Question

A $25\text{kg}$ uniform solid sphere with a $20\text{cm}$ radius is suspended by a vertical wire such that the point of suspension is vertically above the centre of the sphere. A torque of $0.10\text{N-m}$ is required to keep the sphere in an angular position of $1.0\text{rad}$ with respect to equilibrium. If the sphere is then released, its time period of the oscillation will be :

• A. $3\pi$ second
• B. $\pi$ second
• C. $2\pi$ second
• D. $4\pi$ second

Answer 1

The correct option is D $4\pi$ second

For angular SHM
$T_{\text{restoring}}=C\theta$
$0.10\text{N-m}=-C\left[1\text{rad}\right]$
$|C|=0.1$
Now time period
$T=2\pi \sqrt{\frac{I}{C}}$
$I=\text{M.O.I of solid}-\text{sphere}=\frac{2}{5}m{R}^2$
$T=2\pi \sqrt{\frac{\frac{2}{5}\times 25\times (0.2{)}^2}{0.1}}$
$T=4\pi$ second
Hence, option (D) is correct.