Question

A uniform disc of radius $R$ and mass $M$ is free to oscillate about the axis A as shown in the figure. For small oscillations the time period is _______.
(g is acceleration due to gravity)

• A. $2\pi \sqrt{\frac{5R}{4g}}$
• B. $2\pi \sqrt{\frac{2R}{3g}}$
• C. $2\pi \sqrt{\frac{3R}{2g}}$
• D. $2\pi \sqrt{\frac{3R}{g}}$

Answer 1

The Correct Option is C

Step 1: Use physical pendulum formula
\[ T=2\pi \sqrt{\frac{I}{Mgd}} \] Step 2: Find moment of inertia
For disc about center: \[ I_{CM}=\frac{1}{2}MR^2 \] Using parallel axis theorem: \[ I=I_{CM}+MR^2 \] \[ =\frac{1}{2}MR^2+MR^2 \] \[ =\frac{3}{2}MR^2 \] Distance of center from pivot: \[ d=R \] Step 3: Substitute
\[ T=2\pi \sqrt{\frac{\frac{3}{2}MR^2}{MgR}} \] \[ =2\pi \sqrt{\frac{3R}{2g}} \] Final Answer: \[ \boxed{2\pi \sqrt{\frac{3R}{2g}}} \]