A car is safely negotiating an unbanked circular turn at a speed of 18 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a long wet patch in the road decreases the maximum static frictional force to one-ninth of its dry-road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?
m v1^2/r = (9) m v2^2/r
v2^2/v1^2 = 1/9
v2/v1 = 1/3
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To find the speed at which the driver must slow the car, we need to equate the centripetal force required to maintain the circular motion with the maximum static frictional force available.
The centripetal force required can be calculated using the formula:
F_c = (m * v^2) / r
Where:
F_c = Centripetal force
m = Mass of the car
v = Velocity of the car
r = Radius of the circular turn
Since the turn is unbanked, the centripetal force is provided solely by the maximum static frictional force (F_max) available.
Therefore, we have:
F_c = F_max
Given that the maximum static frictional force on the wet patch is one-ninth (1/9) of its dry-road value, we can write the equation as:
(m * v^2) / r = (F_max dry) / 9
Canceling the r terms on both sides, we have:
(m * v^2) = (F_max dry) / 9
Now, we need to find the maximum static frictional force on the dry road.
Since the maximum static frictional force is equal to the coefficient of static friction (μ_s) multiplied by the normal force (N), and the normal force in this case is equal to the weight (mg) of the car, we can write:
F_max dry = μ_s * m * g
Where:
μ_s = Coefficient of static friction
m = Mass of the car
g = Acceleration due to gravity (approximately 9.8 m/s^2)
Substituting this value back into the previous equation, we have:
(m * v^2) = (μ_s * m * g) / 9
Canceling out the mass terms, we get:
v^2 = (μ_s * g) / 9
Finally, solving for the velocity (v):
v = √((μ_s * g) / 9)
Now we can substitute the known values to find the required velocity:
Given that the coefficient of static friction (μ_s) is not provided in the question, we cannot determine the exact required velocity to maintain safe negotiation around the curve.
To solve this problem, we need to use the concept of centripetal force, which is the force required to keep an object moving in a circular path. In this case, the maximum static frictional force acts as the centripetal force to keep the car moving in a circular path.
The formula for the centripetal force is:
F = (m * v^2) / r
where F is the centripetal force, m is the mass of the car, v is the velocity of the car, and r is the radius of the circular path.
In this problem, the maximum static frictional force decreases to one-ninth of its dry-road value, which means the maximum static frictional force on the wet road is now (1/9) times the maximum static frictional force on the dry road. Let's call the maximum static frictional force on the dry road F_d and the maximum static frictional force on the wet road F_w.
So we have:
F_w = (1/9) * F_d
Substituting this into the centripetal force formula, we get:
(1/9) * F_d = (m * v^2) / r
Now, let's solve for v, the velocity of the car on the wet road. To do this, we'll rearrange the equation:
v^2 = (F_w * r) / (m/9)
Now we can substitute the values given in the problem:
F_w = (1/9) * F_d
v^2 = (1/9) * F_d * r / (m/9)
We are given the velocity on the dry road (18 m/s), and we need to find the velocity on the wet road. So let's solve the equation by finding the ratio of velocities:
(18^2) / v_w^2 = (1/9) * F_d * r / (m/9)
Simplifying the equation:
v_w = sqrt((1/9) * 18^2)
v_w ≈ 6 m/s
Therefore, the driver must slow the car down to a speed of approximately 6 m/s to safely negotiate the wet patch on the road and continue around the curve.