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

Answer 1

Total initial kinetic energy of the cylinder,
$K_{i}=\frac{1}{2} M v_{C M}^{2}+\frac{1}{2} I_{C M} \omega^{2}$
$=\frac{1}{2} M v_{C M}^{2}+\frac{1}{2} \times \frac{1}{2} M R^{2} \times \frac{v_{C M}^{2}}{R^{2}}$
$=\frac{1}{2} M v_{C M}^{2}+\frac{1}{4} M v_{C M}^{2}=\frac{3}{4} M v_{C M}^{2}$
Initial potential energy, $U_{i}=0$
Final kinetic energy, $K_{f}=0$
Final potential energy,
$U_{f}=M g h=M g s \sin 30^{\circ}=\frac{1}{2} M g s$
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 $K . E$.
$\frac{1}{2} M g s=\frac{3}{4} M v_{C M}^{2}$
$s=\frac{3 v_{C M}^{2}}{2 g}=\frac{3 \times(5)^{2}}{2 \times 10}=3.75 \,m$
Using equation of motion for the motion up the incline, we get
$0=v_{C M}+a t$
or $a=-\frac{v_{C M}}{t}$
Also,
$0^{2}-v_{C M}^{2}=2 a s$
or $a=-\frac{v_{C M}^{2}}{2 s}$
$\therefore \frac{v_{C M}}{t}=\frac{v_{C M}^{2}}{2 s} $
or $ t=\frac{2 s}{v_{C M}}=\frac{2 \times 3.75}{5}=1.5\, s$
Total time taken in returning to the bottom
$=2 \times 1.5=3.0 \,s$