Question

A car is moving with a constant speed of $20\, m / s$ in a circular horizontal track of radius $40\, m$ 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 $g=10 \, m / s ^2$ )

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{2}$
• C. $\frac{\pi}{4}$
• D. $\frac{\pi}{3}$

Answer 1

The Correct Option is C

$T\sin \theta =\frac{m{l}^2}{R}$
$\tan \theta =\frac{v^2}{Rg}$
$\tan \theta =\frac{20^2}{40\times 10}$
$\tan \theta =1$
$\Rightarrow \theta =\frac{\pi }{4}$

Answer 2

1. The forces acting on the bob are:
- Tension (\(T\)) in the string.
- Centripetal force (\(T \sin \theta = \frac{mv^2}{R}\)).
- Vertical component (\(T \cos \theta = mg\)). 
2. Dividing these equations: \[ \tan \theta = \frac{T \sin \theta}{T \cos \theta} = \frac{\frac{mv^2}{R}}{mg}. \] 
3. Simplify: \[ \tan \theta = \frac{v^2}{gR}. \] 
4. Substituting values (\(v = 20 \, \text{m/s}, \, R = 40 \, \text{m}, \, g = 10 \, \text{m/s}^2\)): \[ \tan \theta = \frac{20^2}{10 \times 40} = 1. \] 
\[ \theta = \tan^{-1}(1) = 45^\circ. \] 
Thus, the angle made by the string is 45°. 

The angle of the string depends on the balance of centripetal force and gravitational force. The tangent of the angle is the ratio of horizontal to vertical forces.