# The rebound height of a mass on a trampoline

When a mass(metal ball) lands on the end of a trampolne bed, the mass is displaced towards the centre of the trampoline. I have read about the energy changes and forces involved in a trampoline but can't find any information about this aspect. I'm guessing it has something to do with a spring always trying to return to equilibrium but can't work out what forces are involved.

In my Advanced Higher Physics project I am changing the mass landing on the edge of the trampoline and measuring how this affects the amount it is displaced towards the centre. Obviously the displacement increases as mass increases. My other investigation is how the rebound height is affeced when I change the distance from the centre a mass dropped. The rebound height increases as I increase the distance. Why is this?

Any suggestions about the physics behind these movements would be much appreciated and very interesting.

When a heavy enough mass $M$ lands near the circumference of a trampoline, the trampoline is deformed in an asymmetric way. The slope of the trampoline on the outer side of the massive object (even closer to the rim) is steeper – more vertical – than the slope on the inner side (closer to the center). That's guaranteed by the fact that the "height" of the trampoline near the rim is more constrained.

So the trampoline is effectively tilted. The normal vector to the "average cloth" surrounding the object goes up but slightly towards the center. When the massive object gets reflected, it throws the object closer to the center because it's just like a light ray reflecting from a mirror that isn't quite horizontal. A picture would probably help but I hope you will be able to draw yours.

Indeed, you are right that there is no equivalence principle because the angle $\gamma$ by which the trampoline's formerly horizontal surface is distorted in average is de facto proportional to the mass $M$, $\gamma\sim C M$, and the direction by which the reflected objects will deviated from the vertical axis is therefore $2\gamma\sim 2CM$. The constant $C$ may actually be estimated from the radius, the distance of the mass from the rim, and "stiffness" of the trampoline.

• I'm reading this because I'm trying to answer this question about trampolines on our game development sister site. I'm getting stuck with the technicalities here. What does that tilde-like "squiggly line" symbol mean? Also, I tried to draw a picture. Am I going in the right direction here?
– Anko
Oct 12, 2014 at 11:21
• The tilde means "is proportional to", $y\sim x$ means $y=kx$ for a constant $k$. I am not sure whether I am able to say whether the picture is "right" or "wrong" without further captions. Oct 12, 2014 at 16:38
• Thanks! I now see that my sketch likely doesn't correspond at all to your words. In any case, here's the same one with a legend. I'm getting stuck on more terminology. What's is an "average cloth"? What does "equivalence principle" mean here? (Do you mean that because the angle keeps changing, an iterative solution is necessary to find the dropped mass' path?) Should I ask these as a separate question? :)
– Anko
Oct 12, 2014 at 18:00
• Yes, I think you may want to ask a new question because the comment above is a combination of requests to explain words you may find by a single simple internet search with some explanations of something I wrote a long time ago which may or may not be relevant to what you really need etc., and the comments are probably not a good place for those.... Oct 12, 2014 at 19:39

G-forces on an competition gymnast can peak at 14 g's

A trampoline isn't even a close approximation of a simple harmonic oscillator. As the mat distorts, the tension vectors of the springs cancel less. The actual upward force as a function of displacement is messy, but I think to first order it increases with the cube. But there is a substantial term due to how much tension there is on the spring when the mat is flat.

The net effect is that as you jump higher, the rebound time (first contact to last contact on the mat) drops. Measuring on my tramp, by filming with my phone at 240 fps, when I "jump" just barely leaving the mat, the bounce takes about 0.78 seconds. When I jump putting my centre of mass about 4 feet in the air, this drops to 0.35 seconds.

The return to centre forces on a cicular trampoline have a good answer here, but there is a second effect:

I've found that what I do for my next trick is critically depend on how I land on this jump.

Jumper lands near edge of mat. Foot closer to the edge gets a somewhat higher upward force. This puts a small torque on the jumper, meaning that his next jump he lands with his centre of mass displaced toward the centre. On his next bounce, force upward is no longer in line with force downward, creating both a net force toward the centre and a stronger rotation.

One of the biggest sources of injury on garden trampolines are heads bumping heads. You can see how kids, tilted toward the centre, and nudged toward the centre can bump.