A heavy particle slides under gravity down the inside of a smooth vertical tube held in vertical plane. It starts from the highest point with velocity $\sqrt{2\mathrm{ag}}$, where a is the radius of the circle. Find the angular position $\mathrm{\theta }$ (as shown in figure) at which the vertical acceleration of the particle is maximum.

At position $\mathrm{\theta }$,

${\mathrm{v}}^2={\mathrm{v}}_0{}^2+2\mathrm{gh}$

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

${\mathrm{v}}^2=(\sqrt{2\mathrm{ag}}{)}^2+2\mathrm{ag}(1-\mathrm{cos\theta })$

$\text{ where, }$

$\text{ or } {\mathrm{v}}^2=2\mathrm{ag}(2-\mathrm{cos\theta })$

$\mathrm{N}+\mathrm{mgcos\theta }=\frac{{\mathrm{mv}}^2}{\mathrm{a}}$

$\text{ or } \mathrm{N}+\mathrm{mgcos\theta }=2\mathrm{mg}(2-\mathrm{cos\theta })$

$\text{ or } \mathrm{N}=\mathrm{mg}(4-3\mathrm{cos\theta })$

$\mathrm{Net} \mathrm{verticalforce},\mathrm{F}=\mathrm{Ncos\theta }+\mathrm{mg}$

$=\mathrm{mg}\left(4\mathrm{cos\theta }-3{\cos }^2\mathrm{\theta }+1\right)$

or $\mathrm{N}=\mathrm{mg}(\cos \theta )(4-3\mathrm{cos\theta })+mg$

This force (or acceleration) will be maximum when $\frac{\mathrm{dF}}{\mathrm{d\theta }}=0$

or $-4\mathrm{sin\theta }+6\mathrm{sin\theta cos\theta }=0$

So, either

$\mathrm{sin\theta }=0,\mathrm{\theta }=0°\text{ or }\mathrm{cos\theta }=\frac{2}{3}\text{, }$

$\mathrm{\theta }={\cos }^{-1}\left(\frac{2}{3}\right)$

$\mathrm{\theta }=0°$ is unacceptable Therefore, the desired position is at

$\mathrm{\theta }={\cos }^{-1}\left(\frac{2}{3}\right)$