Golf Ball Bouncing: What Force Causes It?

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In summary: And when you look at the graph of a potential function, you should recognize in the regions of steep slopes a strong force, repulsive when the curve is decreasing, and attractive when the curve is increasing. So in summary, the strong force at steep slopes is the result of the electric repulsion force between the orbiting electrons.
  • #1
Swapnil
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When you throw a golf ball on the ground it bounces back up slightly after hitting the ground. What kind of force is responsible for this effect?
 
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  • #2
Think of it in terms of energy rather than force. The ball hitting the ground is a two-body collision where one body is immobile. For an elastic collision, the ball's final kinetic energy is the same as its initial KE so it rebounds back to the same height. A Superball bouncing off of a hard springy surface (like a steel anvil or thick steel slab) comes close.

Practical collisions are inelastic because some energy is dissipated as heat (mostly in the ball, some in the ground) and sound. You can measure the inelasticity of the ball by dropping it on an elastic surface and measuring the rebound height. This is called the coefficient of restitution
COR = h_final / h_initial

Sanctioning bodies regulate the properties of balls used in professional sports. The United States Golf Association began specifying the COR of golf balls in 1942 and also limited the COR of golf clubs to a maximum of 0.83 in the 1980's.

I think that the COR of golf balls is around 0.6. If your ball bounced "up slightly after hitting the ground" then you were probably on dirt, which is rather inelastic (the grains move around and out of the way, dissipating energy). The rebound will be even lower on sand. Try dropping your ball on a smooth concrete garage floor, or a steel slab like the 1" thick trench plates used when construction crews dig up a road, and it will come up higher.
 
  • #3
The way I think it works is that basically when something is standing on the ground, the molecules that composes it are in equilibrium with the molecules of the ground.

When you drop something on the ground, the molecules of the body will feel a repulsive force and just how much the object bounces back depends on how elastic it is, i.e. how its atoms are wired together.

But to answer your specific question on the nature of the force itself, it is as you would have guessed electromagnetic (there are only two macroscopic forces: gravity and electromagnetic and it's certainly not gravity). More precisely, the electromagnetic force btw molecules is not coulombian since the molecules are essentially neutral. Rather, the potential is of the Van der Waals type, such as the Liénard-Jones potential:

http://en.wikipedia.org/wiki/Lennard-Jones_potential

It is a potential caracterized by a very strong repulsion ~1/r^12 (!) at close distances and a (quickly fading) attraction past a distance of ~4 Anstrom to the molecule.
 
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  • #4
quasar987 said:
But to answer your specific question on the nature of the force itself, it is as you would have guessed electromagnetic ... More precisely, the electromagnetic force btw molecules is not coulombian since the molecules are essentially neutral. Rather, the potential is of the Van der Waals type, such as the Liénard-Jones potential:
http://en.wikipedia.org/wiki/Lennard-Jones_potential
It is a potential caracterized by a very strong repulsion ~1/r^12 (!) at close distances and a (quickly fading) attraction past a distance of ~4 Anstrom to the molecule.
It seems that the L-J potential doesn't have any theoretical justificaion but it is just an experimental fact. Either way, it is only a potential and it is not even associated with a force! Then how is the L-J potential a type of electromagnetic FORCE?
 
  • #5
To every potential V(x), there is an associated force [tex]F(x)=-\frac{dV}{dx}[/tex]. When one says "potential", you should hear "force". When looking at the graph of a potential function, you should recognize in the regions of steep slopes a strong force, repulsive when the curve is decreasing, attractive when the curve is increasing.
 
  • #6
I see. But what is the origin of that force? If is not Coulomb repulsion force, but it is still electromagnetic in nature then what is it?
 
  • #7
By "not coulombian" I only meant that it is not a potential in 1/r (read "force in 1/r²"!)

As for the nature of the force, I don't really know. The wiki article says that the strong repulsion at short distance is a consequence of the Pauli exclusion principle (two electrons cannot cohabit in exactly the same quantum state). Classically, maybe you can see this as the electric repulsion force btw the orbiting electrons that happens when the molecules get too close together.

I don't know about the 1/r^6 attractive force though. Why is it there and why does it take this form independantly of the structure of the molecule?
 
