Question

Find the moment of inertia of a uniform ring of mass \(M\) and radius \(R\) about a tangent perpendicular to its plane.

• A. \(MR^2\)
• B. \(2MR^2\)
• C. \(3MR^2\)
• D. \(4MR^2\)

Hint:

Parallel axis theorem: \[ I = I_{cm} + Md^2 \] For a ring, \(I_{cm} = MR^2\) about an axis perpendicular to its plane.

Answer 1

The Correct Option is B

Concept: For a uniform ring, \[ I_{\text{centre}} = MR^2 \] about an axis perpendicular to its plane through the centre. Using the parallel axis theorem: \[ I = I_{cm} + Md^2 \] where \(d\) is the distance between the axes.

Step 1: Identify the required axis. The required axis is a tangent perpendicular to the plane of the ring. Distance between the centre and tangent axis: \[ d = R \]

Step 2: Apply the parallel axis theorem. \[ I = MR^2 + M(R)^2 \]

Step 3: Simplify the expression. \[ I = MR^2 + MR^2 \] \[ I = 2MR^2 \] \[ \boxed{2MR^2} \]