Question
A solid sphere rolls without slipping, first horizontally and then up to a point $X$ at height $h$ on an inclined plane before rolling down, as shown in the figure below. The initial horizontal speed of the sphere is
✓
Answer
(a)
Let initial horizontal speed of sphere is $v$.
Then, total kinetic energy of sphere on horizontal part
$=( KE )_{\text {translation }}+( KE )_{\text {rotation }}$
$=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2$
$=\frac{1}{2} m v^2+\frac{1}{2} \times \frac{2}{5} m R^2 \times \frac{v^2}{R^2}=\frac{7}{10} m v^2$
If sphere rises upto height $h$ then, by conservation of energy, we have
$m g h=\frac{7}{10} m v^2$
$\text { or } \quad v=\sqrt{\frac{10}{7} g h}$
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