Vertical circle-direction of a force?

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SUMMARY

The discussion focuses on the forces acting on a car moving in a vertical circle at an amusement park ride. Given a weight of 4.5 kN and a radius of 12 m, two scenarios are analyzed: at a speed of 5.0 m/s, the normal force (N) is 3.59 kN directed upwards, while at 15 m/s, N is -4.11 kN directed downwards. The key conclusion is that the direction of the normal force depends on the relationship between the centripetal force (mv²/r) and the weight (W), where a greater centripetal force results in a downward normal force, indicating a net force direction towards the ground.

PREREQUISITES
  • Understanding of centripetal force and its calculation (mv²/r)
  • Knowledge of forces acting on objects in circular motion
  • Familiarity with the concepts of weight (W) and normal force (N)
  • Basic principles of physics related to motion in vertical circles
NEXT STEPS
  • Study the effects of varying speeds on centripetal force in vertical circular motion
  • Explore the implications of normal force direction in real-world scenarios
  • Learn about the role of centrifugal force in non-inertial reference frames
  • Investigate the dynamics of amusement park rides and safety considerations
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Physics students, mechanical engineers, amusement park ride designers, and anyone interested in the dynamics of circular motion and forces in vertical systems.

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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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