Question

Consider a ring of radius \(R\) which rotates about a horizontal axis as shown (axis is tangent to the ring). Find the time period of small oscillations.

• A. \(2\pi\sqrt{\frac{5R}{g}}\)
• B. \(2\pi\sqrt{\frac{R}{2g}}\)
• C. \(2\pi\sqrt{\frac{3R}{2g}}\)
• D. \(2\pi\sqrt{\frac{R}{g}}\)

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The given setup represents a physical pendulum. A ring of mass \(M\) and radius \(R\) is oscillating about a horizontal axis that is tangent to its topmost point and lies in the plane of the ring.
Step 2: Key Formula or Approach:
The time period \(T\) for small oscillations of a physical pendulum is given by:
\[ T = 2\pi \sqrt{\frac{I}{Mgd}} \]
where:
\(I\) is the moment of inertia of the body about the axis of rotation.
\(d\) is the distance between the center of mass and the axis of rotation.
Step 3: Detailed Explanation:
For a thin ring of mass \(M\) and radius \(R\), the moment of inertia about a diameter (center of mass axis in the plane) is \(I_{cm} = \frac{1}{2}MR^{2}\).
The rotation axis is a horizontal tangent in the plane of the ring, so the distance from the center of mass to this axis is \(d = R\).
Applying the parallel axis theorem, the moment of inertia \(I\) about the tangent axis is:
\[ I = I_{cm} + Md^{2} = \frac{1}{2}MR^{2} + M(R)^{2} = \frac{3}{2}MR^{2} \]
Now, substituting the values into the time period formula:
\[ T = 2\pi \sqrt{\frac{\frac{3}{2}MR^{2}}{MgR}} \]
\[ T = 2\pi \sqrt{\frac{3R}{2g}} \]
Step 4: Final Answer:
The time period of small oscillations is \(2\pi \sqrt{\frac{3R}{2g}}\), which matches option (C).