Rotational Motion Part 2

Given, radius of sphere is a, mass of sphere is $m$, horizontal speed of sphere is $v_{0}$. Moment of inertia of solid sphere, $I=\frac{2}{5}ma^{2}$ Let, $h$ be the height of inclined plane and $/$ be the length of inclined plane.By using law of conservation of energy, Total energy at $1=$ Total energy at 2 $\therefore \; \mathrm{PE}_1 + \mathrm{KE}_1 = \mathrm{PE}_2 + \mathrm{KE}_2$ $\Rightarrow \; 0+\frac{1}{2} m v_{0}^{2}+\frac{1}{2} / \omega^{2}=m g h$. . . (i) As, $\; v_{0}=a \omega \Rightarrow \omega=\frac{v_{0}}{a}$ $\therefore \; \frac{1}{2} m v_{0}^{2} + \frac{1}{2} \left( \frac{2}{5} m a^{2} \right) \frac{v_{0}^{2}}{a^{2}} = m g h$ $\Rightarrow \; \frac{1}{2} m v_{0}^{2} \left(1+\frac{2}{5}\right)=m g h$ $\Rightarrow \; \frac{7 v_{0}^{2}}{10}=g h$ . . . (ii) As, $\; \sin \theta = \frac{h}{l}$ $\Rightarrow \; h = I \sin \theta$ Put this value in Eq. (ii), we get $\Rightarrow$ $\frac{7}{10} v_{0}^{2} = g l \sin \theta \Rightarrow l = \frac{7 v_{0}^{2}}{10 g \sin \theta}$