrac{1}{2}M{R^2} + M{R^2} \\\ \Rightarrow {I_R} = \dfrac{3}{2}M{R^2} \\\
We have determined that $M{R^2} = 4I$
${I_R} = \dfrac{3}{2} \times 4I \\\ \therefore {I_R}= 6I$
Thus, the moment of inertia of the disc about an axis perpendicular to its plane and passing through a plane on its rim is $6I$.

Hence, option B is the right answer.

Note: We have used two theorems: perpendicular axis theorem and parallel axis theorem. Generally, the perpendicular axis theorem is used when we need to determine the moment of inertia about the third axis. Whereas, the parallel axis theorem is applied when we need to find the moment of inertia of the area of a rigid body whose axis is parallel to the axis of the known moment body and it is through the center of gravity of the object.