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

Step 1: Understanding the Question:
We need to find the rotational inertia of a ring about an axis that touches the outer edge (tangent) and is oriented perpendicular to the ring's flat surface.
Step 2: Key Formula or Approach:
1. Moment of inertia of a ring about an axis through center and perpendicular to plane: \(I_{cm} = MR^2\).
2. Parallel Axis Theorem: \(I = I_{cm} + Mh^2\), where \(h\) is the distance between the two parallel axes.
Step 3: Detailed Explanation:
The distance from the center of the ring to the tangent is equal to the radius, \(R\).
So, \(h = R\).
Applying the theorem:
\[ I = I_{cm} + M(R)^2 \]
\[ I = MR^2 + MR^2 \]
\[ I = 2MR^2 \]
Step 4: Final Answer:
The moment of inertia about the specified tangent is \(2MR^2\).