Question

A car is moving in a circular horizontal track of radius $10\text{m}$ with a constant speed of $10\text{m/s}$. A bob is suspended from the roof of the car by a light rigid rod. The angle (in degrees) made by the rod with the vertical is
[Take $g=10{\text{m/s}}^2$]

Answer 1

Given,
Radius of track, $r=10\text{m}$
Speed of car, $u=10\text{m/s}$

Let us suppose tension in rod is $T$, making angle $\theta$ with the vertical and mass of bob is $m$.

From car frame of reference :-
As car is going around a circular horizontal track, it is in accelerated motion. Therefore, we have to apply pseudo force on bob of magnitude $ma$ in the opposite direction of centripetal acceleration of the car.


Acceleration of car is $a$.
$a=\frac{u^2}{r}$ [ Circular Motion ]
$=\frac{{10}^2}{10}$
$=10{\text{m/s}}^2$ (inward).......(1)

F.B.D of bob :


On applying $\sum F=ma$ along horizontal.
$T\sin \theta -ma=0$
$\Rightarrow T\sin \theta =ma$ .......(2)

On applying $\sum F=ma$ along vertical :
$T\cos \theta -mg=0$
$\Rightarrow T\cos \theta =mg$ ........(3)

On dividing equation (2) by (3) :.
$\tan \theta =\frac{a}{g}$
$\Rightarrow \tan \theta =\frac{10}{10}$ [ from equation 1 ]
$\Rightarrow \tan \theta =1$
$\Rightarrow \theta ={45}^{\circ }$

The angle made by the rod with the vertical is ${45}^{\circ }$.