Army Group X Practice Set 1

At the bottom of the p lane, the sphere has both translational and rotational kinetic energies. Hence , total energy $=\frac{1}{2}mv^2+\frac{1}{2}l\omega^2$ Using the law of conservation of energy,we can write $=\frac{1}{2}mv^2+\frac{1}{2}l\omega^2=mgh\dots(i)$ For a solid sphere , the moment of inertia about its centre, $l=\left(\frac{2}{5}\right)mr^2$ From Eq. ( i ) we get $\frac{1}{2}mv^2+\frac{1}{2}(mr^2)\omega^2=mgh$ $\frac{1}{2}v^2+\frac{1}{5}r^2\omega^2=gh$ $v=\sqrt{\frac{10}{7}gh}$ Hence , the velocity of the sphere at thebottom depends only on height h andacceleration due to gravity g. Both thesevalues are constants. Therefore, the velocity at the bottomremains the same from which everinclined plane the sphere is rolled. So , in both c a se s speed w ill be same ,but time of descent will be different.