Question

A cabin (in a horizontal plane) rotates on a smooth horizontal table with a uniform angular speed $\omega$ in a circular path of radius $R$. A smooth groove $AB$ of length $L(<<R$ is made inside the cabin as shown in the figure. If a particle is released from $A$ to reach $B$ along the path $AB$, then find the time taken by the particle to reach the point $B$.

• A. $\sqrt{\frac{L}{R{\omega }^2\cos \theta }}$
• B. $\sqrt{\frac{2L}{R\cos \theta }}$
• C. $\sqrt{\frac{2L}{\cos \theta }}$
• D. $\sqrt{\frac{2L}{R{\omega }^2\cos \theta }}$

Answer 1

The correct option is D $\sqrt{\frac{2L}{R{\omega }^2\cos \theta }}$

Given, Length of groove is very small compared to the radius of circular path ($L<<R$)
Hence, centrifugal force
$F_{\text{centrifugal}}=mR{\omega }^2$ on the particle is assumed to be constant.

Solving from rotating frame of reference:
$F_{\text{centrifugal}}$ will act on the particle in radially outward direction. The component of this force along the path $AB$, provides the particle the force to move along the path $AB$.

$\therefore mR{\omega }^2\cos \theta =ma$
where $m$ is the mass of the particle.
$\Rightarrow a=R{\omega }^2\cos \theta$

Let the length of the groove be $L$.
Then, $L=ut+\frac{1}{2}a{t}^2$
$\because u=0$,
$t=\sqrt{\frac{2L}{a}}=\sqrt{\frac{2L}{R{\omega }^2\cos \theta }}$