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A student of mine is doing a project on this topic, and I have realized that I cannot answer the question.

If we are talking about dropping the ball, and assuming a completely elastic ball, it should bounce back to the original height. Therefore, buy adding pressure, you are increasing the elasticity of the ball. But why does pressure affect elasticity? Is it through the air inside the ball or through the ball material? If it's because of the material, why does it become more elastic when tension is increased?

When kicking a ball, if it's always kicked with the same force, the change of momentum of the ball will be higher if the impact time is longer. If I think about it as an adiabatic gas under an extra pressure $\Delta P$, the new volume would be

$V =V_0 \left(\frac{P_0}{P_0+\Delta P} \right)^{1/\gamma}=V_0 \left(\frac{1}{1+\frac{\Delta P}{P_0}} \right)^{1/\gamma}$

For a small $\Delta P$, $\frac{\Delta P}{P_0} << 1$, I can do a Taylor expansion and I would have:

$V = V_0 \left( 1-\frac{1}{\gamma} \frac{\Delta P}{P_0}\right)$

I will assume that the kicking force (and therefore $\Delta P$) is the same in all cases. Therefore, the higher the initial pressure, the smaller the change in volume, and the impact time is shorter and the change in momentum should be smaller. Does this make sense?

Everything here seems really obvious and I have the feeling that the answer is really stupid. But I cannot find a good way of explaining it. A reference that I can give my student would be much appreciated.

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    $\begingroup$ Re, "...always kicked with the same force..." I would not assume that a person exerts direct control over the contact force when kicking a ball. I think it's much more likely that the person controls the velocity (and therefore, the momentum) of their foot and lower leg, and I would guess that what happens after the foot and ball collide can be mostly explained just by the momentum and the kinetic energy of the foot just at that point of impact. In other words, I'm saying that kicking a ball is much like whacking it with a club, and not so much like throwing it with your arm. $\endgroup$ May 8 '20 at 19:57
  • $\begingroup$ @SolomonSlow: That makes sense. And actually it brings to mind an exercises that I could not understand for the same reason: A golf club is pivoted and the head is raised by a height h. At the bottom of the trajectory it hits a golf ball. Explain how increasing the golf ball will increase the speed at which it is launched. The answer is: "The deformation increases the contact time and at equal force, the change of momentum/velocity is larger" I always thought that in this case, what would remain constant is the change of momentum. $\endgroup$
    – jrglez
    May 9 '20 at 9:19
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When an air filled ball bounces, it is compressed. A flat spot on the floor grows. The reaction force of the floor on the flat spot slows the ball to a stop.

As this happens, the volume of the ball is reduced. Gas pressure increases. The rest of the ball stretches. The kinetic energy is stored as potential energy by these spring like mechanisms.

Then the compressed/stretched springs return to their original shapes, thrusting the ball off the floor.

When pressure inside the ball is high, it takes a smaller deformation to stop the ball. Deformation is accompanied by some loss as heat. A smaller deformation means smaller loss.


Another thought: A low pressure ball develops a bigger flat spot. Air is pushed out of that spot at higher velocity than air that just has to get out of the way as the ball flies. This increases the pressure and slows the ball.

As the ball rebounds, air has to quickly flow into the area uncovered by the flat spot. This reduces air pressure and slows the ball.


As the flat spot develops, pressure inside the ball grows, and the ball walls stretch. Perhaps the flat spot is stretched and slides along the floor. There would be a loss from friction.

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For balls within its elastic limits, (they return to its original form after pressed and released), there should be no difference in the amount of elastic energy accumulated during an impact. So, they should bounce equally.

The difference between two balls in that conditions, but one with more inside pressure is the relation stress x deformation of the membrane. Higher internal pressure means less deformation for a given stress. It is indeed a way to test if a ball is good to play: evaluating the deflection when pressing with a hand.

When the allowable deformation is small (high pressure) the energy loss by membrane waves (and consequentely sound waves) at the moment of the impact is smaller, because energy is proportional to amplitude.

The elastic energy of the membrane at the moment of the impact can be the same, $E = k\sigma \epsilon$, but for the high pressure one, $\sigma$ is greater and $\epsilon$ is smaller.

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