Roller Coaster Loop: Solve for Height and Motion

AI Thread Summary
To determine the height from which a roller coaster must start to stop at the top of a loop, it is concluded that the trolley must begin at a height of at least 2r, where r is the radius of the loop. The discussion emphasizes that for the trolley to reach the top without falling, it must maintain sufficient kinetic energy to counteract gravitational forces. If the trolley reaches the top and stops, it would not be able to stay in contact with the track due to insufficient centripetal acceleration. The conversation highlights that a push-only track cannot provide the necessary force to keep the trolley in motion at the top, leading to the conclusion that it cannot simply stop at the peak. Ultimately, the problem's assumptions and conditions must be clearly defined to arrive at a correct solution.
rbn251
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Homework Statement



Hi, I am stuck on this thought experiment:

A roller coaster of mass m starts on a inclined plane at a certain height, and then enters a circular loop, with radius r. At what height, h on the plane, must the trolley start in order to stop at the exact top of the loop, and then what happens?

Homework Equations



gpe=g*h*m
total energy = KE + PE

The Attempt at a Solution



In the beginning the trolley has only PE which is ghm (for h unknown)
At the top of the loop, again KE is zero, and PE is g(2r)m.

So h=2r (ie same height as the loop), and the trolley falls vertically down.

Is this correct? Thanks,
 
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Looks alright .
 
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What prevents the trolley from falling down before reaching the top point?
 
Before it reaches the top (say at 1 oclock), it is moving in the direction of the track, so it keeps going round. As it moves upwards it losses KE to GPE, I think? Though I am not really sure how centripetal acceleration comes into this?
 
Imagine your are inside a spherical cavity. Can you just walk up to its top?
 
Assuming you have enough kinetic energy to begin with, then yes?
 
No, don't assume that. You are a human, a mere mortal. Can you walk up to the top of say a 2-meter radius sphere, on its internal surface? And can you, a mere mortal, walk up the stairs to a platform that is 4 m higher? Assuming you answer honestly, can explain the difference?
 
I suppose the difference is the direction of the contact force which opposes gravity only in the stairs case?
 
Very well. Now let's say you have a powerful motorcycle. You can make it to the top with that contraption. Why is suddenly the contact force complication no longer a complication?
 
  • #10
You now have some sort of inertia in the upwards direction? sorry I don't know.
 
  • #11
Even if you start at the bottom?
 
  • #12
The track pushes you towards the middle??
 
  • #13
Voko, there are roller coasters that do experience negative g. Those typically are constructed such that the track can give accelerations in both directions. It is here reasonable to assume this is the case, since the problem does not have a solution otherwise - you cannot stop on top without the track providing a force toward the outside of the loop.
 
  • #14
#1 says the trolley falls vertically down in the end. That clearly assumes the track can only push, but not pull.

Either way, the complete answer in #1 is incorrect.
 
  • #15
Agreed, the problem and proposed solution make incompatible assumptions.

It could be reformulated to ask for the minimum height for which the full loop is performed before falling. This was already discussed here (for example)
https://www.physicsforums.com/showthread.php?t=750306
 
  • #16
Ah, so is it not possible that the trolley can get to the exact top and stop? It either keeps going round, or never gets there? For the original question let's assume that the track cannot pull.
 
  • #17
Yes, if the trolley gets to the top, on a push-only track, it will then complete the loop.
 
  • #18
But surely there is a starting height and equivalent speed of entry to the loop at which it just makes it to the top, it's speed drops to zero, and it falls vertically down? Can you explain why this is not possilbe?
 
  • #19
No, this is not possible. I suggest you read the thread I linked above.

If you are at the top and want to continue along the loop, the centripetal acceleration must be v^2/r. Anything bigger than that and you fall. The least acceleration occurs when the force from the track is zero and the acceleration is then due to gravity only, so g. This gives you the minimum velocity the coaster must have at the top to reach the top. Anything less and it falls earlier.
 
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  • #20
Ah, I see now, thanks!
 
  • #21
A body follows a circular path only if at every point of the path the acceleration perpendicular to the direction of motion is the one required for circular motion at the body's velocity at that point and radius of the path. Is that condition if the body's velocity is zero at the top of the track?
 
  • #22
rbn251 said:
But surely there is a starting height and equivalent speed of entry to the loop at which it just makes it to the top, it's speed drops to zero, and it falls vertically down? Can you explain why this is not possilbe?
Suppose it can make it to the top, coming to rest (at least for the moment) as it does so. Then just before that it must have been moving quite slowly. How was it staying in contact?
 
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