Question

6. A tire 0.500 m in radius rotates at a constant rate of 300 revolutions per minute. Find the speed and acceleration of a small stone lodged in the tread of the tire (on its outer edge).

Answer 1

The speed and acceleration of the stone is 15.7 m/s and 31.4 $m/s^{2}$.

What is acceleration?

Acceleration is the rate at which an object's velocity with respect to time changes. They are vector quantities, accelerations. The direction of the net force acting on an object determines the direction of its acceleration.

Calculation of the speed of stone lodged in the tread of the tire:

Given, radius of tire (r) = 0.500 m

$V_{t} =\frac{2\pi r}{T} \\=\frac{2\pi (0.500)}{(60s/300rev)}$

= 15.7 m/s

Calculation of it's acceleration:

$a=\frac{v^{2} }{R}\\\\\= \frac{15.7}{0.500}$

= 31.4 $m/s^{2}$ inward

Hence, the speed and acceleration of the stone is 15.7 m/s and 31.4 $m/s^{2}$

Learn more about velocity here:

brainly.com/question/18084516

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