What velocity does a train need to go up and down the hill

[Jorgen1224]

Homework Statement

There is a train of length d and speed v. It is heading towards a hill with height h and length of each side l. What velocity requirement must be met so that the train can go up and down the hill?

Homework Equations

E_k=m⋅v²/2
E_p=m⋅g⋅h

The Attempt at a Solution

I'm basically stuck at conversation of energy. Train needs to have kintetic energy equivalent to potential energy mgh, but calculating v from this equation seems pointless since it doesn't include either length. I have no idea how to include either of them.

 

[Orodruin]

Train needs to have kintetic energy equivalent to potential energy mgh

Does it now? In what position will the train obtain its maximal potential energy?

 

[Jorgen1224]

Does it now? In what position will the train obtain its maximal potential energy?

When it's at the top of the hill

 

[Orodruin]

When it's at the top of the hill

So where will the train's centre of mass be when it reaches the top of the hill?

 

[Jorgen1224]

Well when the front of the train touches the top of this hill then the center of mass is on the left side.

 

[Orodruin]

Well when the front of the train touches the top of this hill then the center of mass is on the left side.

Is this when the train has its maximal potential energy?

 

[Jorgen1224]

oh, yeah, its' center of mass has to be at the top of the hill for it to reach maximum potential energy

 

[Orodruin]

oh, yeah, its' center of mass has to be at the top of the hill for it to reach maximum potential energy

Is the centre of mass ever going to be on top of the hill? How will the train look when it is in its highest position?

 

[Jorgen1224]

According to the image it would be hanging there unless the top is flat for a distance equal to the length of the train. I don't really see any other option of train's movement, so i'd say that the centre of mass is going to be on top of the hill.

 

[Orodruin]

Is that your drawing or a drawing supplied with the problem?

Note that a train typically is quite flexible (at least in the connections between wagons).

 

[Jorgen1224]

This is the drawing supplied by my teacher. So this train bends and takes shape of the hill while it's on the top?

 

[haruspex]

This is the drawing supplied by my teacher. So this train bends and takes shape of the hill while it's on the top?

Yes. Consider the parts of the train on the uphill and downhill sides separately. You can find the mass centre of each part easily. What does it tell you about the location of the mass centre of the whole train?

 

[Jorgen1224]

That it is in between of both centres of mass meaning beneath the top of the hill?

 

[Orodruin]

That it is in between of both centres of mass meaning beneath the top of the hill?

Yes. The question is: How high?

 

[haruspex]

Yes. The question is: How high?

... at its highest.

 

[Jorgen1224]

Yes. The question is: How high?

I honestly have no idea. Pythagoras theorem doesn't seem to be working here

 

[haruspex]

I honestly have no idea. Pythagoras theorem doesn't seem to be working here

The first step is to figure out when it will be at its highest. Can you decide that?

 

[gneill]

I honestly have no idea. Pythagoras theorem doesn't seem to be working here

Look for some similar triangles and use ratios.

 

[Jorgen1224]

The first step is to figure out when it will be at its highest. Can you decide that?

When the train is divided into two parts each with length d/2

 

[haruspex]

When the train is divided into two parts each with length d/2

Right. And you know where the mass centre of each half is, so where is the mass centre of the whole?

 

[Jorgen1224]

Between them. There's no 90 angle in this triangle then i could separate it in two , but then anyway i know c=d/2, but i don't know a or b(which is some part of the height) so i can't use pythagoras theorem

 

[gneill]

Between them. There's no 90 angle in this triangle then i could separate it in two , but then anyway i know c=d/2, but i don't know a or b(which is some part of the height) so i can't use pythagoras theorem

See my recommendation in post #18. And here's a diagram that may clarify discussion:

 

[DaveC426913]

I am clearly under-thinking this problem - or there is a condition of the problem that is missing - or maybe it's a trick question.

As long as the train is moving at all, it will meet the minimum velocity required. Is there something that prevents the train from moving? For example, is there a condition that the train start coasting at some point?

