Question

A table with smooth horizontal surface is fixed in a cabin that rotates with a uniform angular velocity $\omega$ in a circular path of radius $R$. A smooth groove $A B$ of length $L < < R$ is made on the surface of the table. The groove makes an angle $60^{\circ}$ with the radius $OA$ of the circle in which the cabin rotates. A small particle is kept at the point $A$ in the groove and is released to move along $A B$. The time taken by the particle to reach the point $B$ is a $\sqrt{\frac{L}{\omega^{2} R}}$. Determine a.

Answer 1

As $L < < R$, so the rod $A B$ may be assumed to be moving in a circle of radius $R$.

Now with respect to a person standing on the table a pseudo force $m \omega^{2} R$ acts on the particle and hence acceleration
$=\frac{m \omega^{2} R}{m}=\omega^{2} R$
or acceleration in direction
$ AB =\omega^{2} R \cos \theta$
$S_{\text{rel} }=u_{\text {rel }} t+\frac{1}{2} a_{\text {rel } t^{2}} $
$L=0(t)+\frac{1}{2} \omega^{2} R \cos \theta t^{2} $
$t=\sqrt{\frac{2 L}{\omega^{2} R \cos \theta}}$
$=$ Time taken by particle to reach $B$