Question

Moment of inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is \( I \). If the same rod is bent in the form of a ring, its moment of inertia about the diameter is \( I' \). If \( I_1 = I_2 \), then the value of \( x \) is

• A. \( 2x^2 \)
• B. \( \frac{3}{5} x^2 \)
• C. \( \frac{3}{2} x^2 \)
• D. \( \frac{5}{3} x^2 \)

Hint:

The moment of inertia for a ring differs from that of a rod due to their shape and distribution of mass. The parallel axis theorem allows calculating the moment of inertia for different axes.

Answer 1

The Correct Option is C

Step 1: Moment of inertia for the rod.
The moment of inertia \( I \) of a uniform thin rod of mass \( m \) and length \( L \) about an axis passing through its center and perpendicular to its length is:
\[ I = \frac{1}{12} m L^2. \]

Step 2: Moment of inertia for a ring.
For a ring of mass \( m \) and radius \( R \), the moment of inertia about an axis passing through its center and perpendicular to the plane of the ring is:
\[ I' = m R^2. \]
When the thin rod is bent into a ring, the radius \( R \) of the ring is the same as the length of the rod, i.e., \( R = L \).

Step 3: Moment of inertia about the diameter.
The moment of inertia \( I_{\text{diameter}} \) of the ring about its diameter can be obtained by using the parallel axis theorem. The moment of inertia about the center of mass is \( I' = m R^2 \), and the distance between the center and the diameter is \( R \), so:
\[ I_{\text{diameter}} = I' - m \cdot R^2 = m R^2 - m \cdot R^2 = 0. \]

Step 4: Conclusion.
Thus, the correct value of \( x \) is \( \boxed{\frac{3}{2} x^2} \).