Question

A cylinder rolls up an inclined plane at an angle of ${30}^o$. At the bottom of the inclined plane, the center of mass of the cylinder has speed of $5m/s$. How long will it take to return to the bottom?

• A. $2s$
• B. $3s$
• C. $1.5s$
• D. $4s$

Answer 1

Answer: (B)

$3s$

Retardation of the cylinder, $a=\frac{-g\sin \theta }{1+\frac{I}{m{R}^2}}$

where $\theta ={30}^o$.

Let the cylinder be solid, then

$I=\frac{1}{2}m{R}^2$

$\therefore$ $a=\frac{-g\sin \theta }{1+\frac{1}{2}}=\frac{-2}{3}\times 9.8\times \frac{1}{2}\Rightarrow a=\frac{-9.8}{3}m/s^2$

Using the relation, $v^2-u^2=2as$, we get

$s=\frac{v^2-u^2}{2a}\Rightarrow s=\frac{0-5^2}{2(-\frac{9.8}{3})}\Rightarrow s=3.83m$

Let $T$ be the time taken by the cylinder to return to the bottom. Thus $T=2t$

where $t$ is time of ascending or descending.

Here, initial velocity $u=0$

Using the relation $s=ut+\frac{1}{2}a{t}^2$

$\therefore$ $s=0+\frac{1}{2}a{t}^2$

We get

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2\times 3.83}{\left(\frac{9.8}{3}\right)}}=1.53s$

$⟹T=2\times 1.53=3.06s\approx 3.0s$