Question

A $25kg$ uniform solid sphere with a $20cm$ radius is suspended by a vertical wire such that the point of suspension is vertically above the centre of the sphere. A torque of $0.10N-m$ is required to rotate the sphere through an angle of $1.0rad$ and then the orientation is maintained. If the sphere is then released, its time period of the oscillation will be _____.

Answer 1

$\tau =k\theta \Rightarrow k=\frac{\tau }{\theta }=\frac{0.10}{1}I=\frac{2}{5}M{R}^2=\frac{2}{5}\times 25\times \frac{1}{25}=\frac{2}{5}kg{m}^2$

Now, $T=2\pi \sqrt{\frac{I}{K}}=2\pi \sqrt{\frac{0.4}{0.1}}=4\pi sec$