On a banked race track, the smallest circular path on which cars can move has a radius of 114 m, while the largest has a radius of 170 m, as the drawing illustrates. The height of the outer wall is 17.5 m. Find (a) the smallest and (b) the largest speed at which cars can move on this track without relying on friction.

Question

On a banked race track, the smallest circular path on which cars can move has a radius of 114 m, while the largest has a radius of 170 m, as the drawing illustrates. The height of the outer wall is 17.5 m. Find (a) the smallest and (b) the largest speed at which cars can move on this track without relying on friction.

Answer 1

To find the smallest and largest speeds at which cars can move on the banked race track without relying on friction, we can use the concept of centripetal force.

(a) To find the smallest speed, we need to consider the car moving on the smallest circular path with a radius of 114 m.

From the given information, we know the radius (r = 114 m) and the height of the outer wall (h = 17.5 m). We can use this information to find the angle of inclination (θ) of the banked race track.

First, let's draw a diagram to illustrate the situation:

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In this diagram, the horizontal line represents the ground, the vertical line represents the height of the outer wall, and the slanted line represents the banked track.

The angle of inclination (θ) can be determined using the tangent function:

tan(θ) = h / r

Substituting the values we know:

tan(θ) = 17.5 m / 114 m

Now, we can find the angle (θ) by taking the inverse tangent of both sides:

θ = tan^(-1) (17.5 m / 114 m)

Using a calculator, the angle (θ) is approximately 8.68 degrees.

Next, we can use the concept of centripetal force to find the smallest speed (v) at which cars can move without relying on friction:

Centripetal force (Fc) = m * v^2 / r

In this case, we assume the friction force is zero, so the only force acting on the car is the vertical component of the normal force (n) provided by the banking track.

From the diagram, we can see that the vertical component of the normal force is given by:

n * cos(θ) = m * g

Where:
n = Normal force
m = Mass of the car
g = Acceleration due to gravity (approximately 9.8 m/s^2)

Simplifying, we have:

n = m * g / cos(θ)

Now, substituting the expression for n in the centripetal force equation:

m * g / cos(θ) = m * v^2 / r

Simplifying and rearranging, we can solve for the smallest speed (v):

v = sqrt(r * g * cos(θ))

Substituting the given values:

v = sqrt(114 m * 9.8 m/s^2 * cos(8.68 degrees))

Using a calculator, the smallest speed (v) is approximately 27.87 m/s.

(b) To find the largest speed, we need to consider the car moving on the largest circular path with a radius of 170 m.

Using the same process as above, we can find the largest speed (v) at which cars can move without relying on friction:

v = sqrt(r * g * cos(θ))

Substituting the given values:

v = sqrt(170 m * 9.8 m/s^2 * cos(8.68 degrees))

Using a calculator, the largest speed (v) is approximately 41.09 m/s.

So, the smallest speed at which cars can move on this track without relying on friction is approximately 27.87 m/s, and the largest speed is approximately 41.09 m/s.