Question

A tire 0.650 m in radius rotates at a constant rate of 210 rev/min. Find the speed of a small stone lodged in the tread of the tire (on its outer edge). (Hint: In one revolution, the stone travels a distance equal to the circumference of its path, 2πr.)

Answer 1

Answer:

v = 14.3 m/s

Explanation:

As we know that frequency of rotation is given by

$f = 210 rev/min$

$f = 210\times \frac{1}{60} hz$

$f = 3.5 hz$

now the angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi(3.5 hz)$

$\omega = 7\pi rad/s$

now the tangential speed is given as

$v = r\omega$

$v = (0.650 m)(7\pi) m/s$

$v = 14.3 m/s$