Question

In a loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has a mass m = 286 kg and moves with speed v = 13.82 m/s. The loop-the-loop has a radius of R = 8 m.

1)

What is the magnitude of the normal force on the care when it is at the bottom of the circle? (But as the car is accelerating upward.)

2)

What is the magnitude of the normal force on the car when it is at the side of the circle (moving vertically upward)?

3)

What is the magnitude of the normal force on the car when it is at the top of the circle?

4)

Compare the magnitude of the cars acceleration at each of the above locations:

A) a_bottom = a_side = a_top

B) a_bottom < a_side < a_top

C) a_bottom > a_side > a_top

5)

What is the minimum speed of the car so that it stays in contact with the track at the top of the loop?

Answer 1

Given, m = 286 kg

v = 13.82 m/s

R = 8 m

Centripetal force, Fc = mv^2 /R = 286*(13.82)^2 /8 = 6827.98 N

1) At the bottom of the circle,

F = Fc + mg = 6827.98 + 280*9.81 = 9574.78 N

2) At the side of the circle,

F = Fc = 6827.98 N

3) At the top of the circle,

F = Fc - mg = 6827.98 - 280*9.81 = 4081.18 N

4) a(bottom) = 9574.78/280 = 34.2 m/s^2

a(side) = 6827.98/280 = 24.38 m/s^2

a(top) = 4081.18/280 = 14.57 m/s^2

Option C is correct.

a(bottom) > a(side) > a(top)

5) Minimum speed = sqrt(g*r) = sqrt(9.81*8) = 8.86 m/s