# Thread: Falling from Stilts problem

1. ## Falling from Stilts problem

Apologies if this is in the wrong place.

Heres a problem posed on another forum. Not realistic but a brain teaser.

A man is atop 100m high stilts. He pushes himself forward in such a way that he falls forward but the base of the stilt remains attached to its initial point. He remains attached to the stilts right down to hitting the ground.

Will he cross the finish line of a 100m race before a world record sprinter? (The world record being 9.74 seconds)

Im thinking that because the distance from the origin is constant (100m) due to being on the stilts that the falling mans path is a form of circular motion. Its not uniform circular motion because gravity is always acting and its not acting towards the origin but acts downwards.

Any ideas on how to get either the angle the stilts make with the ground as a function of time or position of the man as a function of time?

Ive got a mess of scrawlings basically:
The tangiential acceleration is
At=alpha r
At is also the radius multiplied by the second derivative of the angle with respect to time. It also appears to be equal to m.g.sin(angle)

From that ive tried to get the angle as a function of time but am stumped? Have i just written rubbish? Any Ideas? (other than advice not to get involved with such insane races!)

2. It depends strongly on the velocity given by the first push, with total time going to infinity as the push goes to 0, so you have to specify how strong the initial push is first.

3. It depends strongly on the velocity given by the first push, with total time going to infinity as the push goes to 0, so you have to specify how strong the initial push is first.
ooops forgot that bit. We pointed that out in the thread. I guess any non zero force would do.

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Do we get to imagine that that stilts have negligible mass compared to the person on top?

I guess the easiest thing to do is to assume that the stilt-walker starts off at some small angle from the vertical. We could then calculate how small that angle has to be before he takes longer than 9.74 seconds to hit the ground.

Grant Hutchison

5. Yes the stilts are negligable mass.

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So the angular acceleration is sin(alpha)*g/r, where alpha is the angle between the vertical and the stilts at any given moment, g is acceleration due to gravity and r is the length of the stilts.

So you have [angular] acceleration as a function of [angular] distance.

There may be a smarter way to do it, but at present I'm thinking there are a couple of integrations involved: first to get velocity as a function of distance, and then again to get the time spent falling.

Grant Hutchison

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As pointed out earlier, the initial 'push' is important. If it imparts a very small initial velocity, the sprinter can cross the finish line before stilt guy moves his first meter. If he starts with an initial velocity of 100 m/s, the unfortunate victor wins nose down in less than 2 seconds.

Is was also suggested that stilt guy starts at a displaced angle. If that is the case, he will have a head start over the sprinter at the base of the stilt. Again, a world of difference between a one degree start and an 89 degree start.

Without information that starts stilt guy, the problem is indeterminate.
Last edited by ggchuck; 2008-Jan-29 at 06:46 PM. Reason: correct typo

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Yes, it's a significant head start if our hero is leaning forward: I make it (assuming I'm doing the integrations correctly) around a 9-degree tilt in order to hit the ground in under 9.74s.

Grant Hutchison

9. Originally Posted by grant hutchison
There may be a smarter way to do it, but at present I'm thinking there are a couple of integrations involved: first to get velocity as a function of distance, and then again to get the time spent falling.
I think you can do it in one integral like this: t = Integral over dt = Integral over dtheta / (dtheta/dt) and just use conservation of energy to get dtheta/dt = Root[2g*(1-Cos(theta))/R]. So the time elapsed from an initial theta_o to theta=Pi/2 is the Integral over dtheta, from theta_o to Pi/2, of 1 / Root[2g*(1-Cos(theta))/R]. I get a 4.5 degree tilt yields 9.74 seconds, did one of us drop a factor of 2 in our conversion from radians to degrees?

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Originally Posted by Ken G
I think you can do it in one integral like this: t = Integral over dt = Integral over dtheta / (dtheta/dt) and just use conservation of energy to get dtheta/dt = Root[2g*(1-Cos(theta))/R]. So the time elapsed from an initial theta_o to theta=Pi/2 is the Integral over dtheta, from theta_o to Pi/2, of 1 / Root[2g*(1-Cos(theta))/R]. I get a 4.5 degree tilt yields 9.74 seconds, did one of us drop a factor of 2 in our conversion from radians to degrees?
No, I get the same as you if I use your formula.
But I have (Cos(theta_o)-Cos(theta)) in my time integral, rather than (1-Cos(theta)). I'm getting that by setting velocity to zero at theta_o, not at theta=0.
I think.

Grant Hutchison

11. Originally Posted by grant hutchison
No, I get the same as you if I use your formula.
But I have (Cos(theta_o)-Cos(theta)) in my time integral, rather than (1-Cos(theta)). I'm getting that by setting velocity to zero at theta_o, not at theta=0.
I think.
Oops, you're right about the zero for the velocity-- I gave the competitor a tiny push from zero but waited for him to get to 4.5 degrees to start the race, so it's a "head start" situation. You did what you claimed-- started him from rest at 9 degrees.

