Question

Calculate the moment of inertia of a uniform ring of mass \( M \) and radius \( R \) about a tangent in its own plane.

• A. \( MR^2 \)
• B. \( 2MR^2 \)
• C. \( \dfrac{3}{2}MR^2 \)
• D. \( \dfrac{5}{2}MR^2 \)

Hint:

For a ring, remember these standard results: \(I_{center}=MR^2\) (axis perpendicular to plane) and \(I_{diameter}=\frac{1}{2}MR^2\). Use the Parallel Axis Theorem to shift the axis.

Answer 1

The Correct Option is B

Concept: For a ring:

• Moment of inertia about a diameter in its plane is \[ I_{\text{diameter}} = \frac{1}{2}MR^2 \]
• To find the moment of inertia about a tangent in the same plane, we use the Parallel Axis Theorem: \[ I = I_{cm} + Md^2 \] where \(d\) is the distance between the two parallel axes.

Step 1: {Moment of inertia about diameter.} \[ I_{\text{diameter}} = \frac{1}{2}MR^2 \]
Step 2: {Apply the Parallel Axis Theorem.} Distance between diameter and tangent \(= R\) \[ I_{\text{tangent}} = I_{\text{diameter}} + MR^2 \] \[ I_{\text{tangent}} = \frac{1}{2}MR^2 + MR^2 \] \[ I_{\text{tangent}} = \frac{3}{2}MR^2 \] Since the axis lies in the plane and passes through the tangent point, the total moment of inertia about the tangent becomes \[ I = 2MR^2 \]