Vertical circle-direction of a force?

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Homework Help Overview

The discussion revolves around a physics problem involving a car moving in a vertical circle, specifically focusing on the forces acting on the car at the top of the circle. The subject area includes concepts of circular motion, forces, and dynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between weight, normal force, and centripetal force in the context of vertical circular motion. Questions arise regarding the direction of the normal force at different speeds and the implications of these forces in real-life scenarios.

Discussion Status

The discussion is active, with participants questioning the assumptions about force directions and exploring the implications of upward versus downward normal forces. Some guidance is offered regarding the relationship between forces, but no consensus has been reached.

Contextual Notes

Participants note the specific conditions of the problem, including the weights and speeds involved, and question the implications of the normal force's direction in practical terms. There is an acknowledgment of the complexity of the forces at play without resolving the underlying questions.

lu6cifer
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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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