# Vertical circle-direction of a force?

1. Dec 4, 2008

### lu6cifer

Vertical circle--direction of a force?

Vertical circle--direction of a force?
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 4.5 kN, and the circle's radius is 12 m.

(a) At the top of the circle, what are the magnitude Fb and direction (up or down) of the force on the car from the boom if the car's speed is v = 5.0 m/s?

(b) At the top of the circle, what are the magnitude Fb and direction (up or down) of the force on the car from the boom if the car's speed is v = 15 m/s?

I have the work and the answers, so I know how this works mathematically. For a, the normal force is positive, so the direction of force is up. But, for b, the normal force is negative, so the direction of force is down. However, since they're at the top for both scenarios, shouldn't there only be one direction of force anyway?

[Note: direction of motion is positive, W = weight, N = normal force]
For a:
W - N = mv2 / r
(v = 5 m/s)
N = 3.59kN

For b:
W - N = mv2 / r
(v = 15 m/s)
N = -4.11kN

2. Dec 4, 2008

### rl.bhat

Re: Vertical circle--direction of a force?

When the car is moving in the vertical circle, centrifugal force acts on it which is away from the centre. When the weight is more than this force, the net force is in the down ward direction. Other wise it is upwards.

3. Dec 4, 2008

### lu6cifer

Re: Vertical circle--direction of a force?

So, when the centripetal force, ie, the mv2/r part of the equation is less than W, that means that the normal force is directed upwards, and if it's more than W, the normal force is directed downwards?

And, what would an upward normal force mean as opposed to a downward one if this situation were in real life?

4. Dec 4, 2008

### rl.bhat

Re: Vertical circle--direction of a force?

In the real life, a person sitting in the car will be thrown to the roof of the car if the normal force is upwards.