A 765-kg car is travelling north and makes a gradual turn to the east at a constant speed of 15 m/s. The radius of the turn is 112 m.
I calculated the angular velocity to be 0.135 m/s and the friction force needed to be 1410.86 N.
What is the smallest radius for which the turn could be designed so that the car doesn't slip at this speed? Assume the coefficient of static friction is 0.65 and the road is level.
omega = v/R = 15 m/s / 112 m = 0.134 radians (not meters) /s
m Ac = m v^2/R = 0.65 * m *g
so at slip
v^2/R = .65 g
225/R = .65 g
R = 225/ .65 g = 225 / (.65*9.81) = 35.3 meters
Thank you so much
To determine the smallest radius for which the car doesn't slip at this speed, we need to consider the maximum static friction force that can act on the car to provide the necessary centripetal force.
The centripetal force required to keep the car moving in a circle is given by the equation:
F_c = (m * v^2) / r
where F_c is the centripetal force, m is the mass of the car, v is the speed of the car, and r is the radius of the turn.
We can rearrange this equation to solve for the radius:
r = (m * v^2) / F_c
Given that m = 765 kg, v = 15 m/s, and F_c = 1410.86 N, we can substitute these values into the equation to find the smallest radius:
r = (765 kg * (15 m/s)^2) / 1410.86 N
r ≈ 193.8 m
Therefore, the smallest radius for which the turn could be designed so that the car doesn't slip at this speed is approximately 193.8 meters.
To find the smallest radius for which the car doesn't slip, we need to consider the maximum force of static friction available. The car will not slip as long as the centripetal force required for the turn is less than or equal to the maximum force of static friction.
The maximum force of static friction can be calculated using the formula:
F_max = μ_s * N
Where:
F_max is the maximum force of static friction,
μ_s is the coefficient of static friction,
and N is the normal force exerted on the car.
In this case, since the road is level and there is no vertical acceleration, the normal force N is equal to the weight of the car, which can be calculated as:
N = m * g
Where:
m is the mass of the car,
and g is the acceleration due to gravity.
Given that the mass of the car is 765 kg, and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the normal force:
N = 765 kg * 9.8 m/s^2
Next, let's calculate the maximum force of static friction:
F_max = 0.65 * (765 kg * 9.8 m/s^2)
Now, let's find the centripetal force required for the turn. The centripetal force is given by:
F_c = (m * v^2) / r
Where:
F_c 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 turn.
Given that the mass of the car is 765 kg, and the velocity is 15 m/s, we can calculate the centripetal force required for the turn using the given radius of 112 m:
F_c = (765 kg * (15 m/s)^2) / 112 m
To ensure that the car doesn't slip, the centripetal force should be less than or equal to the maximum force of static friction:
F_c ≤ F_max
Now, let's substitute the values and solve for the smallest radius:
(765 kg * (15 m/s)^2) / r ≤ 0.65 * (765 kg * 9.8 m/s^2)
Simplifying the equation:
(765 kg * (15 m/s)^2) ≤ 0.65 * (765 kg * 9.8 m/s^2) * r
(765 kg * 225 m^2/s^2) ≤ (0.65 * 765 kg * 9.8 m/s^2) * r
(765 kg * 225 m^2/s^2) / (0.65 * 765 kg * 9.8 m/s^2) ≤ r
225 m^2/s^2 / (0.65 * 9.8 m/s^2) ≤ r
Now, we can calculate the value of r:
r = 225 m^2/s^2 / (0.65 * 9.8 m/s^2)
By plug in the numbers:
r ≈ 34.63 m
Therefore, the smallest radius for which the turn could be designed so that the car doesn't slip at this speed is approximately 34.63 meters.