Rotational Motion Question 18

Question: If a cyclist turns around a right-angled corner at a speed of 16 feet/sec, making a turn in 0.8 seconds, then he must lean over at an angle of nearly:

Options:

A) 45°

B) 90°

C) 15°

D) More than one of the above

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Answer:

Correct Answer: A

Solution:

  • [a]Given speed = 16 ft/sec

The cyclist makes a turn in 0.8 seconds.

Angular velocity $(\omega = \frac{\pi}{1.6})$ rad/s

Centripetal acceleration $(\alpha = v\omega = 10\pi)$ ft/s²

Angle of leaning (\theta) is given by $(\tan \theta = \frac{a}{g} = 10\pi/32 = 0.98175)$

$(\theta = \tan^{-1}$(0.98175)$ \approx 45°)$

Therefore, the cyclist must lean over at an angle of nearly 45°1



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