A tire 0.500m in radius rotates at a constant rate of 200rev/min. Find the speed and acceleration of a small stone lodged in the tread of the tire (on its outer edge )
speed = 2 * π * r * 200/60 m/s
acceleration = [(speed)^2] / r
80,000
To find the speed and acceleration of the stone lodged in the tread of the tire, we can use the following steps:
Step 1: Convert the tire's rotational speed from revolutions per minute (rev/min) to radians per second (rad/s). Since 1 revolution is equal to 2π radians, we can use the following formula:
Angular speed (ω) = (2π * rotational speed) / 60
Plugging in the given values: ω = (2π * 200) / 60 = 20.94 rad/s
Step 2: Calculate the linear speed of the stone on the outer edge of the tire. The linear speed is the distance traveled per unit of time. It can be found by multiplying the radius of the tire (0.500m) by the angular speed (ω).
Linear speed (v) = radius * angular speed
Plugging in the given values: v = 0.500m * 20.94 rad/s = 10.47 m/s
Therefore, the speed of the stone lodged in the tread is 10.47 m/s.
Step 3: Calculate the acceleration of the stone. Since the stone is moving in a circle, it experiences centripetal acceleration towards the center of the circle. The centripetal acceleration (a) can be found using the following formula:
Centripetal acceleration (a) = (angular speed)^2 * radius
Plugging in the given values: a = (20.94 rad/s)^2 * 0.500m = 219.31 m/s^2
Therefore, the acceleration of the stone lodged in the tread is 219.31 m/s^2.