A solid sphere rolls without slipping, first horizontally and then up to a point X at height h on an

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6.System of Particles and Rotational Motion

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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

A

$\sqrt{10 g h / 7}$

B

$\sqrt{7 g h / 5}$

C

$\sqrt{5 g h / 7}$

D

$\sqrt{2 g h}$

(KVPY-2013)

Solution

(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

$m g h=\frac{7}{10} m v^2$

$\text { or } \quad v=\sqrt{\frac{10}{7} g h}$

Standard 11

Physics

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