Provincial Exam: Circular Motion

AI Thread Summary
To determine the height from which a 3.0 kg roller coaster cart should be released to complete a loop with a radius of 6.0 m, the minimum velocity at the top of the loop must be calculated. The forces acting on the cart at the loop's peak include gravitational force and normal force, with the latter being zero when the cart just loses contact with the track. Applying Newton's second law and the equation for centripetal acceleration, the relationship between height and velocity can be established through energy conservation principles. The discussion emphasizes the need to find the normal force and its implications on the cart's motion. Ultimately, understanding these forces is crucial for solving the problem effectively.
Senjai
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Homework Statement



A frictionless 3.0 kg roller coaster cart rolls down an incline, and then "loops the loop."

At what height, h, should the cart be released so that it does not fall off the circular track.

The loop has a radius of 6.0m and a height of 12.0 m.

Homework Equations



a_c = \frac{v^2}{r}

The Attempt at a Solution



This is an old practice exam, so if this has anything to do with escape velocities, it has been taken out of our curriculum.

I don't know quite where to start.

I know that you would require a certain velocity, conceptually, to be able to make it around the loop.

And v \propto h because of the law of energy conservation. but i don't know what minimum velocity i would need?

I could kinda see a relation of using a centripetal acceleration to find out the velocity required and therefore the height, and find out the centripetal accelleration using Newtons Second, i just don't know how i can work it out.
 
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Senjai said:
I know that you would require a certain velocity, conceptually, to be able to make it around the loop.
Figure out the minimum speed at the top of the loop by examining the forces acting. Apply Newton's 2nd law.
 
Hi Doc Al, haven't seen you for a while.

At the top of the loop, their are two forces, both the normal force and the gravitational force are in the same direction.

But i don't know the normal force? I would assume that their is one because something has to keep it in its circular path. But i have no way to calculate it.

Once i have the normal force.. i would assume

F_c = F_N + F_g

m\frac{v^2}{r} = F_N + mg

Again though, i am unsure how to find the normal force, two unknowns and i only have one equation. Is their another equation I am neglecting?
 
What will the normal force be when the cart just barely loses contact with the track?
 
zero, so then gravity will F centrepetal? and calculate from there?
 
Exactly.
 
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