Question

A vehicle is moving with a constant speed of $10\text{ m/s}$ in a circular horizontal track of radius $20\text{ m}$ . A bob is suspended from the roof of a vehicle by a massless string. The angle made by the string with the vertical will be (acceleration due to gravity, $g = 10\text{ m/s}^2$ )}

• A. $\tan^{-1}(0.5)$
• B. $\tan^{-1}(0.6)$
• C. $\tan^{-1}(0.7)$
• D. $\tan^{-1}(0.8)$

Hint:

For a bob in a turning vehicle: \[ \tan\theta=\frac{v^2}{rg} \] This is a standard circular motion result.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A suspended bob in a turning vehicle experiences a pseudo-force (centrifugal force) radially outward. The string aligns along the vector sum of gravity and this pseudo-force.
Step 2: Key Formula or Approach:
Equilibrium angle $\theta$ is given by: $\tan \theta = \frac{v^2}{rg}$.
Step 3: Detailed Explanation:
Given values: $v = 10\text{ m/s}$, $r = 20\text{ m}$, $g = 10\text{ m/s}^2$.
\[ \tan \theta = \frac{10^2}{20 \times 10} \]
\[ \tan \theta = \frac{100}{200} = 0.5 \]
\[ \theta = \tan^{-1}(0.5) \]
Step 4: Final Answer:
The angle with the vertical is $\tan^{-1}(0.5)$.