A car moves with a speed v on a horizontal circular highway turn of radius is R = 100. Assume the height of the car’s center of mass above the ground is h = 1 m, and the separation between its inner and outer wheels (car's width) is w = 2 m. The car does not skid (static friction works between tires and road surface).
With what maximum speed (in mph) can the car negotiate the turn with without losing equilibrium and starting to overturn? Use g = 10 m/s^2.
Hint 1: car is in circular motion with constant speed. Hint 2: car starts to rotate about its center of gravity when net torque about its center of gravity becomes non-zero. Hint 3: draw an extended free body diagram and identify two torques counteracting each other: torque due to friction force on the inner tire versus torque due to normal force on the outer tire (assume that at the moment the car lifts its inner tires, the normal force becomes equal to the car’s weight: N = mg). Hint 4: friction force is in perpendicular direction to the direction of car’s velocity (it is towards the center of the circle). So, Newton’s 2nd law: F_net = mv2/R. Hint 5: friction force is the only force acting on the car in horizontal direction: F_net = F_fr.
Given
Radius , r = 100 m
Height , h = 1 m
Width ,w = 2 m
We will find angle of banking
sinθ = 1 / 2
θ = 300
The speed of the car is
v = gr tan30
= 10 m/s^2 * 100 m *0.5773
v = 24 m/s
53.686471 mph
That answer is wrong, it has to be in MPH btw, please help me.
To find the maximum speed of the car in mph, we can follow the given hints and use the concept of torque.
First, we need to determine the angle of banking (θ) for the turn. The hint mentions that the car does not skid, so it implies that the friction force between the tires and the road provides the centripetal force required for circular motion. We can use the value of sinθ given in the question.
sinθ = h / r
sinθ = 1 / 100
θ = arcsin(1 / 100)
θ ≈ 0.573°
Now, we can calculate the maximum speed (v) at which the car can negotiate the turn without losing equilibrium. The hint suggests that at this speed, the net torque acting on the car about its center of gravity becomes zero.
The torque due to the friction force (τ_friction) on the inner tire acts in the opposite direction to the torque due to the normal force (τ_normal) on the outer tire.
τ_friction = w/2 * μ_friction * mg * r ----(1)
τ_normal = w/2 * mg * R ----(2)
Here, μ_friction is the coefficient of friction between the tires and the road, and g is the acceleration due to gravity. The normal force (mg) equals the car's weight.
Setting τ_friction equal to τ_normal:
w/2 * μ_friction * mg * r = w/2 * mg * R
μ_friction * r = R
μ_friction = R / r
μ_friction = 100 / 100
μ_friction = 1
Since the car is not skidding, the friction coefficient (μ_friction) equals 1. This means the maximum static friction force is equal to the weight of the car.
Now, we can use the second hint to find the net force acting on the car.
F_net = mv² / R ----(3)
Here, m is the mass of the car, and v is the velocity at which the car negotiates the turn.
Since F_net is provided by the friction force (F_friction):
F_friction = m * v² / R ----(4)
We can equate F_friction with the maximum static friction force (μ_friction * mg).
m * v² / R = μ_friction * mg
Since m cancels out:
v² = μ_friction * g * R
Substituting the values:
v² = 1 * 10 * 100
v² = 1000
v = √1000
v = 31.62 m/s
Now, we can convert the speed to mph:
1 m/s = 2.23694 mph
v (in mph) = 31.62 * 2.23694
v ≈ 70.74 mph
Therefore, the maximum speed at which the car can negotiate the turn without overturning is approximately 70.74 mph.