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Question

A cabin rotates on a smooth horizontal table with a uniform angular velocity ω in a circular path of radius R (cabin is in a horizontal plane). A smooth path AB is made inside the cabin as shown in 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 point B, considering length of path AB =L.


A
ln(1+RcosθL)ω
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B
ln(1+LcosθR)ωsinθ
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C
ln(1+LcosθR)ωcosθ
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D
ln(1+RcosθL)ωcosθ
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Solution

The correct option is C ln(1+LcosθR)ωcosθ
FBD of the particle:


Since, the cabin is rotating with uniform angular velocity velocity ω about point O, it is a case of accelerated frame (centripetal acceleration ac ).

So, there will be a centrifugal force Fcentrifugal=mxω2 in radially outward direction for particle.

Let us take particle at a distance x from point O, so from FBD:-
Component of Fcentrifugal along groove =mxω2cosθ

If the acceleration of the particle along the groove is a, then net force acting on the particle
=ma=mvdvdx
mxω2cosθ=mvdvdx

Integrating both sides,
v0vdv=ω2cosθx0xdx
v2=ω2cosθ x2
v=xωcosθ

Expressing v as dxdt:
dxdt= xωcosθ
Integrating both sides,
R+LcosθR1xdx= ωcosθt0dt
lnR+LcosθR= ωcosθ×t
t= ln(1+LcosθR)ωcosθ

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