Question

On a banked race track, the smallest circular path on which cars can move has a radius of 104 m, while the largest has a radius of 171 m, as the drawing illustrates. The height of the outer wall is 19.0 m. Find (a) the smallest and (b) the largest speed at which cars can move on this track without relying on friction.

Answer 1

v_smallest = 289 m/s

v_largest = 475.23 m/s

Step-by-step explanation:

I've attached a free body diagram to illustrate this problem.

The angle of incline of the track would be given by:

tanθ = (19)/(171 - 104) = 0.2836

θ = tan^(-1)0.2836

θ = 15.83°

Now, when going around the bend, the centrifugal force pushing the car horizontally to the outside of the track is given by:

F_c = mv²/r

The component of this force pushing up the slope is given by:

F_c(up slope) = mv²/(rcosθ) - - - eq1

The component of gravity pulling the car down the slope is given as:

F_g(down slope) = mg•sinθ -- eq(2)

Now, to find the maximum or minimum speed, let's equate eq1 to eq2.

Thus;

mv²/(rcosθ) = mg•sinθ

This gives;

v² = r•g•tan(θ)

Smallest speed will be at smallest radius, thus;

v_smallest = 104 x 9.8 x tan 15.83

v_smallest = 104 x 9.8 x 0.2836

v_smallest = 289 m/s

Similarly largest speed will be at largest radius. Thus;

v_largest = 171 x 9.8 x tan 15.83

v_largest = 171 x 9.8 x 0.2836 =

v_largest = 475.23 m/s

Answer 2

A) v_smallest = 289 m/s

B) v_largest = 475.23 m/s

Step-by-step explanation:

I've attached a free body diagram to illustrate this problem.

The angle of incline of the track would be given by:

tanθ = (19)/(171 - 104) = 0.2836

θ = tan^(-1)0.2836

θ = 15.83°

Now, when going around the bend, the centrifugal force pushing the car horizontally to the outside of the track is given by:

F_c = mv²/r

The component of this force pushing up the slope is given by:

F_c(up slope) = mv²/(rcosθ) - - - eq1

The component of gravity pulling the car down the slope is given as:

F_g(down slope) = mg•sinθ -- eq(2)

Now, to find the maximum or minimum speed, let's equate eq1 to eq2.

Thus;

mv²/(rcosθ) = mg•sinθ

This gives;

v² = r•g•tan(θ)

Smallest speed will be at smallest radius, thus;

v_smallest = 104 x 9.8 x tan 15.83

v_smallest = 104 x 9.8 x 0.2836

v_smallest = 289 m/s

Similarly largest speed will be at largest radius. Thus;

v_largest = 171 x 9.8 x tan 15.83

v_largest = 171 x 9.8 x 0.2836 =

v_largest = 475.23 m/s