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

[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".