A solid cylinder is released from rest from the top of an inclined plane of inclination 30 ° and length…

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

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

So, $\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2=mgl\sin \theta$

$\Rightarrow \frac{1}{2}\left(\frac{{\displaystyle m{r}^2}}{2}\right){\omega }^2+\frac{1}{2}m{v}^2=mgl\sin 30°$

For pure rolling, $v=r\omega$

$\Rightarrow \frac{1}{2}\left(\frac{{\displaystyle m{r}^2}}{2}\right){\left(\frac{v}{r}\right)}^2+\frac{1}{2}m{v}^2=mgl\sin 30°$

$$\begin{gathered} \Rightarrow \frac{1}{4}m{v}^2+\frac{1}{2}m{v}^2=mgl\sin 30° \\ \Rightarrow \frac{3}{4}m{v}^2=\frac{mgl}{2} \\ \Rightarrow v=\sqrt{\frac{4gl}{6}} \end{gathered}$$

Putting the values, we have

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