Kinematics

JEE Main level, single correct, medium

283 of 308

Question 283 of 308

A car moving with a speed of 5454 km/h takes a turn of radius 2020 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=10g=10 m/s2^2)

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Answer: (C) tan⁡−1(1.125)\tan^{-1}(1.125)

Speed of the car, v=54×518=15v = 54 \times \dfrac{5}{18} = 15 m/s.

In the frame of the car, the pendulum bob experiences a pseudo force (centrifugal force) horizontally outwards and the gravitational force downwards.

Let TT be the tension in the string and θ\theta be the angle it makes with the vertical. For equilibrium in the car's frame, the horizontal and vertical forces balance:

Tsin⁡θ=mv2rT \sin \theta = \dfrac{mv^2}{r}

Tcos⁡θ=mgT \cos \theta = mg

Dividing the two equations, we get:

tan⁡θ=v2rg\tan \theta = \dfrac{v^2}{rg}

Substituting the given values v=15v = 15 m/s, r=20r = 20 m, and g=10g = 10 m/s2^2:

tan⁡θ=(15)220×10=225200=1.125\tan \theta = \dfrac{(15)^2}{20 \times 10} = \dfrac{225}{200} = 1.125

θ=tan⁡−1(1.125)\theta = \tan^{-1}(1.125)

Answer: tan⁡−1(1.125)\tan^{-1}(1.125)

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