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Question Regarding Circular Motion and Normal Forces

  • Thread starter Chan M
  • Start date
  • #1
Chan M

Homework Statement


A roller coaster car has a mass of 500 kg when fully loaded with passengers. The path of the coaster from its initial point involves only up and down motion with no motion to the left or right. (A) If the vehicle has a speed of 20 m/s at the bottom of the first dip which follows a circular path with a radius of 10 m, what is the force exerted by the track on the car at this point?

(B) What is the maximum speed the roller coaster can have when it reaches the next hill which follows a circular path of 15 m and still remain on the track?

Homework Equations


Ac = (v^2) / r
Net force in y direction is equal to the normal force minus the gravitational force
(ny - Fg = 0)

The Attempt at a Solution


Ac = (20 m/s)^2 / 10m
Ac = 40 m/s^2


The answer to part A is the normal force the track exerts onto the car. This force is equal to gravity plus something else. The track exerts more normal force because the car is moving in a circular path. However, in my Free-Body Diagram, I have only the normal force and the force of gravity, both of which are equal and opposite. The question is not asking for the Fg since Fg does not equal the normal.

I am thinking there must be another downward force to cause the normal to be greater than Fg. I don't need much help with the question, just these concepts so I can apply them to other questions.

EDIT:

So I have been looking at other websites looking for a adequate answer. I keep seeing that the solution to part A is "centripetal force + m*g" or "Ac*m + Fg".

The thing is... my teacher has only talked about Ac and not Ac*m so....
Yeah, new "concept." So Ac*m is the force an object experiences due to centripetal acceleration? And this force always points toward the center of rotation or away from it? Can someone give me a rundown of how centripetal acceleration is related to centripetal force, Fg, and the normal force in this equation? Thanks!
 
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Answers and Replies

  • #2
Orodruin
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both of which are equal and opposite
No they are not. Why would they be? They do not form a third-law pair.
You seem to be thinking of a static situation. The car is not static, it is accelerating.
 
  • #3
CWatters
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The answer to part A is the normal force the track exerts onto the car. This force is equal to gravity plus something else. The track exerts more normal force because the car is moving in a circular path. However, in my Free-Body Diagram, I have only the normal force and the force of gravity, both of which are equal and opposite.
They are equal because you have only drawn the forces for a stationary car. As Orodruin says.. in this case the car is accelerating towards the centre of the curve. There is a name for this force which I'm sure you know.
 

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