On a banked race track, the smallest circular path on which cars can move has a radius of 119 m, while the largest has a radius 155 m, as the drawing illustrates. The height of the outer wall is 18 m.
(a) Find the smallest speed at which cars can move on this track without relying on friction.
m/s
(b) Find the largest speed at which cars can move on this track without relying on friction.
m/s
To find the smallest speed at which cars can move on the track without relying on friction, we can use the concept of centripetal force.
(a) Smallest speed without relying on friction:
The centripetal force required to keep the cars moving in a circular path is provided by the normal force and the component of the car's weight acting toward the center of the circle.
First, let's find the normal force on the car when it is at the top of the track. At the top, the car is in equilibrium, so the net force on it is zero. This means the normal force is equal in magnitude and opposite in direction to the gravitational force acting on the car. The weight of the car can be calculated using the formula W = mg, where m is the mass of the car and g is the acceleration due to gravity.
The normal force is equal to the weight of the car minus the downward component of the weight due to gravity, since it is acting towards the center of the circle. The downward component of the weight can be found using the equation mg*sin(θ), where θ is the angle of the slope.
At the top of the track, the angle of the slope is given by the equation tan(θ) = h / (R - r), where h is the height of the outer wall (18 m), R is the radius of the largest path (155 m), and r is the radius of the smallest path (119 m).
tan(θ) = 18 / (155 - 119)
tan(θ) = 18 / 36
θ = tan^(-1)(0.5)
θ ≈ 26.565 degrees
Now, we can find the normal force using the equation:
Normal force = mg - mg*sin(θ)
Let's assume the mass of the car is 1000 kg:
Normal force = 1000 kg * 9.8 m/s^2 - 1000 kg * 9.8 m/s^2 * sin(26.565 degrees)
Normal force ≈ 9482.17 N
The centripetal force required to keep the car moving in a circular path can be calculated using the formula F = m * v^2 / r, where m is the mass of the car, v is the velocity of the car, and r is the radius of the path.
Since the normal force is providing the centripetal force, we can equate the two:
Normal force = m * v^2 / r
Re-arranging the formula, we get:
v = √(Normal force * r / m)
Using the values we obtained:
v = √(9482.17 N * 119 m / 1000 kg)
v ≈ √(1128.98 m^2/s^2)
v ≈ 33.6 m/s
So, the smallest speed at which cars can move on this track without relying on friction is approximately 33.6 m/s.
(b) Largest speed without relying on friction:
Using a similar approach, we can find the largest speed at which cars can move on this track without relying on friction.
The angle of the slope at the bottom of the track can be found using the same equation:
tan(θ) = h / (R - r)
tan(θ) = 18 / (155 - 119)
tan(θ) = 18 / 36
θ = tan^(-1)(0.5)
θ ≈ 26.565 degrees
Since the normal force is equal to the weight of the car plus the upward component of the weight due to gravity:
Normal force = mg + mg*sin(θ)
Using the equation:
v = √(Normal force * r / m)
We can calculate the largest speed:
v = √((mg + mg*sin(θ)) * r / m)
Assuming the same mass of the car (1000 kg):
v = √((1000 kg * 9.8 m/s^2 + 1000 kg * 9.8 m/s^2 * sin(26.565 degrees)) * 119 m / 1000 kg)
v ≈ √(18078.3 N * 119 m / 1000 kg)
v ≈ √(2154107.7 m^2/s^2 / 1000)
v ≈ √(2154.11 m^2/s^2)
v ≈ 46.4 m/s
Therefore, the largest speed at which cars can move on this track without relying on friction is approximately 46.4 m/s.