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

In summary, the homework statement asks for an equation to be found for the train's kinetic energy as it moves up and down a hill. The equation requires taking into account the length and speed of the train, but does not include either of these values. Once the kinetic energy equation is found, it is found that the train's center of mass will be at the top of the hill.
  • #1
Jorgen1224
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0

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?
fiz.JPG

Homework Equations


Ek=m⋅v2/2
Ep=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.
 

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  • #2
Jorgen1224 said:
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?
 
  • #3
Orodruin said:
Does it now? In what position will the train obtain its maximal potential energy?
When it's at the top of the hill
 
  • #4
Jorgen1224 said:
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?
 
  • #5
Well when the front of the train touches the top of this hill then the center of mass is on the left side.
 
  • #6
Jorgen1224 said:
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?
 
  • #7
oh, yeah, its' center of mass has to be at the top of the hill for it to reach maximum potential energy
 
  • #8
Jorgen1224 said:
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?
 
  • #9
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.
 
  • #10
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).
 
  • #11
This is the drawing supplied by my teacher. So this train bends and takes shape of the hill while it's on the top?
 
  • #12
Jorgen1224 said:
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?
 
  • #13
That it is in between of both centres of mass meaning beneath the top of the hill?
 
  • #14
Jorgen1224 said:
That it is in between of both centres of mass meaning beneath the top of the hill?
Yes. The question is: How high?
 
  • #15
Orodruin said:
Yes. The question is: How high?
... at its highest.
 
  • #16
Orodruin said:
Yes. The question is: How high?
I honestly have no idea. Pythagoras theorem doesn't seem to be working here
 
  • #17
Jorgen1224 said:
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?
 
  • #18
Jorgen1224 said:
I honestly have no idea. Pythagoras theorem doesn't seem to be working here
Look for some similar triangles and use ratios.
 
  • #19
haruspex said:
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
 
  • #20
Jorgen1224 said:
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?
 
  • #21
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
 
  • #22
Jorgen1224 said:
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:
upload_2018-11-13_18-16-19.png
 

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  • #23
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.
 
  • #24
DaveC426913 said:
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.
 
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  • #25
I wonder if the OP could wow his teacher by providing the answer v>0...
 
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  • #26
DaveC426913 said:
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.
 
  • #27
I like this problem. The answer's clear when reached, but (for me) there's been a bit of bouncing around getting there.
 
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  • #28
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"
 
  • #29
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.
 
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  • #30
PeroK said:
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.
 
  • #31
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.
 
  • #32
Jorgen1224 said:

Homework Equations


Ek=m⋅v2/2
Ep=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.
 
  • #33
mrsmitten said:
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.
 
  • #34
Jorgen1224 said:
So yeah, the answer i got is v > sqrt(g⋅h(2-d/2l)).
Right.
 
  • #35
Orodruin said:
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.
 

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