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  • #8
quasar987 said:
To every potential V(x), there is an associated force [tex]F(x)=-\frac{dV}{dx}[/tex]. When one says "potential", you should hear "force". When looking at the graph of a potential function, you should recognize in the regions of steep slopes a strong force, repulsive when the curve is decreasing, attractive when the curve is increasing.

Actually, this isn't quite right.

Only when one says potential gradient would there be a force. Remember that a constant potential will not result in a force, as in the equation you wrote since the gradient would be zero, but not V.

Secondly, the last part of what you said is a bit confusing. Usually, we denote an attractive potential with having a negative value. Thus, in electrostatics for example, the attractive potential is given as V proportional to - 1/r.

Zz.
 
  • #9
What I wanted to emphasize is that when one says "potential", you should hear "force".

One can talk about what the force looks like or one can talk about what the potential looks like and convey the same information. In particular, a statement such as "the force is zero inside the conductor" is equivalent to "the electric potential is constant inside the conductor".
 
  • #10
I think its the elastic energy. When a ball hits the ground, it takes the elastic energy in and then expels it out by bouncing up. This is what I think.
 
  • #11
How much the golf ball rebounces has to depend on the type of surface it hits. For hard surfaces, sound produced means a loss of energy, thus the final energy of the rebouncing golf ball is small than its initial energy and it rebounces to a smaller height compared to its initial height. For soft surfaces, most of the energy of golf ball is absorbed by the surface and there is negligible energy loss through sound, hence the golf ball rebounces little or no.
 
  • #12
Hi Swapnil, I didn't realize you were interested in the atomic-level origins of elasticity. Elasticity in most solids comes from the chemical bonds between atoms, with the repulsive force due to overlap of electron wavefunctions. This is not the dominant mechanism in polymers like the rubber that makes up most of your golf ball, however. In soft rubbers entropy effects can dominate over classic springiness.

The simplest polymer chain looks like a string of balls separated by rods with free joints, with Brownian motion wiggling all the balls randomly. The chain tends to collapse into a clump with a mean size ("radius of gyration")that maximizes the entropy. Pulling the chain straight or compressing it gives an unfavorable lower entropy state. A rubber solid is a three-dimensional version of this (due to cross-linking of the chains). You can find the effective "entropic force" by differentiating the Helmholtz free energy with respect to displacement. It's a little long to explain in a post, but not hard to follow. Take a look at books on polymers. de Gennes has a nice one, the title is something like Scaling Behavior of polymers. I'll take a look for a web site, too.
 
  • #13
guys..lets not go deep into this...the force making the ball go up is the normal force exerted by the grounds...caused by electrostatic replusions...blah blah blah
 

Related to Golf Ball Bouncing: What Force Causes It?

1. What is the force that causes a golf ball to bounce?

The force that causes a golf ball to bounce is an elastic force, also known as a restoring force. This force is created when the golf ball compresses upon impact with a surface, and then quickly expands back to its original shape, propelling the ball upwards.

2. What factors can affect the height of a golf ball's bounce?

Several factors can affect the height of a golf ball's bounce, including the material and construction of the ball, the surface it is bouncing on, and the angle and force of impact. Additionally, air pressure, temperature, and humidity can also play a role in the ball's bounce.

3. Why do some golf balls bounce higher than others?

The height of a golf ball's bounce depends on its physical characteristics, such as the material and construction. Some balls are designed to have a higher elasticity, which allows them to compress and expand more efficiently, resulting in a higher bounce. Additionally, the surface the ball is bouncing on can also affect its bounce height.

4. Does the force of gravity affect a golf ball's bounce?

Yes, the force of gravity plays a significant role in a golf ball's bounce. As the ball is propelled upwards by the elastic force, gravity pulls it back down to the ground. The height of the bounce is determined by the balance between these two opposing forces.

5. Can the force of a golf swing affect the ball's bounce?

Yes, the force and angle of a golf swing can significantly impact the height and direction of a golf ball's bounce. A more powerful swing will result in a higher initial velocity and a higher bounce, while a more angled swing may cause the ball to bounce in a different direction.

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