Question

The moment of inertia of a circular ring is $I$ about an axis perpendicular to its plane and passing through its center. About an axis passing through the tangent of the ring in its plane, its moment of inertia is

Answer 1

Step 1: Given data

1. The moment of inertia of a circular ring is $I$ about an axis perpendicular to its plane and passing through its center.
2. In the figure, a moment of inertia of the ring through the axis perpendicular to the plane is $I$.
3. Moment of inertia of ring, $I=M{R}^2$.
4. We have to find a moment of inertia of the ring about an axis passing through the tangent of the ring

Step 2: Formula and theorem used

Perpendicular axis theorem- The moment of inertia of a body about an axis perpendicular to a plane is equal to the sum of the moment of inertia about any two perpendicular axes in the plane of the body.

$I_Z=I_X+I_Y$

where $I_Z$, $I_X$ and $I_Y$ are the three perpendicular axes.

Parallel axis theorem- The moment of inertia of a body about an axis parallel to the axis passing through the center of it is equal to the sum of the moment of inertia of the body through the center and the product of the mass of the body and the square of the distance between them.

'$I\text{'}=I+M{R}^2$

where $I$ is the moment of inertia through any axis passing through the center of the body and $I\text{'}$ is the moment of inertia of an axis parallel to $I$, $M$ is mass of the body, and $R$ is the distance between these two axis

Step 3: Calculating Moment of Inertia

Here, by perpendicular axis theorem, we can say that,

moment of inertia about the axis passing through the center and perpendicular to the plane,$I$ = moment of inertia about any two perpendicular axes passing through the diameter($I\text{'}$)

$\therefore I=2I\text{'}$

where $I\text{'}$ is the moment of inertia of the ring through any diametrical axis and $I$ is the moment of inertia about the axis passing through the center and perpendicular to the plane.

$I\text{'}=\frac{I}{2}$

Let $I"$ be the moment of inertia of the ring about an axis passing through the tangent in its plane, then

by parallel axis theorem,

$I"=I\text{'}+M{R}^2$

$I"=\frac{I}{2}+M{R}^2$

We know, a moment of inertia of a ring is $M{R}^2$.

$$\begin{gathered} \because I=M{R}^2 \\ \Rightarrow I"=\frac{M{R}^2}{2}+M{R}^2 \\ \Rightarrow I"=\frac{3}{2}M{R}^2 \\ \Rightarrow I"=\frac{3}{2}I \end{gathered}$$

The moment of inertia of a ring about an axis passing through the tangent of the ring in its plane is $\frac{3}{2}I$