Question

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

• A. $\frac{ln(1+\frac{R\cos \theta }{L})}{\omega }$
• B. $\frac{ln(1+\frac{L\cos \theta }{R})}{\omega \sqrt{\sin \theta }}$
• C. $\frac{ln(1+\frac{L\cos \theta }{R})}{\omega \sqrt{\cos \theta }}$
• D. $\frac{ln(1+\frac{R\cos \theta }{L})}{\omega \sqrt{\cos \theta }}$

Answer 1

The correct option is C $\frac{ln(1+\frac{L\cos \theta }{R})}{\omega \sqrt{\cos \theta }}$

FBD of the particle:


Since, the cabin is rotating with uniform angular velocity velocity $\omega$ about point $\text{O}$, it is a case of accelerated frame (centripetal acceleration $a_c$ ).

So, there will be a centrifugal force $F_{\text{centrifugal}}=mx{\omega }^2$ in radially outward direction for particle.

Let us take particle at a distance $x$ from point O, so from FBD:-
Component of $F_{\text{centrifugal}}$ along groove $=mx{\omega }^2\cos \theta$

If the acceleration of the particle along the groove is $a$, then net force acting on the particle
$=ma=mv\frac{dv}{dx}$
$\therefore mx{\omega }^2\cos \theta =mv\frac{dv}{dx}$

Integrating both sides,
${\int }_0^{v}vdv={\omega }^2\cos \theta {\int }_0^{x}xdx$
$v^2={\omega }^2\cos \theta x^2$
$\therefore v=x\omega \sqrt{\cos \theta }$

Expressing $v$ as $\frac{dx}{dt}$:
$\frac{dx}{dt}=x\omega \sqrt{\cos \theta }$
Integrating both sides,
${\int }_R^{R+L\cos \theta }\frac{1}{x}dx=\omega \sqrt{\cos \theta }{\int }_0^{t}dt$
$\Rightarrow \ln \frac{R+L\cos \theta }{R}=\omega \sqrt{\cos \theta }\times t$
$\Rightarrow t=\frac{\ln (1+\frac{L\cos \theta }{R})}{\omega \sqrt{\cos \theta }}$