Homework Help: Roller Coaster Loop

1. Aug 17, 2014

rbn251

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. 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,

2. Aug 17, 2014

Vibhor

Welcome to PF!

Looks alright .

Last edited: Aug 17, 2014
3. Aug 17, 2014

voko

What prevents the trolley from falling down before reaching the top point?

4. Aug 17, 2014

rbn251

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?

5. Aug 17, 2014

voko

Imagine your are inside a spherical cavity. Can you just walk up to its top?

6. Aug 17, 2014

rbn251

Assuming you have enough kinetic energy to begin with, then yes?

7. Aug 17, 2014

voko

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?

8. Aug 17, 2014

rbn251

I suppose the difference is the direction of the contact force which opposes gravity only in the stairs case?

9. Aug 17, 2014

voko

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. Aug 17, 2014

rbn251

You now have some sort of inertia in the upwards direction? sorry I don't know.

11. Aug 17, 2014

voko

Even if you start at the bottom?

12. Aug 17, 2014

rbn251

The track pushes you towards the middle??

13. Aug 17, 2014

Orodruin

Staff Emeritus
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. Aug 17, 2014

voko

#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. Aug 17, 2014

Orodruin

Staff Emeritus
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)

16. Aug 17, 2014

rbn251

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 lets assume that the track cannot pull.

17. Aug 17, 2014

voko

Yes, if the trolley gets to the top, on a push-only track, it will then complete the loop.

18. Aug 17, 2014

rbn251

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. Aug 17, 2014

Orodruin

Staff Emeritus
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.

20. Aug 17, 2014

rbn251

Ah, I see now, thanks!

21. Aug 17, 2014

voko

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. Aug 17, 2014

haruspex

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?