# Do different length ramps violate conservation of energy?

• Aperture
Sorry, I am still not clear on this. Did the original problem ask about the final speed or the time taken?f

#### Aperture

Homework Statement
Say you have 2 different lamps that start and end at the same heights. One ramp is steeper and longer than the other, while the second ramp is the shortest line from the top to the bottom. If you released two identical balls, one on each ramp, from rest, you would find that the ball on the steeper ramp has a greater final velocity. How is this possible if mgh=1/2m(v^2) and m, g, and h are the same for both cases?
Relevant Equations
mgh=(1/2)(m)(v^2)
mgh=(1/2)(m)(v^2)
gh=(1/2)v^2
sqrt(2gh)=v
Should have the same v, but this is not the case based on the answer and real-life experiments.

When are you measuring the final velocity? At the end of the ramp, or once the ball is rolling horizontally after leaving the ramp?

When are you measuring the final velocity? At the end of the ramp, or once the ball is rolling horizontally after leaving the ramp?
end of the ramp, when h = 0 so only 1/2mv^2 remains.

Homework Statement: Say you have 2 different lamps that start and end at the same heights. One ramp is steeper and longer than the other, while the second ramp is the shortest line from the top to the bottom. If you released two identical balls, one on each ramp, from rest, you would find that the ball on the steeper ramp has a greater final velocity. How is this possible if mgh=1/2m(v^2) and m, g, and h are the same for both cases?
Relevant Equations: mgh=(1/2)(m)(v^2)

mgh=(1/2)(m)(v^2)
gh=(1/2)v^2
sqrt(2gh)=v
Should have the same v, but this is not the case based on the answer and real-life experiments.
Can you show the two ramps in a drawing? I cannot imagine a ramp that is steeper and longer than the other if they start and end at the same heights.

Also, why do you say "If you released two identical balls, one on each ramp, from rest, you would find that the ball on the steeper ramp has a greater final velocity." Did you do an experiment?

On Edit: Here is a convincing demonstration that should clarify what's going on. Be sure to watch the video in its entirety.

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• DrClaude, Aperture, Lnewqban and 1 other person
why do you say "If you released two identical balls, one on each ramp, from rest, you would find that the ball on the steeper ramp has a greater final velocity."
The way I read post #1, the OP is there quoting the given answer. Indeed, the OP believes they should end with the same speed.
Of course, if the nonlinear path is always below the linear one then, ignoring losses, the ball will arrive sooner, though at the same speed.
If we take losses into account, but no slipping, then the straight line should produce the greatest final speed.
Edit: I need to check that; it's not as obvious as I thought.

If the gradient is gentle until a final descent too steep to maintain rolling then that may give the greatest final speed.

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Yes, this involves a combination of rolling and slipping. I missed that this is a homework problem posed as an energy non-conservation paradox.

Can you show the two ramps in a drawing? I cannot imagine a ramp that is steeper and longer than the other if they start and end at the same heights.

Also, why do you say "If you released two identical balls, one on each ramp, from rest, you would find that the ball on the steeper ramp has a greater final velocity." Did you do an experiment?

On Edit: Here is a convincing demonstration that should clarify what's going on. Be sure to watch the video in its entirety.

The video cleared things up. Thanks.

• Drakkith
The video cleared things up. Thanks.
Well that's great, but can you please still clarify why you said this:
Should have the same v, but this is not the case based on the answer and real-life experiments.

Was it just a misunderstanding of some data or something? Thanks.

Well that's great, but can you please still clarify why you said this:

Was it just a misunderstanding of some data or something? Thanks.
Yes. I failed to consider that a faster total time does not mean a faster final speed.

• Drakkith
Yes. I failed to consider that a faster total time does not mean a faster final speed.
Sorry, I am still not clear on this. Did the original problem ask about the final speed or the time taken?

The way I read post #1, the OP is there quoting the given answer. Indeed, the OP believes they should end with the same speed.
Of course, if the nonlinear path is always below the linear one then, ignoring losses, the ball will arrive sooner, though at the same speed.
If we take losses into account, but no slipping, then the straight line should produce the greatest final speed.
Edit: I need to check that; it's not as obvious as I thought.

If the gradient is gentle until a final descent too steep to maintain rolling then that may give the greatest final speed.
Fwiw, I considered descending in two straight lines, first at angle ##\theta_1##, by ##h_1## vertically, then at ##\theta_2##, by ##h_2## vertically. Drag per unit mass is ##kv^2##.
I get a final speed given by ##\alpha_2v^2=1-e^{-2\alpha_2gh_2}(1-(\frac{\alpha_2}{\alpha_1})(1-e^{-2\alpha_1gh_1}))##, where ##h_1+h_2=Y##, ##h_1\cot(\theta_1)+h_2\cot(\theta_2)=X## and ##\alpha_i=k/(g\sin(\theta_i))##.
Maximisation wrt ##h_1, \theta_1## would be the next step, but I might not get to it.

An alternative, of course, is the full calculus of variations approach. Haven't tried it.