$A$ solid cylinder rolls up an inclined plane of angle of inclination $30^{\circ}$. At the bottom of the inclined plane,the centre of mass of the cylinder has a speed of $5 \; m/s$.
$(a)$ How far will the cylinder go up the plane?
$(b)$ How long will it take to return to the bottom?

(N/A) solid cylinder rolling up an inclination is shown in the figure.
Initial velocity of the solid cylinder,$v = 5 \; m/s$.
Angle of inclination,$\theta = 30^{\circ}$.
$(a)$ Let $h$ be the maximum height reached by the cylinder.
Total energy at the bottom = Translational kinetic energy + Rotational kinetic energy
$E_{bottom} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Since $I = \frac{1}{2}mr^2$ and $v = r\omega$,we have:
$E_{bottom} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$
At the maximum height,the cylinder momentarily comes to rest,so total energy is potential energy:
$E_{top} = mgh$
By conservation of energy,$\frac{3}{4}mv^2 = mgh$
$h = \frac{3v^2}{4g} = \frac{3 \times 5^2}{4 \times 9.8} = \frac{75}{39.2} \approx 1.91 \; m$
Distance along the plane,$d = \frac{h}{\sin 30^{\circ}} = \frac{1.91}{0.5} = 3.82 \; m$.
$(b)$ The acceleration of a body rolling down an inclined plane is $a = \frac{g \sin \theta}{1 + K^2/R^2}$.
For a solid cylinder,$K^2/R^2 = 1/2$,so $a = \frac{g \sin 30^{\circ}}{1 + 0.5} = \frac{g(0.5)}{1.5} = \frac{g}{3} = \frac{9.8}{3} \approx 3.27 \; m/s^2$.
Using $d = \frac{1}{2}at^2$,the time to return is $t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2 \times 3.82}{3.27}} = \sqrt{2.337} \approx 1.53 \; s$.