A 1200-kg automobile is stolen by the notorious Physics joyriders known as “the centripetal gang”. The gang tests
their grasp on Newtonian physics by cruising around a banked circular track of radius 1025 m. From previous
‘experiments’ the gang has discovered the car’s coefficient of static friction to be μS =0.260, and that the sharp bank is
29.5°.
2.00 m
35.0°
(i) The gang performs a dangerous experiment to find the maximum speed the car can travel the track without
skidding. Use a free-body diagram and Newton’s equations to calculate this speed.
(ii) A conscientious member of the gang states that it might be safer to instead test for the minimum speed for the carto not slip into the bank. Solve for this speed (Note – you should not have to start from scratch.)
To calculate the maximum speed the car can travel without skidding, we need to find the centripetal force required to keep the car moving in a circular path. The maximum static friction force between the car's tires and the road will provide this centripetal force.
First, let's calculate the maximum static friction force. The formula for static friction is:
fs ≤ μS * N
Where fs is the static friction force, μS is the coefficient of static friction, and N is the normal force.
The normal force acting on the car can be calculated as the component of the car's weight perpendicular to the surface of the track. Since the track is banked, we can calculate the normal force using the following formula:
N = m * g * cos(θ)
Where m is the mass of the car, g is the acceleration due to gravity, and θ is the angle of the bank (29.5°).
Next, we need to calculate the maximum speed the car can travel without skidding. The centripetal force required to maintain a circular path can be calculated using the formula:
Fc = m * v^2 / r
Where Fc 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 track.
Finally, we can equate the maximum static friction force to the centripetal force and solve for the maximum velocity:
fs = Fc
μS * N = m * v^2 / r
Now, we can plug in the given values and solve for the maximum speed.
(i) Maximum speed without skidding:
Given:
m = 1200 kg
r = 1025 m
θ = 29.5°
μS = 0.260
g = 9.8 m/s^2
Calculate N:
N = m * g * cos(θ)
N = 1200 kg * 9.8 m/s^2 * cos(29.5°)
Calculate fs:
fs = μS * N
fs = 0.260 * N
Calculate Fc:
Fc = m * v^2 / r
Fc = 1200 kg * v^2 / 1025 m
Equating fs and Fc:
μS * N = m * v^2 / r
Substitute the values of N and Fc:
0.260 * (1200 kg * 9.8 m/s^2 * cos(29.5°)) = 1200 kg * v^2 / 1025 m
Solve for v:
v = sqrt((0.260 * (1200 kg * 9.8 m/s^2 * cos(29.5°)) * 1025 m) / 1200 kg)
(ii) Minimum speed without slipping:
The minimum speed at which the car will not slip into the bank is when the friction force is equal to the gravitational force pulling the car downhill. In this case, we can calculate the minimum speed using the same formula as before, but with a friction force equal to the gravitational force:
fs = mg * sin(θ)
Substituting this into the equation for the centripetal force, we have:
mg * sin(θ) = m * v^2 / r
Solve for v:
v = sqrt((g * r * sin(θ))
Now, you can substitute the given values of r, θ, and g to find the minimum speed without slipping.