A tire 0.500 m in radius rotates at a constant rate of 200 rev/min. Find the speed and acceleration of a small stone lodged in the tread of the tire ( on its outer edge ).

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To find the speed and acceleration of the stone lodged in the tread of the tire, we can use some basic principles of circular motion.

First, let's find the circumference of the tire. The circumference of a circle is given by the formula C = 2πr, where r is the radius of the circle. In this case, the radius is given as 0.500 m, so the circumference is:

C = 2π(0.500 m) = 3.1416 m

Since the tire rotates at a constant rate of 200 revolutions per minute, we can find the angular velocity (ω) of the tire. The angular velocity is defined as the angle rotated per unit of time. In this case, one revolution corresponds to 2π radians, so the angular velocity can be calculated as:

ω = 2π × 200 rev/min = 400π rad/min

Next, let's convert the angular velocity to radians per second. Since there are 60 seconds in a minute, we can divide the angular velocity by 60:

ω = (400π rad/min) ÷ 60 = 20π rad/s

Now, the speed (v) of the stone can be found by multiplying the circumference of the tire by the angular velocity:

v = C × ω = 3.1416 m × 20π rad/s = 62.832 m/s

Therefore, the speed of the stone is approximately 62.832 m/s.

To find the acceleration (a) of the stone, we can use the formula for centripetal acceleration:

a = ω^2 × r

Plugging in the values for ω and r, we get:

a = (20π rad/s)^2 × 0.500 m = 200π^2 m/s^2

Therefore, the acceleration of the stone is approximately 200π^2 m/s^2.

the solution of it.

C = pi * 2r = 3.14 * 1.0 = 3.14 m. =

Circumference.

V = 200rev/min * 3.14m/rev * 1min/60s =
10.5 m/s.