Question

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

Answer 1

(a) Let the cylinder roll up to height $h$.

From conservation of energy between the initial and final states,

$\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=mgh$

and, $I=\frac{1}{2}m{r}^2$

$⟹\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{1}{2}m{r}^2\right){\omega }^2=mgh$

Also for rolling, $v=r\omega$

Thus $h=\frac{3v^2}{4g}=1.913m$

If s is distance it goes up the plane, $sin\theta =\frac{h}{s}$

Thus $s=\frac{h}{sin\theta }=3.826m$

(b) Time taken to return the bottom=$\sqrt{\frac{2s(1+\frac{K^2}{r^2})}{g\sin \theta }}=\sqrt{\frac{2\times 3.826\times (1+\frac{1}{2})}{9.8\times \frac{1}{2}}}=1.53s$