A stone has a mass of 6.01 g and is wedged into the tread of an automobile tire, as the drawing shows. The coefficient of static friction between the stone and each side of the tread channel is 0.948. When the tire surface is rotating at 14.7 m/s, the stone flies out of the tread. The magnitude FN of the normal force that each side of the tread channel exerts on the stone is 2.09 N. Assume that only static friction supplies the centripetal force, and determine the radius r of the tire (in terms of m).
r = (FN/μs) / (mv^2/r)
r = (2.09 N/0.948) / (6.01 g * (14.7 m/s)^2/r)
r = 0.0223 m
To solve this problem, we can use the following steps:
Step 1: Identify the known values from the problem.
- Mass of the stone (m) = 6.01 g = 0.00601 kg
- Coefficient of static friction (μs) = 0.948
- Speed of the tire surface (v) = 14.7 m/s
- Magnitude of the normal force (FN) = 2.09 N
Step 2: Set up the equation for the centripetal force.
The centripetal force required to keep the stone in circular motion is given by the equation:
Fc = m * v^2 / r
where Fc is the centripetal force, m is the mass of the stone, v is the speed of the tire surface, and r is the radius of the tire.
Step 3: Determine the centripetal force.
Since only static friction (Fs) supplies the centripetal force, we can write:
Fc = Fs = μs * FN
where Fs is the force of static friction and FN is the normal force.
Step 4: Substitute the known values into the equation.
Substituting the values from the problem into the last equation, we get:
μs * FN = m * v^2 / r
Step 5: Solve for the radius (r).
Re-arranging the equation to solve for r, we have:
r = m * v^2 / (μs * FN)
Plugging in the known values, we have:
r = 0.00601 kg * (14.7 m/s)^2 / (0.948 * 2.09 N)
Step 6: Calculate the radius (r) using the formula.
Evaluating the expression, we find:
r ≈ 0.00601 kg * 215.49 m^2/s^2 / (0.948 * 2.09 N)
r ≈ 3.139 m
Therefore, the radius of the tire is approximately 3.139 m.
To find the radius of the tire, we need to start by using the concept of centripetal force.
In this scenario, the only force providing the centripetal force is the static friction between the stone and the tread of the tire. The centripetal force can be calculated using the following equation:
Fc = m * v² / r
Where Fc is the centripetal force, m is the mass of the stone, v is the velocity of the tire, and r is the radius of the tire.
From the given information, we know that the stone flies out of the tread when the tire is rotating at 14.7 m/s. We can assume that at this point, the static friction force reaches its maximum value. Therefore, the maximum static friction force is equal to the centripetal force:
Fs_max = Fc
To find the maximum static friction force, we can use the coefficient of static friction (μs) and the normal force (FN) acting on the stone. The maximum static friction force can be calculated using the equation:
Fs_max = μs * FN
Given that the coefficient of static friction (μs) is 0.948 and the magnitude of the normal force (FN) is 2.09 N, we can substitute these values into the equation to find Fs_max.
Fs_max = 0.948 * 2.09 N = 1.98092 N
Now, we have Fs_max, which is equal to Fc. So we can rewrite the equation as:
1.98092 N = m * v² / r
We are given the mass of the stone (m) as 6.01 g, but we need to convert it to kilograms for consistency in the equation. 1 gram is equal to 0.001 kilograms, so:
m = 6.01 g * 0.001 kg/g = 0.00601 kg
Substituting the known values into the equation, we have:
1.98092 N = 0.00601 kg * (14.7 m/s)² / r
Simplifying the equation:
1.98092 N = 0.00601 kg * 215.49 m²/s² / r
Now, we can solve for r:
r = 0.00601 kg * 215.49 m²/s² / 1.98092 N
r ≈ 0.0655 m
Therefore, the radius of the tire is approximately 0.0655 meters or 65.5 mm.