Seesaw Saga - Velocity transfer on a seesaw

[K^2]

I was watching an old episode of MythBusters, and the myth they tackled was the Seesaw Saga. It's about a sky diver hitting a seesaw and launching a girl on the other end high into the air. I immediately noticed that MythBusters had no idea how momentum transfer on a seesaw works, which isn't surprising, but then I quickly realized that my initial intuition about it was all wrong as well.

I don't think of it as a hard problem, once you do it properly, but I found it interesting because of how wrong the intuition about it is. And having done a quick search for seesaw problems on this forum, I've found a lot of posts with similar mistakes!

So I present the problem purely as a mental exercise. A massless, frictionless, perfectly rigid seesaw has a mass M placed on the end of one arm. Another body of mass 3M strikes the end of the other arm going 120mph straight down.

1) At what speed does the lighter body leave the seesaw?

2) Can different placement of the lighter body or the impact point increase the outgoing speed? If so, what is the optimal placement and maximum speed?

Assume perfectly elastic collisions between seesaw and both bodies. Further assume that at the moment of impact the seesaw is horizontal and motionless for simplicity.

Feel free to post your solutions. If nobody posts anything similar to my approach, I will post it later. In either case, I suspect there will be some disagreement and discussion, which will, hopefully, resolve to everyone's benefit.

 

[McLaren Rulez]

Ok I will try.

1) The skydiver comes down at 120 mph which is around 50m/s. The initial angular momentum is therefore 3M*50*r where r is the length of each arm.

This is equal to the final total angular momentum (change in angular momentum due to gravity can be ignored). They both must have the same velocity when they are on the seesaw because the seesaw has some fixed angular velocity and both are at equal distance from the pivot. That is, Mvr+3Mvr = 150Mr

This gives v=37.5 m/s and that is the velocity with which the kid leaves the seesaw.

2) Placing her in a different spot, say r', should optimize her velocity. Let me think about that though because right now, I am getting an answer which suggests that placing her pretty much on the pivot i.e. r'=0 maximizes her velocity. Is that right?

 

[K^2]

They both must have the same velocity when they are on the seesaw because the seesaw has some fixed angular velocity and both are at equal distance from the pivot.

That was actually my mistake. I mean, obviously, the constraint is valid for the seesaw itself, but something more interesting happens to the bodies it comes in contact with.

I can tell you that the reason this isn't a factor with real seesaw is because real seesaw flexes. But even with idealized, perfectly rigid seesaw, there is a solution that violates the above constraint.

You have the right idea with angular momentum, though.

 

[McLaren Rulez]

Hmm I still don't see why that constraint doesn't hold but I'm certainly interested in finding out when you eventually post the solution :)

 

[K^2]

Think of what happens if a 3M body slams into 1M on a horizontal, frictionless surface.

 

[256bits]

So the answer is 180 mph then for M1 and 60 mph for M2

 

[K^2]

So the answer is 180 mph then for M1 and 60 mph for M2

Yes. Same as for linear collision. Except in linear collision, that's the maximum transfer possible. So the question remains, what about the seesaw? Can you transfer more of the incoming body's energy?

 

[McLaren Rulez]

Wait why is it the same as a linear collision? Since no one else is answering, why don't you explain it K^2? I want to know the answer to this one.

 

[AlephZero]

Yes. Same as for linear collision. Except in linear collision, that's the maximum transfer possible. So the question remains, what about the seesaw? Can you transfer more of the incoming body's energy?

Yes, by making the two arms of the seesaw different lengths. For the optimum length ratio, you can transfer ALL the kinetic energy to the second body.

Once you know (or guess) that is the answer, it's easy to work out the length ratio of the arms that you need to achieve it. Or you can tie yourself in knots doing it from first principles (use Lagrangian mechanics for extra credit) if you are a phyicist not an engineer

 

[K^2]

Solving rigid collision with Lagrangian formalism? I mean, it's doable, but I think there is a clause in Geneva Convention against forcing students to do that.

But yeah. It's all about momentum and energy conservation, just like with linear collisions. Except, linear momentum isn't conserved due to the pivot, so angular momentum is used instead.

Long story short, the general solution for masses M and m that allows transfer of all energy from one body to another is that the distances must have ratio r²/R²=M/m, where r corresponds to position of m, and R to M. For special case of r=R, the energy and momentum fractions transferred are the same as in 1D collision. And for the case of M=3m striking seesaw at 120 mph, as described in OP, holding R=1, and varying r, the resulting velocities of m (blue) and M (red) are presented bellow.

Wait why is it the same as a linear collision? Since no one else is answering, why don't you explain it K^2? I want to know the answer to this one.

It's because the conservation of angular momentum condition is MRV_i+mrv_i = MRV_f+mrv_f, which in case of R=r simplifies to the momentum conservation constraint in linear collision.

 

[haruspex]

I'm not getting the bit about transferring ALL the energy. There are losses, energy isn't conserved. If you mean transferring all the momentum, that requires M1 to come to a halt instantly, which it is not going to do.
Yes, the distance ratios should be the inverse square root of the mass ratios, but I get that M1's momentum is halved.
M1's speed before impact = U
M1's speed after impact = U/2 (down)
M2's speed after impact = U sqrt(M1/M2) / 2 (up)
Must say, this doesn't feel right, but I don't see an error.

 

[AlephZero]

I'm not getting the bit about transferring ALL the energy. There are losses, energy isn't conserved.

The OP said

A massless, frictionless, perfectly rigid seesaw ...

Assume perfectly elastic collisions between seesaw and both bodies. Further assume that at the moment of impact the seesaw is horizontal and motionless for simplicity

Of course you are right that in reality, mechanical energy will not be conserved.

 

[haruspex]

Sorry, I misread it - thought it said INelastic.
If perfectly elastic then I agree - all energy transferred.
Sorry for the noise.

 

[guyver]

This exact problem is being asked for my physics final. Thank you for the break down.

 

[Drakkith]

Within the constraints of the OP's post, would the lighter object ever reach maximum velocity before the heavier object impacts the ground and loses the rest of it's momentum and energy? Or is this like the balls in a Newton's cradle where momentum is transferred nearly instantly?

 

[meldraft]

Hmmm... what I think is going to actually happen is that due to the impact the second body will take off at a higher velocity transverse velocity than the first body is going down. I don't have time to do the math right now, but depending on the arm/mass/velocity ratio we can have the following scenarios:

1. The second mass takes off at the moment of impact and does not stay in contact to the seesaw at any moment afterwards.

2. The second mass takes off at the moment of impact but the arm ratio and angular velocity are such that the mass wants to move transversely slower than the tip of the seesaw is going up. If the seesaw is flexible, it will bend and store the energy like a spring. If it's perfectly rigid, this will cause the angular velocity of the system to drop.

 

[K^2]

Or is this like the balls in a Newton's cradle where momentum is transferred nearly instantly?

Exactly. The actual transfer time will depend on how much flex the seesaw has. In idealized case of perfectly rigid seesaw, the transfer is instantaneous.