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

Concept:
For a physical pendulum performing small oscillations, \[ T = 2\pi\sqrt{\frac{I}{mgd}} \] where \(I\) = moment of inertia about the axis of rotation, \(d\) = distance between pivot and centre of mass. For a ring, the moment of inertia about its centre is \[ I_{cm} = mR^2 \] Using the parallel axis theorem, \[ I = I_{cm} + md^2 \] Step 1: Find distance of centre of mass from pivot. Since the axis is tangent to the ring, \[ d = R \] Step 2: Find moment of inertia about the tangent axis. \[ I = I_{cm} + mR^2 \] \[ I = mR^2 + mR^2 \] \[ I = 2mR^2 \] Step 3: Substitute into time period formula. \[ T = 2\pi\sqrt{\frac{2mR^2}{mgR}} \] \[ T = 2\pi\sqrt{\frac{2R}{g}} \] \[ T = 2\pi\sqrt{\frac{3R}{2g}} \]