Question

A pendulum of mass $m$ hangs from a support fixed to a trolley. The inclination of the string (i.e. angle $\theta$) when the trolley moves up a plane of inclination $\alpha$ with acceleration $a$ is-

• A. Zero
• B. ${\tan }^{-1}\alpha$
• C. ${\tan }^{-1}\left(\frac{a+g\sin \theta }{g\cos \alpha }\right)$
• D. ${\tan }^{-1}\left(\frac{a}{g}\right)$

Answer 1

The correct option is C ${\tan }^{-1}\left(\frac{a+g\sin \theta }{g\cos \alpha }\right)$

Given,
Mass of pendulum = $m$
Inclination of plane = $\alpha$
Acceleration of trolley = $a$
From trolley frame of reference, pseudo force will act on the pendulum as shown.
In trolley frame of refrence, pendulum is at rest.
FBD of pendulum:


Along x - axis
$\sum F=0$ [Particle is at rest]
$-ma-mg\sin \alpha +T\sin \theta =0$
$T\sin \theta =ma+mg\sin \alpha ----\left(1\right)$

Along y -axis
$\sum F=0$ [Particle is at rest]
$-mg\cos \alpha +T\cos \theta =0$
$T\cos \theta =mg\cos \alpha ----\left(2\right)$

On dividing equation (1) with (2)
$\tan \theta =\frac{a+g\sin \alpha }{g\cos \alpha }$
$\Rightarrow \theta ={\tan }^{-1}\left[\frac{a+g\sin \alpha }{g\cos \alpha }\right]$