Physics - Rotation Question with Solution | TestHub

Drawing free body diagram of the cylinder with respect to plank.

$2\mu mg=ma\Rightarrow a=2\mu g$

$\left(\mu mgR\right)=\frac{1}{2}m{R}^2\alpha \mathrm{ };\alpha -\frac{2\mu g}{r}$

Acceleration of point of contact with respect to plank is$4\mu g$

Velocity of pure rolling starts,

$-\mathrm{v}+4\mathrm{\mu }\mathrm{g}\mathrm{t}=0$

$\mathrm{t}=\frac{20}{4\times 0.5\times 10}=1\mathrm{s}$

Distance traveled by cylinder with respect to plank in 1s is

${\mathrm{S}}^{\mathrm{\prime }}=-\mathrm{v}\mathrm{t}+\frac{1}{2}\left(2\mathrm{\mu }\mathrm{g}\right){\mathrm{t}}^2=-15\mathrm{m}$

At$\mathrm{t}=1\mathrm{s}$, the velocity of cylinder with respect to plank is

$\mathrm{t}=1\mathrm{s}$

${\mathrm{v}}_{\mathrm{r}\mathrm{e}\mathrm{l}}=-\mathrm{v}+2\mathrm{\mu }\mathrm{g}\mathrm{t}=-20+2\times 0.5\times 10=-10$m/s

Remaining 10 m will be travelled in time${\mathrm{t}}^{\mathrm{\prime }}=\frac{10}{10}=1\mathrm{s}$

$\therefore$Total time = 2s