Question

A car is moving in a circular path of radius \(20\,m\) with speed \(54\,km/hr\). A pendulum is hanging from the roof of the car. Find the angle made by pendulum with vertical. (Take \(g=10\,m/s^2\)).

• A. \( \tan^{-1}\left(\frac{9}{8}\right) \)
• B. \( \tan^{-1}\left(\frac{8}{9}\right) \)
• C. \( \tan^{-1}\left(\frac{4}{3}\right) \)
• D. \( \tan^{-1}\left(\frac{3}{4}\right) \)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In the non-inertial frame of the turning car, the pendulum bob experiences a centrifugal force radially outward and gravity acting downward.
The string aligns itself along the resultant of these two forces to maintain equilibrium in that frame.
Step 2: Key Formula or Approach:
The angle $\theta$ made with the vertical is given by $\tan\theta = \frac{F_c}{F_g} = \frac{mv^2/r}{mg} = \frac{v^2}{rg}$.
Step 3: Detailed Explanation:
First, convert the speed from km/hr to m/s:
\[ v = 54 \times \frac{5}{18} = 15 \text{ m/s} \]
Given values: $r = 20 \text{ m}$ and $g = 10 \text{ m/s}^2$.
Substitute these into the formula:
\[ \tan\theta = \frac{v^2}{rg} = \frac{(15)^2}{20 \times 10} \]
\[ \tan\theta = \frac{225}{200} \]
Simplifying the fraction by dividing by 25:
\[ \tan\theta = \frac{9}{8} \]
Taking the inverse tangent:
\[ \theta = \tan^{-1}\left(\frac{9}{8}\right) \]
Step 4: Final Answer:
The angle made by pendulum with vertical is $\tan^{-1}\left(\frac{9}{8}\right)$.