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 = 280 kg and moves with speed v = 14.48 m/s. The loop-the-loop has a radius of R = 8.9 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)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?

Answer 1

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

For #1 I calculated the normal force to be 2744, which is incorrect because the car is accelerating upward the loop, I don't understand how to find normal force with acceleration.

For #2,3,and 4 I tried using many different approaches, ie. Number #3 I thought the magnitude of the normal force at the top of the circle would just be mg, but thats incorrect.

If anyone could help that would be awesome.

Answer 2

1. At the bottom of the loop, the normal force is the greatest: \[F _{gravity}+F _{centripetal}\]
2.On the sides, force of gravity is orthogonal to the normal force, so \[F _{normal}=F _{centripetal}.\]
3. At the top of the loop, \[F _{normal}=F _{centripetal}-F _{gravity}.\]
4. At the top of the loop, the car must be travelling at a threshold speed, v, in order to maintain a normal force, and \[F _{gravity}=F _{centripetal}. Then solve for v _{minimum}.\]
I hope this helps.