Question

A bus is moving with a constant acceleration $a=\frac{3g}{4}$ towards right. In the bus, a ball is tied to a rope of length $l$ and it is rotated in a vertical circle as shown in figure. At what angle $\theta$ w.r.t the horizontal will the tension in the rope be minimum?

• A. $\theta ={37}^{\circ }$
• B. $\theta ={53}^{\circ }$
• C. $\theta ={30}^{\circ }$
• D. $\theta ={90}^{\circ }$

Answer 1

The correct option is B $\theta ={53}^{\circ }$

In a normal vertical circular motion, when a particle is projected with speed $u$ from the bottom most position, tension in the string will be minimum at the highest point $\&$ maximum at the lowest point.
$\therefore T_{max}\&T_{min}$ will be at the diametrically opposite ends.

$T_{max}-mg=\frac{m{u}^2}{l}$
$\Rightarrow T_{max}=mg+\frac{m{u}^2}{l}...\left(i\right)$
$T_{min}+mg=\frac{m{u}^2}{l}$
$\Rightarrow T_{min}=\frac{m{u}^2}{l}-mg...\left(ii\right)$


In this case, pseudo force will act in opposite direction of acceleration of the bus since we are dealing from the frame of reference of bus.
$\Rightarrow T_{min}\&T_{max}$ positions will be at the diametrically opposite ends, as shown in figure.


The bob will be in equilibrium from bus frame, so from the vector triangle:
$\tan \alpha =\frac{ma}{mg}=\frac{m\left(\frac{3g}{4}\right)}{mg}=\frac{3}{4}$
$\therefore \alpha ={37}^{\circ }$
Hence, from the right angled triangle in figure,
$\theta ={90}^{\circ }-{37}^{\circ }={53}^{\circ }$ is the angle w.r.t horizontal at which tension in the string will be the minimum.