A solid sphere of mass $2 \mathrm{~kg}$ rolls on a smooth horizontal surface at $10 \mathrm{~m} / \mathrm{s}$. It then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. Then what is the height (in meter) attained by the sphere before it stops ?

Question

Aicharya · Accepted Answer

If a body rolls on a horizontal surface, it possesses both translational and rotational kinetic energies. The net kinetic energy is given by
where $\mathrm{K}$ is the radius of gyration. So from law of conservation of energy,
$\frac{1}{2} m v^{2}\left(1+\frac{K^{2}}{R^{2}}\right)=m g h$
where $\mathrm{h}$ is the height attained by the sphere.
i.e., $\frac{1}{2} \times 2 \times(10)^{2}\left(1+\frac{2}{5}\right)=2 \times 9.8 \times \mathrm{h}$.
i.e., $\frac{1}{2} \times 100 \times\left(\frac{7}{5}\right)=9.8 \mathrm{~h}$
or $\quad \mathrm{h}=\frac{700}{98}=7.1 \mathrm{~m}$

,

A solid sphere of mass $2 \mathrm{~kg}$ rolls on a smooth horizontal surface at $10 \mathrm{~m} /…

Solution