Question

A car moving with a speed of 54 km/h takes a turn of radius 20 m. A simple pendulum is suspended from the ceiling of the car. Determine the angle made by the string of the pendulum with the vertical during the turning. (Take \(g = 10\) m/s²)

• A. \(\tan^{-1}(0.5) \)
• B. \(\tan^{-1}(0.75) \)
• C. \(\tan^{-1}(1.125) \)
• D. \(\tan^{-1}(0.25) \)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When the car turns, it undergoes circular motion and experiences a centripetal acceleration. In the non-inertial frame of the car, the pendulum experiences a pseudo force (centrifugal force) horizontally outwards, causing it to deflect from the vertical until equilibrium is reached with gravity and tension.
Step 2: Key Formula or Approach:
1. Convert velocity from km/h to m/s.
2. Centripetal acceleration: $a_c = \frac{v^2}{r}$.
3. Equilibrium condition angle: $\tan \theta = \frac{F_{pseudo}}{F_{gravity}} = \frac{m a_c}{m g} = \frac{v^2}{rg}$.
Step 3: Detailed Explanation:
Given values:
Speed $v = 54\text{ km/h}$
Radius of the turn $r = 20\text{ m}$
Acceleration due to gravity $g = 10\text{ m/s}^2$
First, convert the speed into standard SI units (m/s):
$v = 54 \times \frac{5}{18}\text{ m/s} = 3 \times 5\text{ m/s} = 15\text{ m/s}$.
Now calculate the centripetal acceleration required to take the turn:
$a_c = \frac{v^2}{r} = \frac{15^2}{20} = \frac{225}{20} = 11.25\text{ m/s}^2$.
Let $\theta$ be the angle the string makes with the vertical.
Balancing the forces in the frame of the car:
Horizontal force: $T \sin \theta = m a_c$
Vertical force: $T \cos \theta = m g$
Dividing the two equations gives:
$\tan \theta = \frac{m a_c}{m g} = \frac{a_c}{g}$.
Substitute the values:
$\tan \theta = \frac{11.25}{10} = 1.125$.
Therefore, $\theta = \tan^{-1}(1.125)$.
Step 4: Final Answer:
The angle made by the string is $\tan^{-1}(1.125)$.