91Ó°ÊÓ

The torque is the product of the moment of inertia and the square of the distance.For this problem, first, find the angular acceleration of the solid sphere and then find the torque.

The initial angular velocity of the solid sphere is \({\omega _1} = 0\).

The diameter of the solid sphere is \(d = 0.72\;{\rm{m}}\).

The time taken by the solid sphere is \(t = 15.0\;{\rm{s}}\).

The torque on the sphere is \(\tau = 10.8\;{\rm{m}} \cdot {\rm{N}}\).

The angular displacement is \(\theta = 160\;{\rm{revolution}} = \left( {160 \times 2\pi } \right)\;{\rm{rad}}\).

Let m be the mass of the sphere and \(\alpha \) be the angular acceleration of the sphere.

The radius of the solid sphere is \(r = \frac{d}{2}\).

The moment of inertia of the sphere is

\(\begin{align}I &= \frac{2}{5}m{r^2}\\ &= \frac{2}{5}m{\left( {\frac{d}{2}} \right)^2}\end{align}\)

Now, for the rotational motion of the sphere,

\(\begin{align}\theta &= {\omega _1}t + \frac{1}{2}\alpha {t^2}\\\theta &= \frac{1}{2}\alpha {t^2}\;{\rm{as}}\;{\omega _1} &= 0\\\alpha &= \frac{{2\theta }}{{{t^2}}}\end{align}\).

Now, the torque is

\(\begin{align}\tau &= I\alpha \\\tau &= \frac{2}{5}m{\left( {\frac{d}{2}} \right)^2}\frac{{2\theta }}{{{t^2}}}\\\tau &= m \times \frac{2}{5}\frac{{{d^2}}}{4}\frac{{2\theta }}{{{t^2}}}\\m &= \frac{{5\tau {t^2}}}{{{d^2}\theta }}\end{align}\)

Now, substituting the values in the above equation,

\(\begin{align}m &= \frac{{5 \times \left( {10.8\;{\rm{m}} \cdot {\rm{N}}} \right) \times {{\left( {15.0\;{\rm{s}}} \right)}^2}}}{{{{\left( {0.72\;{\rm{m}}} \right)}^2} \times \left( {160 \times 2\pi } \right)\;{\rm{rad}}}}\\ &= 23.33\;{\rm{kg}}\end{align}\).

Hence, the mass of the solid sphere is \(23.33\;{\rm{kg}}\).