Question 294 of 308

A car is moving with a constant speed of $20m{s}^{-1}$ in a circular horizontal track of radius $40m$. A bob is suspended from the roof of the car by a massless string. The angle made by the string with the vertical will be : (Take $g=10m{s}^{-2}$)

A$\dfrac{\pi }{6}$B$\dfrac{\pi }{2}$C$\dfrac{\pi }{4}$D$\dfrac{\pi }{3}$

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Answer: (C) $\dfrac{\pi }{4}$

Let the frame of reference to be attached to the car with x-axis along horizontal direction and y-axis along vertical direction. Since the frame of reference is a rotating frame, a centrifugal force term should be added while applying Newton's second law.
Let $\theta$ be the angle made by the pendulum with the vertical.
Applying Newton's second law for horizontal direction:
$T\sin \theta -\dfrac{m{v}^{2}}{R}=0$ (since the acceleration of the pendulum in the rotating frame is 0).
$∴T\sin \theta =\dfrac{m{v}^{2}}{R}$ ..........(1)
Applying Newton's second law for vertical direction:
$T\cos \theta -mg=0$
$\Rightarrow T\cos \theta =mg$ ..........(2)
Dividing equation 1 by equation 2:
$\tan \theta =\dfrac{{v}^{2}}{Rg}=\dfrac{{\left(20\right)}^{2}}{\left(40\right)\left(10\right)}=1 \Rightarrow \theta =\dfrac{\pi }{4}$