- #1
JamieLT
- 6
- 0
Hi!
I'm trying to help my 5th grader with her science fair project - she's comparing how a trampoline would work on Earth vs. the moon.
The trick is in the force of the jump I think - what the jumper herself puts into it. The larger the downward force she exerts on the mat the higher she goes. On the moon, the downward force you can get is less because gravity is less – you wouldn't be able to push down on it as much right? If you do just the simple energy balance
½ k x^2 = mgh
PE at the bottom = PE at the top.
you get a smaller x, smaller x^2 - and since x goes by the square, and g is not squared, seems like it would be harder to jump higher on the moon? (even though without the tramp, just jumping on the surface of the moon, you could go higher)
F = kx = mg at rest
not at rest F = kx + leg force ...
leg force is constant, just the mg that is not.
LOL, this should be simple, but for some reason I think I'm missing something.
For the experiment we're just going to drop different weighted balls onto the tramp and measure the height and deflections... then we'll have a jumper with and without weights strapped on to measure the different heights (same legs lifting different amounts of weight)... we also thought about jumping in a pool (astronauts train in pools) but the drag would mess it up. Any other way to test it out? I want to get the math right too though.
Air resistance - the moon doesn't have much air, but is it a negligible effect?
I'm trying to help my 5th grader with her science fair project - she's comparing how a trampoline would work on Earth vs. the moon.
The trick is in the force of the jump I think - what the jumper herself puts into it. The larger the downward force she exerts on the mat the higher she goes. On the moon, the downward force you can get is less because gravity is less – you wouldn't be able to push down on it as much right? If you do just the simple energy balance
½ k x^2 = mgh
PE at the bottom = PE at the top.
you get a smaller x, smaller x^2 - and since x goes by the square, and g is not squared, seems like it would be harder to jump higher on the moon? (even though without the tramp, just jumping on the surface of the moon, you could go higher)
F = kx = mg at rest
not at rest F = kx + leg force ...
leg force is constant, just the mg that is not.
LOL, this should be simple, but for some reason I think I'm missing something.
For the experiment we're just going to drop different weighted balls onto the tramp and measure the height and deflections... then we'll have a jumper with and without weights strapped on to measure the different heights (same legs lifting different amounts of weight)... we also thought about jumping in a pool (astronauts train in pools) but the drag would mess it up. Any other way to test it out? I want to get the math right too though.
Air resistance - the moon doesn't have much air, but is it a negligible effect?