Question

A car is accelerating horizontally with constant acceleration $a=10m/s$${}^2$. One end of a light string is attached to the roof of the ceiling and there is a small bob at another end of the string.The bob is given an initial velocity such that it continues to move in uniform circular motion with respect to an observer inside the car.The bob moves such that the maximum vertical separation between two points of its path is $h=1m$ . The length of the string is $\sqrt{\frac{5}{2}}$ m and acceleration due to gravity $g=10m/s$ ${}^2$. If the angular speed of the bob in rad/s is $\sqrt{2x}$, find the value of $x$.

Answer 1

First, we will derive a result for conical pendulum in stationary

T cos $\theta$ = mg

T sin $\theta$ = m${\omega }^2$r = m${\omega }^2$ $l$ sin $\theta$

From the above two equations we get

$\omega =\sqrt{\frac{g}{lcos\theta }}=\frac{g}{H}$

Now in accelerating the car , with respect to an inside observer, there will be a pseudo force acting on the particle opposite to the acceleration of the car, and hence perpendicular to the weight. Since a = g, the effective gravity will be

$g_{eff}=\sqrt{a^2+g^2}$ = $g\sqrt{2}$, inclined at an angle of ${45}^{\circ }$ with the vertical axis.

So now $\omega =\sqrt{\frac{g_{eff}}{H}}$

From the fig.

$2r=h$ $\sqrt{2}$ and $H^2=l^2-r^2$

From these two conditions $H$ = $\sqrt{2}$ m

$\therefore \omega =\sqrt{10}\Rightarrow$ $x=5$