Vertical circle-direction of a force?

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In a vertical circle, the direction of the force on a car from the boom depends on the car's speed and the balance of forces acting on it. At the top of the circle, if the speed is 5.0 m/s, the normal force is 3.59 kN directed upwards, indicating that the weight is greater than the centripetal force. Conversely, at 15 m/s, the normal force becomes -4.11 kN, directed downwards, meaning the centripetal force exceeds the weight. The discussion highlights that when the centripetal force is less than the weight, the normal force acts upwards, potentially causing riders to feel pushed against the ceiling of the car. Understanding these dynamics is crucial for safety in amusement park rides.
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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
 
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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.
 


rl.bhat said:
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.

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?
 


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.
 
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