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
Show Answer
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
BETA