IOW:
Q: "What velocity requirement must be met...?"
A: That v > 0.

Addendum: OK, that has to be the unwritten assumption - that the train is coasting.

 

[haruspex]

I am clearly under-thinking this problem - or there is a condition of the problem that is missing - or maybe it's a trick question.

As long as the train is moving at all, it will meet the minimum velocity required. Is there something that prevents the train from moving?

IOW:
Q: "What velocity requirement must be met...?"
A: That v > 0. Full stop.

It is not made clear, but I think we have to read the problem as that the engine generates negligible power during the ascent. It has to get there on initial KE alone.

 

[DaveC426913]

I wonder if the OP could wow his teacher by providing the answer v>0...

 

[haruspex]

I wonder of the OP could wow his teacher by providing the answer v>0...

Maybe not. We do not necessarily have the original statement of the problem, only the OP's rendition of it.

 

[hmmm27]

I like this problem. The answer's clear when reached, but (for me) there's been a bit of bouncing around getting there.

 

[Jorgen1224]

Ah yes, I'm terribly sorry for not stating that at the beginning. It should be "what initial velocity does the train need to go up and down the hill without a drive"

 

[PeroK]

It seems to me that the train in this case is doing a sort of Fosbury flop. A technique by which a high jumper clears a bar without their centre of gravity ever exceeding the height of the bar.

 

[DaveC426913]

It seems to me that the train in this case is doing a sort of Fosbury flop. A technique by which a high jumper clears a bar without their centre of gravity ever exceeding the height of the bar.

Ah. I see!

It's also how a siphon works! As long as there is more water on the 'down' side than the 'up' side, the water will flow.

 

[Jorgen1224]

So yeah, the answer i got is v > sqrt(g⋅h(2-d/2l) from similar triangles and their ratios. New height on which the center of mass is, is h((hl-d)/4l) so then we plug it as height in equation mv²/2 > mgh and rearrange it to get velocity.

 

[mrsmitten]

Homework Equations

E_k=m⋅v²/2
E_p=m⋅g⋅h

The Attempt at a Solution

I'm basically stuck at conversation of energy. Train needs to have kintetic energy equivalent to potential energy mgh, but calculating v from this equation seems pointless since it doesn't include either length. I have no idea how to include either of them.

if there is no friction then those are the only 2 equations that you need, because the length of l does not matter.

 

[Orodruin]

if there is no friction then those are the only 2 equations that you need, because the length of l does not matter.

As has been thoroughly discussed in this thread already, what you just wrote is not correct.

 

[haruspex]

So yeah, the answer i got is v > sqrt(g⋅h(2-d/2l)).

Right.

 

[mrsmitten]

As has been thoroughly discussed in this thread already, what you just wrote is not correct.

sorry I was thinking of the train as a rigid body.

i see what you are talking about now.

 

[jbriggs444]

So yeah, the answer i got is v > sqrt(g⋅h(2-d/2l)

Right.

With the proviso that the train is not long enough to span the entire hill. That is, as long as d < 2l.

As drawn, the train appears to only be long enough to span about half of the hill, so the proposed solution above appears to be the intended one.

If the train is longer than two hill-spans (d > 4l ), the proposed solution goes really wonky and predicts an imaginary required speed. [If you draw it out, that's because scenario assumed by the formula would have the front and back ends of the train dangling underground. The center of gravity would be below ground when the midpoint of the train hits the peak of the hill].

 

[DaveC426913]

osed solution goes really wonky and predicts an imaginary required speed.

(At the risk of derailing thread,is that literally an imaginary number? As in i?)

 

[jbriggs444]

(At the risk of derailing thread,is that literally an imaginary number? As in i?)

Yes.

A sufficiently long train has less potential energy when draped over the top of a hill than when it is flat at a specific level below the hilltop. To end up with zero kinetic energy you therefore have to start with negative kinetic energy. Which means an imaginary starting velocity. The square root of minus 1 sort of "imaginary".