Question

If the moment of inertia of a thin circular ring about an axis passing through its edge and perpendicular to its plane is I, then the moment of inertia of the ring about its diameter is

• A. $I/4$
• B. $4I$
• C. $I/2$
• D. $2I$

Hint:

Standard M.I. for Ring: Center, perp: $MR^2$. Diameter: $MR^2/2$. Tangent, in plane: $3/2 MR^2$. Tangent, perp: $2MR^2$.

Answer 1

The Correct Option is A

Step 1: Relate 'I' to Mass and Radius:
Moment of Inertia of a ring about an axis passing through the center and perpendicular to the plane is $I_{CM} = MR^2$. Using the Parallel Axis Theorem, the moment of inertia about an axis through the edge (tangent) and perpendicular to the plane is: \[ I_{edge} = I_{CM} + MR^2 = MR^2 + MR^2 = 2MR^2 \] Given $I_{edge} = I$. So, $I = 2MR^2 \implies MR^2 = \frac{I}{2}$.
Step 2: Calculate Moment of Inertia about Diameter:
Using the Perpendicular Axis Theorem ($I_z = I_x + I_y$), for a ring lying in the x-y plane: $I_z = I_{CM} = MR^2$. By symmetry, $I_x = I_y = I_{dia}$. So, $MR^2 = 2 I_{dia}$. \[ I_{dia} = \frac{1}{2} MR^2 \]
Step 3: Express in terms of I:
Substitute $MR^2 = I/2$ into the expression for $I_{dia}$: \[ I_{dia} = \frac{1}{2} \left( \frac{I}{2} \right) = \frac{I}{4} \]