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A solid cylinder rolls up an inclined plane of angle of inclination $30^{\circ}$. At the bottom of the inclined plane, the C.M. of the cylinder has a speed of $5 \mathrm{~m} / \mathrm{s}$
(a) How far will the cylinder go up the plane?
(b) How long will it take to return to the bottom?
(a) How far will the cylinder go up the plane?
(b) How long will it take to return to the bottom?
Solution:
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Verified Answer
$\theta=30^{\circ}, v=5 \mathrm{~m} / \mathrm{s}$
Let $h$ be the height on the plane upto which the cylinder will go up.
$\therefore \quad$ From conservation of energy,
Total K.E $=$ P.E $\Rightarrow \frac{1}{2} m v^2+\frac{1}{2} I \omega^2=\mathrm{mgh}$
$$
\begin{aligned}
&\Rightarrow \frac{1}{2} m v^2+\frac{1}{2}\left(\frac{1}{2} m r^2\right) \omega^2=m g h\left[I_{\mathrm{cyl}}=\frac{1}{2} m r^2\right] \\
&\left.\Rightarrow \frac{3}{4} m v^2=m g h \text { [using } v=r \omega\right] \\
&\Rightarrow h=\frac{3 v^2}{4 g}=\frac{3 \times 5^2}{4 \times 9.8}=1.913 \mathrm{~m} .
\end{aligned}
$$
Let, $s=$ distance moved up by the cylinder on the inclined plane.

$$
\therefore \sin \theta=\frac{h}{s} \Rightarrow s=\frac{h}{\sin \theta}=\frac{1.913}{\sin 30^{\circ}}=3.826 \mathrm{~m}
$$
Time taken to return to the bottom $=t$
$$
=\sqrt{\frac{2 s\left(1+\frac{k^2}{r^2}\right)}{g \sin \theta}}=\sqrt{\frac{2 \times 3.826\left(1+\frac{1}{2}\right)}{9.8 \sin 30^{\circ}}}=1.53 \mathrm{~s} .
$$
Let $h$ be the height on the plane upto which the cylinder will go up.
$\therefore \quad$ From conservation of energy,
Total K.E $=$ P.E $\Rightarrow \frac{1}{2} m v^2+\frac{1}{2} I \omega^2=\mathrm{mgh}$
$$
\begin{aligned}
&\Rightarrow \frac{1}{2} m v^2+\frac{1}{2}\left(\frac{1}{2} m r^2\right) \omega^2=m g h\left[I_{\mathrm{cyl}}=\frac{1}{2} m r^2\right] \\
&\left.\Rightarrow \frac{3}{4} m v^2=m g h \text { [using } v=r \omega\right] \\
&\Rightarrow h=\frac{3 v^2}{4 g}=\frac{3 \times 5^2}{4 \times 9.8}=1.913 \mathrm{~m} .
\end{aligned}
$$
Let, $s=$ distance moved up by the cylinder on the inclined plane.

$$
\therefore \sin \theta=\frac{h}{s} \Rightarrow s=\frac{h}{\sin \theta}=\frac{1.913}{\sin 30^{\circ}}=3.826 \mathrm{~m}
$$
Time taken to return to the bottom $=t$
$$
=\sqrt{\frac{2 s\left(1+\frac{k^2}{r^2}\right)}{g \sin \theta}}=\sqrt{\frac{2 \times 3.826\left(1+\frac{1}{2}\right)}{9.8 \sin 30^{\circ}}}=1.53 \mathrm{~s} .
$$
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