When a ball hits the ground, where in the ball is potential energy = 0

[1nvictus]

This is a question that me and a friend have been wondering. He says that when the ball hits the ground and is at rest for a very brief moment, potential energy will be 0 in the center of the ball, while I thought that it doesn't matter as the entire ball's potential energy is at 0. Which answer is correct - or are we both wrong?

 

[Simon Bridge]

Welcome to PF;
There is no absolute zero for potential energy.
Whichever part of the ball is at the height you have marked off as "zero" on your PE scale - that is the part that has zero (gravitational) PE.

If the entire ball is stationary, then the entire ball has zero kinetic energy.

When you do gravitational PE questions, you use the change in PE between different positions - not any absolute PE.

 

[russ_watters]

This is a question that me and a friend have been wondering. He says that when the ball hits the ground and is at rest for a very brief moment, potential energy will be 0 in the center of the ball, while I thought that it doesn't matter as the entire ball's potential energy is at 0. Which answer is correct - or are we both wrong?

You thought what doesn't matter? I don't see two clear choices here.

 

[1nvictus]

I thought that when the ball hits the ground, at that exact moment the potential energy throughout the entire ball is 0. After seeing Simon's answer (thank you, by the way!) however, it seems that PE is 0 at the bottom of the ball.

 

[Simon Bridge]

If you defined PE=0 at the ground, then the part of the ball in contact with the ground has zero PE.

BUT: the potential energy of the whole ball at a position is the amount of work needed to get the ball to that position from some reference position. It so happens that when you do the calculation, you get the same result that you would if all the mass of the ball were concentrated at it's center of mass.

If the reference position for the ball is "on the ground" then the ball, as a unit, has zero PE.
Each part of the ball may have a different PE.

This also means that your reference height is the ball's radius above the ground... i.e. at the ball's center of mass.

Hence - you and your friend are both correct :)

You don't normally need to worry about it since it is the change in PE that matters.

 

[mikeph]

potential energy is not a spatially varying quantity, it's a global property of the ball itself. you are effectively integrating rho*g*h over its volume; that integral is a single number at every instant in time.

 

[.Scott]

This is a question that me and a friend have been wondering. He says that when the ball hits the ground and is at rest for a very brief moment, potential energy will be 0 in the center of the ball, while I thought that it doesn't matter as the entire ball's potential energy is at 0. Which answer is correct - or are we both wrong?

There are two sources of potential energy. One is the elevation above the ground. In that sense, only the bottom has zero PE.

However, what will propel the bounce of the ball is the elastic compression or distortion. Only the very top tip of the ball is not under some sort of elastic distortion, so only that point will not contribute to the bounce.

 

[AlephZero]

I assume the OP was asking about gravitational potential energy, not the internal strain energy in the ball when it is deformed by hitting the ground.

Ignoring the deformation of the ball when it bounces, and assuming it is not rotating, you can look at this two ways:

1. The "top" of the ball always has a different amount of GPE from the "bottom", because they are at different heights above the ground. But since the difference in heights is always the same, you might as well simplify the problem and ...

2. Measure the GPE of the complete ball using the height of its center of mass above the ground. And since you can choose the height at which GPE = 0, you might as well choose the position of the CM when the ball is touching the ground, not at the actual ground level.

 

[mikeph]

It doesn't make sense to me to say the top has more potential energy than the bottom. There is no potential field, and the units are J, not J/m^3 so you can't treat it as if it were a material property.

When you say the top has more potential energy, is that just because it's higher up? But higher up from what? The bottom can fall exactly the same distance as the top, unless somehow the ball completely flattens itself upon landing.

 

[mfb]

It doesn't make sense to me to say the top has more potential energy than the bottom. There is no potential field, and the units are J, not J/m^3 so you can't treat it as if it were a material property.

If you divide the ball into an upper and a lower part (both with half the mass of the ball), the upper part will have a larger potential energy. This follows directly from its definition as "upper part".
It does not make sense to say "the top has more potential energy", but the different parts of the ball lead to different contributions to the total potential energy.

 

[sophiecentaur]

It would probably be more strict to say that the top parts of the ball lose more GPE than the bottom bits, because of the strain / distortion of the ball whilst it is bouncing (the top bits move further down than the bottom bits, in contact with the (rigid) ground. The higher the modulus of the ball material, the less difference there will be. The potential (stored) energy due to the strain will be greater for a softer ball. So will the losses due to hysteresis, which is a reason why steel balls and 'super balls' bounce higher than soft rubber ones.

 

[Simon Bridge]

It doesn't make sense to me to say the top has more potential energy than the bottom. There is no potential field, and the units are J, not J/m^3 so you can't treat it as if it were a material property.

Close to the Earth's surface a particle of mass m has gravitational potential energy U=mgy ... where y is the vertical distance from some reference height... the ground is a common choice but it does not have to be.

Particle B higher than particle A has more potential energy than particle A.

The ball is made of particles.

When we say that the top part of the ball has more potential energy than the lower part, we are referring to the constituent particles at the different places in the ball.

All-together, the ball, as a unit, has potential energy at some position equal to the work needed to bring the whole ball to that position. In that sense, we can assign a PE to the whole ball as a unit.

That will be the sum of the work needed to bring each individual particle to the appropriate location to end up with a ball.

When you say the top has more potential energy, is that just because it's higher up?

Yes.

But higher up from what?

... higher up from the reference position where zero potential energy is defined.

The bottom can fall exactly the same distance as the top, unless somehow the ball completely flattens itself upon landing.

That is correct - but now you are thinking of the change in position of the different parts of the ball - which will be the same, and lead to the same value for the change in potential energy. But the start and end potential energies will still be different.

If the bottom of the ball falls a distance h to the ground, then it ends at y=0 and starts at postion y=h. Its change in potential energy is (final minus initial) $\Delta U = mg(0)-mgh)=-mgh$. [1]

At the same time, the same amount of mass at the top of the ball starts out at y=h+d (d=diameter of the ball) and ends up at y=d. This is a change in potential energy of $\Delta U = mgd-mg(h+d) = -mgh$ ... i.e. exactly the same. [2]

However, at all times through the drop, the (the particle at the) top of the ball had $mgd$ more potential energy than the (the particle at the) bottom part.[3]



This sort of stuff gets glossed over in beginning physics classes because it gives students (and their teachers) headaches that they don't need right then.


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[1] assuming y=0 at the floor, and +y is "upwards". If I put y=0 at the center of the Earth, and U=0 on the ground (y=R), then the equation becomes U=mg(y-R). Of course, I can put U=0 anywhere I like!

[2] ... but what happens if the field is not uniform?

[3] ... making the usual high-school idealizations. IRL: there would be more to it.