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 $AB$ 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 $AB$. The time taken by the particle to reach the point $B$ is a $\sqrt{\frac{L}{{\omega }^{2}R}}$. Determine a.

Answer 1

Answer: 2.00

As $L<<R$, so the rod $AB$ 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
$AB={\omega }^{2}R\cos \theta$
$S_{\text{rel}}=u_{\text{rel }}t+\frac{1}{2}{a}_{\text{rel }{t}^2}$
$L=0(t)+\frac{1}{2}{\omega }^{2}R\cos \theta t^2$
$t=\sqrt{\frac{2L}{{\omega }^{2}R\cos \theta }}$
$=$ Time taken by particle to reach $B$