A solid cylinder is released from rest from the top of an inclined plane of inclination $30 °$ and length $60 cm$ . If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is ______ $m s^{- 1}$ . (Given $g = 10 m s^{- 2}$ )

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TestHub · Accepted Answer

Let speed of cylinder upon reaching the bottom of the inclined plane be $v$ .

Here, gain in kinetic energy $=$ loss in potential energy

So, $\frac{1}{2} I ω^{2} + \frac{1}{2} m v^{2} = m g l \sin θ$

$\Rightarrow \frac{1}{2} (\frac{m r^{2}}{2}) ω^{2} + \frac{1}{2} m v^{2} = m g l \sin 30 °$

For pure rolling, $v = r ω$

$\Rightarrow \frac{1}{2} (\frac{m r^{2}}{2}) {(\frac{v}{r})}^{2} + \frac{1}{2} m v^{2} = m g l \sin 30 °$

$\Rightarrow \frac{1}{4} m v^{2} + \frac{1}{2} m v^{2} = m g l \sin 30 ° \Rightarrow \frac{3}{4} m v^{2} = \frac{m g l}{2} \Rightarrow v = \sqrt{\frac{4 g l}{6}}$

Putting the values, we have

$\Rightarrow v = \sqrt{\frac{4 \times 10 \times 0.6}{6}} = 2 m s^{- 1}$

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