Theoretical expression for relationship between height and force

[Simon Bridge]

I don't think there's any way you'd be expected to do this - included just to show you what you were asking about. Note: you could have used a more "hand-wavey" argument... but I don't know what level you needed it for.

When the mass hits the force pad, it remains in contact for a short time T, then it lifts off (because it deforms during the impact, and bounces) If the ball and the pad are pretty hard then the force-time graph should be well modeled by an inverted parabola. Something like:

$F(t)=4F_{max}(T-t)t/T$

This is related to the change in momentum of the dropped mass as:
$$\Delta p = \int_0^T F(t)\text{d}t = \frac{2}{3}T^2F_{max}$$
... $T$ depends on the material properties of the dropped mass and the plate - it's probably safe to call T a constant for the setup.

Thus $\Delta p \propto F_{max}$

The momentum of the object just before impact (t=0 in the above equation) is mv, given from the equations of motion for an object falling under gravity ... there's air resistance to consider... but if the distances are less than about three meters for objects of order of centimeters then air resistance can be neglected and if the experiment were conducted within a km of sea level with no big mountains or anything close by then the acceleration due to gravity will be a constant g=9.8m/s/s directly downwards.

By conservation of energy - an object falling distance h from rest, without air resistance, will hit with momentum $p=mv=m\sqrt{2gh}$.

The final momentum we don't know - it depends on the materials involved but it is usually a proportion of the initial momentum ... if the object hits exactly dead-on, then the initial momentum is down and the final momentum is up.

If $p_f=-ep: 0<e<1$ [1]
... then $\Delta p = (e+1)m\sqrt{2gh}$

From there: $$(e+1)m\sqrt{2gh}=\frac{2}{3}T^2 F_{max}\\ \qquad \implies F_{max} = \frac{3(e+1)m\sqrt{2g}}{2T^2}\sqrt{h}\\ \qquad \implies F_{max}\propto \sqrt{h}$$ ... sooooo: if the assumptions (remember what they were) hold, the graph of $F_{max}$ vs $h$ should be a square-root function.

If your graph is indistinguishable from a line, then you may not have had enough data for lower heights.

Approximating the square-root function as a 1st-order Taylor series about h=1m yields a line with slope = $$s=\frac{3(e+1)m\sqrt{2g}}{4T^2}$$
It should be possible to look up "typical" values for T and e that are close to the experimental setup. i.e. I have found some measurements[2] suggesting values: e=0.65/0.77=0.84 and T=0.15ms - for a ball bearing off a pretty hard surface. But I think you have used a softer surface/object because these values get max forces in the order of kilo-Newtons where you had values of tens of Newtons.


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[1] ... $e$ is called the "coefficient of restitution".
[2] Cross R. (2006) The bounce of a ball - table p224
http://www.physics.usyd.edu.au/~cross/PUBLICATIONS/BallBounce.pdf

 

[wreckemtech]

I don't think there's any way you'd be expected to do this - included just to show you what you were asking about. Note: you could have used a more "hand-wavey" argument... but I don't know what level you needed it for.

When the mass hits the force pad, it remains in contact for a short time T, then it lifts off (because it deforms during the impact, and bounces) If the ball and the pad are pretty hard then the force-time graph should be well modeled by an inverted parabola. Something like:

$F(t)=4F_{max}(T-t)t/T$

This is related to the change in momentum of the dropped mass as:
$$\Delta p = \int_0^T F(t)\text{d}t = \frac{2}{3}T^2F_{max}$$
... $T$ depends on the material properties of the dropped mass and the plate - it's probably safe to call T a constant for the setup.

Thus $\Delta p \propto F_{max}$

The momentum of the object just before impact (t=0 in the above equation) is mv, given from the equations of motion for an object falling under gravity ... there's air resistance to consider... but if the distances are less than about three meters for objects of order of centimeters then air resistance can be neglected and if the experiment were conducted within a km of sea level with no big mountains or anything close by then the acceleration due to gravity will be a constant g=9.8m/s/s directly downwards.

By conservation of energy - an object falling distance h from rest, without air resistance, will hit with momentum $p=mv=m\sqrt{2gh}$.

The final momentum we don't know - it depends on the materials involved but it is usually a proportion of the initial momentum ... if the object hits exactly dead-on, then the initial momentum is down and the final momentum is up.

If $p_f=-ep: 0<e<1$ [1]
... then $\Delta p = (e+1)m\sqrt{2gh}$

From there: $$(e+1)m\sqrt{2gh}=\frac{2}{3}T^2 F_{max}\\ \qquad \implies F_{max} = \frac{3(e+1)m\sqrt{2g}}{2T^2}\sqrt{h}\\ \qquad \implies F_{max}\propto \sqrt{h}$$ ... sooooo: if the assumptions (remember what they were) hold, the graph of $F_{max}$ vs $h$ should be a square-root function.

If your graph is indistinguishable from a line, then you may not have had enough data for lower heights.

Approximating the square-root function as a 1st-order Taylor series about h=1m yields a line with slope = $$s=\frac{3(e+1)m\sqrt{2g}}{4T^2}$$
It should be possible to look up "typical" values for T and e that are close to the experimental setup. i.e. I have found some measurements[2] suggesting values: e=0.65/0.77=0.84 and T=0.15ms - for a ball bearing off a pretty hard surface. But I think you have used a softer surface/object because these values get max forces in the order of kilo-Newtons where you had values of tens of Newtons.


----------------------

[1] ... $e$ is called the "coefficient of restitution".
[2] Cross R. (2006) The bounce of a ball - table p224
http://www.physics.usyd.edu.au/~cross/PUBLICATIONS/BallBounce.pdf

Much better explanation. You're good!

 

[Simon Bridge]

I'm glad you pointed out the conservation of energy bit though - saved me some typing.
The F=ma from earlier was needed for OP to understand whether the acceleration was a constant - OP expressed confusion about that.

Acceleration under gravity is only a approximately a constant - for falling distances that are small compared with the radius of the Earth (as in this case) and where the speeds are too small for air resistance to matter.

The apparently constant slope of the F vs h data suggests that air resistance was not a big factor compared with the experimental uncertainties.

Although, I suspect that if the data were split in half, and a best fit line found for the top (large h) half, and also for the bottom half, the bottom half slope would be bigger than the top half slope ... unless those error-bars are bigger than I'm imagining or there's fewer than, say, ten data points.

There's not really enough information to give any proper help.


Aside: I did try modelling the outcomes of an experiment and found that to get peak forces around 50N required impact times and restitution coefficients more appropriate hard rubber - this sort of experiment is normally done with metal masses and the force-plates are usually high-impact plastic, so I wonder if the force plate was sitting on hard rubber feet, maybe on lino or vinyl flooring.

That link I gave in [2], prev post, - the author's homepage is a lot of fun.

 

[wreckemtech]

True. I didn't really think about it, but g will vary inversely with the distance squared between the center mass of the 2 bodies. But like you say, negligible for this problem. Those were both great links!