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Originally Posted by Ken G
Oops, you're right about the zero for the velocity-- I gave the competitor a tiny push from zero but waited for him to get to 4.5 degrees to start the race, so it's a "head start" situation. You did what you claimed-- started him from rest at 9 degrees.
Ah, right.
Taking your integral, adding a v_o² inside the Root brackets, and integrating from 0 to π/2, gives us the situation where our hero starts from vertical with an initial non-zero angular velocity, v_o.
I'm getting the equivalent of about 5 m.s-1: so he needs to start off with a moderate degree of oomph if he is to beat the sprinter.

Grant Hutchison

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It seems only fair that the sprinter should be allowed to start at the same distance from the finish line as the stilt guy, not at the base of his stilts. The speed of the sprinter would then have to be assumed constant.

Probably a more realistic start would be to calculate the speed required of stilt guy from his push off at zero angle... an 'off the blocks' speed kind of thing.

14. I agree ggchuck, and that's what Grant calculated just above. He found that it takes an initial speed of 5 m/s for the stilt guy to make it a tie. As that is a pretty hefty speed (about half the sprinter's speed), it is hardly a "fair" start. Also, it should perhaps be mentioned that the stilt guy, with or without such a big initial push, arrives at the ground at a highly problematic 45 m/s speed, so he needs to worry about surviving the 100-foot fall!

15. The impact speed was something i noticed too. Painful for sure!

It seemed like a simple problem when set on the other board but it theres more to it than it appeared. Though its not a evil a thread topic as the aeroplane on the treadmill 'classic'!

16. Originally Posted by ggchuck
It seems only fair that the sprinter should be allowed to start at the same distance from the finish line as the stilt guy, not at the base of his stilts.
You want the sprinter to be moved farther back??

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Originally Posted by Ken G
...it should perhaps be mentioned that the stilt guy, with or without such a big initial push, arrives at the ground at a highly problematic 45 m/s speed, so he needs to worry about surviving the 100-foot fall!
No pain... no gain.

=======
What is the speed out of the blocks for a sprinter anyway? Does anybody know?
Last edited by ggchuck; 2008-Jan-30 at 09:31 PM. Reason: Added a question.

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Originally Posted by hhEb09'1
You want the sprinter to be moved farther back??
Good point... I guess I should have lobbied for the sprinter to be allowed to start at the same distance from the finish plane as the stilt guy.

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Stilt-guy of course wins if the distances travelled are the same: 100m arc to the finish line versus 100m horizontal track.

Grant Hutchison

20. What's his time?

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Originally Posted by hhEb09'1
What's his time?
Looks like 5.6 seconds, if he starts from rest.

Grant Hutchison

22. Originally Posted by jumbo
Though its not a evil a thread topic as the aeroplane on the treadmill 'classic'!
What problem is that?

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Originally Posted by Ken G
What problem is that?
The inexplicably protracted Will It Fly? thread.

Grant Hutchison

24. The first page was pretty interesting, and I think Atraxani's corrections to tofu's insights pretty much resolved everything, underscored by your point that if a treadmill moving backward at the aircraft takeoff speed could really keep the plane in place, then the plane could never take off in the first place (since Atraxani pointed out the drag doesn't increase with velocity)! But I note the thread went on a long time after that-- "inexplicable" is right! Maybe there were further insights, I didn't check.

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Originally Posted by Ken G
Maybe there were further insights
The fact that the mode of propulsion (propeller vs drive train) isn't relevant to the problem was an insight I missed until well after the first page.

26. OK, I'll look for that, that is surprising isn't it.

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Originally Posted by Ken G
...that [mode of propulsion not relevant] is surprising isn't it.
Perhaps. I have to admit, I haven't convinced anyone of that fact.

28. I think the key issue if you want to compare a car to a plane on a conveyor belt is to have a model for how the wheels interact with the belt. So your view might, for example, work if the wheels are like cogs that interlace with the belt (where you can control the force between them), and fail if the wheels are like actual wheels on a roadway (where that force is limited by the rules of friction).

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Originally Posted by Ken G
I think the key issue if you want to compare a car to a plane on a conveyor belt is to have a model for how the wheels interact with the belt. So your view might, for example, work if the wheels are like cogs that interlace with the belt (where you can control the force between them), and fail if the wheels are like actual wheels on a roadway (where that force is limited by the rules of friction).
Thanks for the advice. Actually I did mention the rack and pinion concept when the thread was active last year. This year, the issue of traction hasn't been discussed and I hope it remains that way. It doesn't directly relate to the puzzle and tends to further muddy the waters.

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