Question

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 $\sqrt{2ag}$ where a is the radius of the circle. Find the angular position $\theta$ (as shown in figure) at which the vertical acceleration of the particle is maximum.

• A. $\theta ={\cos }^{-1}\left(\frac{1}{3}\right)$
• B. $\theta ={\cos }^{-1}\left(\frac{2}{3}\right)$
• C. $\theta ={\cos }^{-1}\left(\frac{2}{5}\right)$
• D. $\theta ={\cos }^{-1}\left(\frac{3}{4}\right)$

Answer 1

The correct option is B $\theta ={\cos }^{-1}\left(\frac{2}{3}\right)$

Le the angle for maximum vertical acceleration be $\theta$ (as shown in figure), and speed at that point be $v$.

By conservation of mechanical energy,

$\frac{1}{2}mv(\sqrt{2ag}{)}^2+mga=\frac{1}{2}m{v}^2+mgacos\theta$

$⟹m{v}^2=2mga(2-cos\theta )$

Let the normal reaction on particle, in radially inward direction be $N$.

Thus centripetal force is given as,

$N+mgcos\theta =\frac{m{v}^2}{a}=2mg(2-cos\theta )$

$⟹N=mg(4-3cos\theta )$

Net vertical force on the particle is,

$Ncos\theta +mg=mg(4cos\theta -3co{s}^2\theta +1)$

Net vertical acceleration is,

$a_v=g(4cos\theta -3co{s}^2\theta +1)$

To get angle for maximum vertical acceleration, we differentiate this equation w.r.t. $cos\theta$ and equate to zero.

So, $g(4-6cos\theta )=0⟹\theta =co{s}^{-1}(\frac{2}{3})$