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

v = 14.3 m/s

Step-by-step explanation:

As we know that frequency of rotation is given by

`f = 210 rev/min`

`f = 210* (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`