A smooth circular tube of radius R is fixed in a vertical plane. A particle is projected from its lowest point with a velocity just sufficient to carry it to the highest point. Show that the time taken by the particle to reach the end of the horizontal diameter is $\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}\ln (1+\sqrt{2})$.

 

Minimum velocity of particle at the lowest position to complete the circle should be $\sqrt{4\mathrm{gR}}$ inside a tube.

So.

$\mathrm{u}=\sqrt{4} \mathrm{gR}$

$\mathrm{h}=\mathrm{R}(1-\mathrm{cos\theta })$

${\mathrm{v}}^2={\mathrm{u}}^2-2\mathrm{gh}$

${\mathrm{v}}^2=4\mathrm{gR}-2\mathrm{gR}(1-\mathrm{cos\theta })$

=2 g R(1+cos $\theta$)

$\mathrm{Or} v^2=2\mathrm{gR}\left(2{\cos }^2\frac{\mathrm{\theta }}{2}\right)$

or $\mathrm{v}=2\sqrt{\mathrm{gR}}\cos \frac{\mathrm{\theta }}{2}$

From $\mathrm{ds}=\mathrm{v}·\mathrm{dt}$, we get

$\mathrm{Rd\theta }=2\sqrt{\mathrm{gR}}\cos \frac{\mathrm{\theta }}{2}·\mathrm{dt}$

or ${\int }_0^1\mathrm{dt}=\frac{1}{2}\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}{\int }_0^{\mathrm{\pi }/2}\sec \left(\frac{\mathrm{\theta }}{2}\right)\mathrm{d\theta }$

or $\mathrm{t}=\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}{\left[\ln \left(\sec \frac{\mathrm{\theta }}{2}+\tan \frac{\mathrm{\theta }}{2}\right)\right]}_0^{\mathrm{\pi }/2}$

or $\mathrm{t}=\sqrt{\frac{\mathrm{R}}{\mathrm{g}}}\ln (1+\sqrt{2})$

Hence proved.