a tire 0.500 m in radius rotates at a constant rate of 200 rev/min.find the speed and accelation of a small stone lodge in the thread(on its outer edge).
C = pi * 2r = 3.14 * 1 = 3.14 m.
V=200rev/min * 3.14m/rev * 1min/6=10.5 m/s.
To find the speed and acceleration of a small stone lodged in the tread of a rotating tire, we can start by calculating the linear speed of a point on the outer edge of the tire.
The linear speed of a point on the outer edge of the tire can be calculated as the circumference of the tire multiplied by the number of revolutions per minute (RPM). The circumference of a circle is given by the formula C = 2πr, where r is the radius of the circle.
Given that the radius of the tire is 0.500 m, we can calculate the circumference of the tire as follows:
C = 2π(0.500) = 3.1416 m
Next, we need to convert the revolutions per minute (RPM) to revolutions per second (RPS) since acceleration is usually expressed in terms of seconds.
To convert from RPM to RPS, divide the RPM by 60:
RPS = 200 rev/min / 60 min/s = 3.333 rev/s
Now, we can calculate the linear speed (v) of a point on the outer edge of the tire using the formula:
v = C * RPS
v = 3.1416 m * 3.333 rev/s = 10.471 m/s
Therefore, the speed of a small stone lodged in the tread on the outer edge of the tire is 10.471 m/s.
To calculate the acceleration, we need to consider that the stone is undergoing circular motion and experiences centripetal acceleration.
The centripetal acceleration (a) can be calculated using the formula:
a = v² / r
where v is the linear speed and r is the radius of the circle.
Substituting the given values, we have:
a = (10.471 m/s)² / 0.500 m = 218.209 m/s²
Therefore, the acceleration of the small stone lodged in the tread of the tire is 218.209 m/s².