A car moving with a speed of $54\ \text{km/h}$ 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 $g = 10\ \text{m/s}^2$)
Answer: (C) $\tan^{-1}(1.125)$
Concept. In the car's frame (rotating with it), the bob is at rest under its weight $mg$ downward, the centrifugal force $\dfrac{mv^2}{r}$ outward, and the tension. The string lines up with the resultant of weight and centrifugal force.
Step 1. Convert the speed.
$$v=54\ \text{km h}^{-1}=54\times\frac{5}{18}=15\ \text{m s}^{-1}$$
Step 2. Balance on the bob. With the string at angle $\theta$ to the vertical:
$$T\cos\theta=mg,\qquad T\sin\theta=\frac{mv^2}{r}$$
Step 3. Divide.
$$\tan\theta=\frac{v^2}{rg}=\frac{15^2}{20\times10}=\frac{225}{200}=1.125$$
Answer. $\theta=\tan^{-1}(1.125)$, option C. The bob leans outward, away from the centre of the turn.
Solution by Sreeraj P, M.Sc Physics
