A tire 0.650 m in radius rotates at a constant rate of 240 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, 2¥ðr.)
What is the acceleration of the stone?
and its direction?
a)away from the center of the path
b)toward the center of the path
c)tangent to the path
Angular speed of wheel = 240*(2 pi)/60 = 25.13 rad/s = w
Speed of stone (relative to axle) = R*w
The speed of the stone relative to the ground depends upon its height above the ground.
Centripetal acceleration of stone = R*w^2 = 411 m/s^2, , toward the center.
To find the speed of the small stone lodged in the tread of the tire, we need to consider the relationship between the rotation rate of the tire and the linear speed of the stone.
1. First, let's convert the rotation rate from rev/min to radians/second. Since one revolution is equal to 2π radians, the conversion factor is:
240 rev/min * (2π radians/1 rev) * (1 min/60 sec) = 8π radians/sec
2. The linear speed of the stone is equal to the circumference of its path, which is given by the formula C = 2πr, where r is the radius of the tire.
C = 2π * 0.650 m = 1.3π m
3. Therefore, the speed of the stone is equal to the linear speed divided by the time taken for one revolution, which is the rotation period. The rotation period can be calculated using:
T = 1 / (rotation rate)
T = 1 / (8π radians/sec) = 1 / (8π) sec
4. Finally, the speed of the stone is given by:
Speed = (linear speed) / (rotation period) = (1.3π m) / (1 / 8π sec) = 10.4 m/s
Now let's determine the acceleration of the stone and its direction:
The acceleration of an object moving in a circular path is given by the centripetal acceleration formula:
a = v^2 / r
Where:
a is the acceleration
v is the velocity (speed)
r is the radius of the path
In this case, the stone is moving in a circular path, so the acceleration is directed toward the center of the path. Therefore, the answer is option (b) - toward the center of the path.
Substituting the values we calculated earlier:
a = (10.4 m/s)^2 / 0.650 m = 169.07 m/s^2
So, the acceleration of the stone is 169.07 m/s^2 toward the center of the path.