Question

A small box of mass $m$ is placed on the outer surface of a smooth fixed sphere of radius $R$ at a point where the radius makes an angle $\varphi$ with the vertical. The box is released from this position. Find the distance travelled by the box before it leaves contact with the sphere.

• A. $R[{\cos }^{-1}\left(\frac{2}{3}\sin \varphi \right)-\varphi ]$
• B. $R[{\cos }^{-1}\left(\frac{2}{3}\cos \varphi \right)-\varphi ]$
• C. $R[{\cos }^{-1}\left(\frac{1}{3}\sin \varphi \right)-\varphi ]$
• D. $R[{\cos }^{-1}\left(\frac{1}{3}\cos \varphi \right)-\varphi ]$

Answer 1

The correct option is B $R[{\cos }^{-1}\left(\frac{2}{3}\cos \varphi \right)-\varphi ]$


The situation is shown in figure. It starts falling along the circular path on the sphere and breaks off from the surface when normal reaction becomes zero.
Solving from the frame of reference of the box, there is a centrifugal force.
$N+\frac{m{v}^2}{R}=mg\cos \theta$
$N=0\Rightarrow \frac{m{v}^2}{R}=mg\cos \theta$
or $v=\sqrt{Rg\cos \theta }$
where $v$ is the instantaneous velocity of the box at an angle $\theta$ from the vertical. It can be obtained by
$v=\sqrt{2gh}$ [As from rest it falls a distance $h$]
or $v=\sqrt{2gR(\cos \varphi -\cos \theta )}$

Using the above equations, we get
$\sqrt{Rg\cos \theta }=\sqrt{2gR(\cos \varphi -\cos \theta )}$
or $3\cos \theta =2\cos \varphi$
or $\theta ={\cos }^{-1}\left(\frac{2}{3}\cos \varphi \right)$

Thus, distance travelled by the box before leaving contact with the sphere is
$s=R(\theta -\varphi )$
or $s=R[{\cos }^{-1}\left(\frac{2}{3}\cos \varphi \right)-\varphi ]$