Question

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

Answer 1

Let the separation between the two atoms be $2r$

and mass of each oxygen atom be $m/2$

Hence moment of inertia of the oxygen molecule=$\left(\frac{m}{2}\right)r^2+\left(\frac{m}{2}\right)r^2=m{r}^2$

$\therefore$ the radius of rotation of the molecules is

$r=\sqrt{\frac{I}{M}}$

$\Rightarrow \sqrt{\frac{1.94\times {10}^{-46}}{5.3\times {10}^{-26}}}=6.1\times {10}^{-11}m$

It is given that $K{E}_{rot}=\frac{2}{3}K{E}_{trans}$

$⟹\frac{1}{2}I{\omega }^2=\frac{2}{3}\times \frac{1}{2}m{v}^2$

$⟹r^2{\omega }^2=\frac{2}{3}{v}^2$

$⟹\omega =\sqrt{\frac{2}{3}}\frac{v}{r}$

$\omega =\sqrt{\frac{2}{3}}\frac{500}{6.1\times {10}^{-11}}$

$=6.80\times {10}^{12}rad/s$