A solid sphere of mass ' $m$ ' and radius ' $r$ ' is allowed to roll without slipping from the highe

Question

A solid sphere of mass ' $m$ ' and radius ' $r$ ' is allowed to roll without slipping from the highest point of an inclined plane of length ' $L$ ' and makes an angle $30^{\circ}$ with the horizontal. The speed of the particle at the bottom of the plane is $v_{1}$. If the angle of inclination is increased to $45^{\circ}$ while keeping $L$ constant. Then the new speed of the sphere at the bottom of the plane is $v_{2}$. The ratio $v_{1}^{2}: v_{2}^{2}$ is\

Professor PC Thomas Classes · Answer

(d) : By using energy conservation,\ $m g h=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$\ $\left(I=\frac{2}{5} m r^{2}, \omega=\frac{v}{r}ight)$\ $\Rightarrow m g h=\frac{7 m v^{2}}{10} \Rightarrow v^{2}=\frac{10 g h}{7}$\ So $v^{2} \propto h$\ $	herefore \frac{v_{1}^{2}}{v_{2}^{2}}=\frac{h_{1}}{h_{2}}=\frac{l \sin 30^{\circ}}{l \sin 45^{\circ}}=\frac{1}{\sqrt{2}}$