A stone has a mass of 3.40 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 0.916. When the tire surface is rotating at 18.8 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 2.50 N. Assume that only static friction supplies the centripetal force, and determine the radius r of the tire (in terms of m).
Friction force F(fr) = μ•N = 0.916•2.5 = 2.29 N.
F(total) = 2•F(fr) = 2•2.29 = 4.58 N.
F(total) = F(centripet) = mv^2/R.
R = mv^2/F(total) = 3.4•10^-3•(18.8)^2/4.58 = 0.262 m
To solve for the radius of the tire, we can start by finding the centripetal force acting on the stone using the given information.
The centripetal force is given by the formula:
Fc = (m * v^2) / r
Where:
Fc is the centripetal force
m is the mass of the stone (converted to kg)
v is the velocity of the tire surface
r is the radius of the tire
Given:
Mass of the stone, m = 3.40 g = 0.0034 kg
Velocity of the tire surface, v = 18.8 m/s
Substituting the given values into the formula:
Fc = (0.0034 kg * (18.8 m/s)^2) / r
Next, we need to determine the maximum static friction force, which is given by:
Fs_max = μ * FN
Where:
Fs_max is the maximum static friction force
μ is the coefficient of static friction
FN is the normal force (the force exerted by the sides of the tread channel)
Given:
Coefficient of static friction, μ = 0.916
Normal force, FN = 2.50 N
Substituting the given values into the formula:
Fs_max = 0.916 * 2.50 N
Now, since the stone flies out of the tread, the maximum static friction force becomes equal to the centripetal force:
Fs_max = Fc
Setting these two equations equal to each other:
0.916 * 2.50 N = (0.0034 kg * (18.8 m/s)^2) / r
Simplifying the equation:
2.29 N = (0.0034 kg * 353.44 m^2/s^2) / r
Rearranging the equation to solve for r:
r = (0.0034 kg * 353.44 m^2/s^2) / 2.29 N
Calculating the value of r:
r = 0.522 m
Therefore, the radius of the tire, in terms of meters, is approximately 0.522 m.
To find the radius of the tire, we can use the concept of centripetal force. In this case, the static friction between the stone and the tread channel is providing the centripetal force.
The centripetal force is given by the formula:
Fc = (mv^2) / r
where Fc is the centripetal force, m is the mass of the stone, v is the velocity, and r is the radius of the tire.
First, let's convert the mass of the stone from grams to kilograms:
Mass (m) = 3.40 g = 3.40 * 10^-3 kg
Next, we need to find the centripetal force. The centripetal force is equal to the static friction force. The static friction force is equal to the coefficient of static friction multiplied by the normal force:
Fs = μs * FN
where Fs is the static friction force, μs is the coefficient of static friction, and FN is the normal force.
Given that the coefficient of static friction (μs) is 0.916 and the normal force (FN) is 2.50 N, we can substitute these values in the equation:
Fs = (0.916)(2.50 N) = 2.29 N
Now, we can equate the centripetal force to the static friction force:
Fc = Fs
(mv^2) / r = Fs
Substituting the values we have:
(3.40 * 10^-3 kg)(18.8 m/s)^2 / r = 2.29 N
Simplifying the equation:
(3.40 * 10^-3 kg)(18.8 m/s)^2 = 2.29 N * r
Solving for r:
r = (3.40 * 10^-3 kg)(18.8 m/s)^2 / 2.29 N
Calculating this expression will give you the value of the radius (r) of the tire in meters.