Question

A thin uniform wire of mass 'm' and linear density '\(\rho\)' is bent in the form of a circular ring. The moment of inertia of the ring about a tangent parallel to its diameter is

• A. \( \frac{3m^3}{8\pi^2\rho^2} \)
• B. \( \frac{8m^3}{3\pi^2\rho^2} \)
• C. \( \frac{8\pi^2m^3}{3\rho^2} \)
• D. \( \frac{3\pi^2m^3}{8\rho^2} \)

Hint:

The Parallel Axis Theorem is crucial for finding the moment of inertia about any axis, given the moment of inertia about a parallel axis through the center of mass. Remember the formula: \(I = I_{cm} + Md^2\).

Answer 1

The Correct Option is A

Let the radius of the circular ring be R.
The total mass m, linear density \(\rho\), and length L (circumference) are related by \( m = \rho L \).
The length of the wire is the circumference of the ring, \( L = 2\pi R \).
So, \( m = \rho (2\pi R) \). We can express the radius R in terms of m and \(\rho\): \( R = \frac{m}{2\pi\rho} \).
The moment of inertia of a ring about its diameter is \( I_{dia} = \frac{1}{2}MR^2 \). In our case, \( I_{dia} = \frac{1}{2}mR^2 \).
We need to find the moment of inertia about a tangent that is parallel to a diameter.
By the Parallel Axis Theorem, the moment of inertia \(I\) about an axis parallel to an axis through the center of mass is \( I = I_{cm} + Md^2 \), where \(I_{cm}\) is the moment of inertia about the center of mass axis and d is the distance between the two axes.
Here, the axis through the center of mass is the diameter. So, \( I_{cm} = I_{dia} = \frac{1}{2}mR^2 \).
The parallel axis is a tangent, so the distance \(d\) between the diameter and the tangent is the radius R.
Therefore, the moment of inertia about the tangent is \( I_{tangent} = I_{dia} + mR^2 = \frac{1}{2}mR^2 + mR^2 = \frac{3}{2}mR^2 \).
Now, we substitute the expression for R in terms of m and \(\rho\).
\( I_{tangent} = \frac{3}{2}m \left( \frac{m}{2\pi\rho} \right)^2 = \frac{3}{2}m \left( \frac{m^2}{4\pi^2\rho^2} \right) \).
\( I_{tangent} = \frac{3m^3}{8\pi^2\rho^2} \).