Question

A particle slides on surface of a fixed smooth sphere starting from topmost point. The angle rotated by the radius through the particle, when it leaves contact with the sphere, is

• A. $\theta =co{s}^{-1}\left(\frac{1}{3}\right)$
• B. $\theta =co{s}^{-1}\left(\frac{2}{3}\right)$
• C. $\theta =ta{n}^{-1}\left(\frac{1}{3}\right)$
• D. $\theta =si{n}^{-1}\left(\frac{4}{3}\right)$

Answer 1

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

Keeping in mind that, when it leaves the contact the particle is no more under the influence of the Normal force. Using the conservation of energy till the last point of contact (ie. $\theta$) we get:

$K_f-K_i=U_i-U_f$

$\frac{1}{2}m{v}^2-0=mgr(1-\cos \theta )$....(1)

where $r$ is the radius of the sphere and $m$ is the mass of the particle. Now using the force balance equation at the point when the particle has acquired this velocity $v$ and lose its point of contact;

$mg\cos \theta =\frac{m{v}^2}{r}$ ..... (2)

Using both equations (1) and (2), we finally get:

$\cos \theta =2-2\cos \theta$

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