A solid cylinder P rolls without slipping fro | vedclass

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Question

(b)

In presence of friction, cylinder rolls and by energy conservation, we have

Initial potential energy $=$ Kinetic energy at bottom

$m g h

Vedclass · Accepted Answer

$\sqrt{\frac{3}{2}}$

Vedclass · Answer

(b)

In presence of friction, cylinder rolls and by energy conservation, we have

Initial potential energy $=$ Kinetic energy at bottom

$m g h =( KE )_{	ext {translation }}+( KE )_{	ext {rotation }}$

$=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2$

$=\frac{1}{2} m v^2+\frac{1}{2}\left(\frac{1}{2} m r^2ight) \omega^2$

$=\frac{3}{4} m v^2 \quad[	herefore v=r \omega]$

So, $\quad v_p=\sqrt{\frac{4}{3} g h}$

In absence of friction, cylinder slips without rolling.

Hence, $\frac{1}{2} m v^2=m g h$

$\Rightarrow \quad v_q=\sqrt{2 g h}$

So, ratio of $v_q / v_p=\frac{\sqrt{2 g h}}{\sqrt{\frac{4}{3} g h}}=\sqrt{\frac{3}{2}}$

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