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A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination. Will it reach the bottom with the same speed in each case? Will it take longer to roll down one plane than the other? If so, which one and why?
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Let $v=$ speed of the solid sphere at the bottom of the inclined plane.
$\therefore \quad$ P.E. at the top of inclined plane $=$ Total K.E. at the bottom of the plane
$M g h=\frac{1}{2} M v^2+\frac{1}{2} I \omega^2$;
M.I. of solid sphere $=\frac{2}{5} \mathrm{MR}^2$
$$
\begin{aligned}
\therefore \quad M g h &=\frac{1}{2} M v^2+\frac{1}{2} \times \frac{2}{5} M R^2 \omega^2 \\
&=\frac{1}{2} M v^2+\frac{1}{5} M v^2[\because v=r \omega]
\end{aligned}
$$
$$
M g h=\frac{7}{10} M v^2 \Rightarrow v=\sqrt{\frac{10}{7} g h}
$$
$\because$ The velocity is independent of the angle of inclination
$\therefore \quad$ Both the spheres will reach the bottom at the same speed.
As the height is same, so the time taken by them will be same.
$\therefore \quad$ P.E. at the top of inclined plane $=$ Total K.E. at the bottom of the plane
$M g h=\frac{1}{2} M v^2+\frac{1}{2} I \omega^2$;
M.I. of solid sphere $=\frac{2}{5} \mathrm{MR}^2$
$$
\begin{aligned}
\therefore \quad M g h &=\frac{1}{2} M v^2+\frac{1}{2} \times \frac{2}{5} M R^2 \omega^2 \\
&=\frac{1}{2} M v^2+\frac{1}{5} M v^2[\because v=r \omega]
\end{aligned}
$$
$$
M g h=\frac{7}{10} M v^2 \Rightarrow v=\sqrt{\frac{10}{7} g h}
$$
$\because$ The velocity is independent of the angle of inclination
$\therefore \quad$ Both the spheres will reach the bottom at the same speed.
As the height is same, so the time taken by them will be same.
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