Question

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 $5m/s$. How long (in seconds) will it take to return to the bottom? [Take $g=10m/s^2$ ]

Answer 1

Answer: 3

Total initial kinetic energy of the cylinder,
$K_i=\frac{1}{2}M{v}_{CM}^2+\frac{1}{2}{I}_{CM}{\omega }^2$
$=\frac{1}{2}M{v}_{CM}^2+\frac{1}{2}\times \frac{1}{2}M{R}^2\times \frac{v_{CM}^2}{R^2}$
$=\frac{1}{2}M{v}_{CM}^2+\frac{1}{4}M{v}_{CM}^2=\frac{3}{4}M{v}_{CM}^2$
Initial potential energy, $U_i=0$
Final kinetic energy, $K_f=0$
Final potential energy,
$U_f=Mgh=Mgs\sin 30^{\circ }=\frac{1}{2}Mgs$
where $s$ is the distance travelled up the incline and $h$ is the vertical height covered above the bottom.
Gain in $P.E:=$ Loss in $K.E$.
$\frac{1}{2}Mgs=\frac{3}{4}M{v}_{CM}^2$
$s=\frac{3v_{CM}^2}{2g}=\frac{3\times (5{)}^2}{2\times 10}=3.75m$
Using equation of motion for the motion up the incline, we get
$0=v_{CM}+at$
or $a=-\frac{v_{CM}}{t}$
Also,
$0^2-v_{CM}^2=2as$
or $a=-\frac{v_{CM}^2}{2s}$
$\therefore \frac{v_{CM}}{t}=\frac{v_{CM}^2}{2s}$
or $t=\frac{2s}{v_{CM}}=\frac{2\times 3.75}{5}=1.5s$
Total time taken in returning to the bottom
$=2\times 1.5=3.0s$