Energy made available by Change in CoG

  • Thread starter Str1pe
  • Start date
  • #1
12
0

Main Question or Discussion Point

I would like to discuss a physics problem. It's not homework or coursework. It's just an area of interest.

I know there are models with which one may calculate a center of gravity for different 2d or 3d objects..

and

.. I know that there are models to calculate available energy..

but

I don't know enough maths in order to combine the two. So, say a seesaw has a kid standing in the middle of it who jumps a meter down the beam. The seesaw-chair will hit the ground with X amount of energy. But the same kid jumping two meters down the beam will see the chair hit the ground with significantly more energy. How much more?

I'd like to learn more about how to do the math to solve these sorts of problems. It's not homework - just a friendly discussion. :)

Thanks.
Stripe.
 
Last edited:

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
249
Welcome to PF!

Hi Str1pe! Welcome to PF! :smile:
… a seesaw has a kid standing in the middle of it who jumps a meter down the beam. The seesaw-chair will hit the ground with X amount of energy. But the same kid jumping two meters down the beam will see the chair hit the ground with significantly more energy. How much more?
This has very little to do with centre of gravity (centre of mass), it's all about conservation of energy …

ie PE + KE = constant.

The potential energy (PE) is mass times g times difference in height.

The seesaw will go the same distance in either case, and its change in potential energy is zero anyway (because it's balanced at the pivot, which doesn't move).

So the only difference is the difference in height of the kid … obviously, if he's eg twice as far away from the pivot, then his difference in height is twice as much, and so the kinetic energy (KE) of the kid and the seesaw will be twice as much. :smile:
 
  • #3
12
0


Hi Str1pe! Welcome to PF! :smile:
:smile:

This has very little to do with centre of gravity (centre of mass), it's all about conservation of energy … ie PE + KE = constant. The potential energy (PE) is mass times g times difference in height. The seesaw will go the same distance in either case, and its change in potential energy is zero anyway (because it's balanced at the pivot, which doesn't move). So the only difference is the difference in height of the kid … obviously, if he's eg twice as far away from the pivot, then his difference in height is twice as much, and so the kinetic energy (KE) of the kid and the seesaw will be twice as much. :smile:
Oh, OK.

The same calculation could be done knowing only the distance the center of gravity changed and the weight of the beam and kid, right?

Thanks for your time.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
249
The same calculation could be done knowing only the distance the center of gravity changed and the weight of the beam and kid, right?
s'right! :biggrin:
 
  • #5
12
0
Cool. Now how about a large scale exercise...?

The Center of Gravity of the moon is currently located slightly to the near side from the center of mass for the moon. Assuming the moon was originally perfectly spherical with the CoG located at the geometric center, how might we discover energy made available from a theoretical shift in CoG? Energy is provided by gravity which will work to mold the mass of the moon into a sphere around its CoG.

I'm assuming the simple answer provided for the see saw example will not work here.

Thanks again for your time.
 
  • #6
tiny-tim
Science Advisor
Homework Helper
25,832
249
The Center of Gravity of the moon is currently located slightly to the near side from the center of mass for the moon.
Is it? Do you have a reference for this? :confused:

Anyway, I don't think there's any such thing as the centre of gravity … see http://en.wikipedia.org/wiki/Centre_of_gravity" [Broken] …
Symon, in his textbook Mechanics, shows that the center of gravity of an extended body must always be defined relative to an external point, at which location resides a point mass that is exerting a gravitational force on the object in question. In fact, as Symon says:
"For two extended bodies, no unique centers of gravity can in general be defined, even relative to each other, except in special cases, as when the bodies are far apart, or when one of them is a sphere....The general problem of determining the gravitational forces between bodies is usually best treated by means of the concepts of the field theory of gravitation..."
Even when considering tidal forces on planets, it is sufficient to use centers of mass to find the overall motion. In practice, for non-uniform fields, one simply does not speak of a "center of gravity"​
 
Last edited by a moderator:
  • #7
12
0
Is it? Do you have a reference for this? :confused:
Yeah .. http://exploration.grc.nasa.gov/education/rocket/moon.html" [Broken] so I figure its CoG is as well.

Anyway, I don't think there's any such thing as the centre of gravity … see http://en.wikipedia.org/wiki/Centre_of_gravity" [Broken] …Symon, in his textbook Mechanics, shows that the center of gravity of an extended body must always be defined relative to an external point, at which location resides a point mass that is exerting a gravitational force on the object in question. In fact, as Symon says: "For two extended bodies, no unique centers of gravity can in general be defined, even relative to each other, except in special cases, as when the bodies are far apart, or when one of them is a sphere....The general problem of determining the gravitational forces between bodies is usually best treated by means of the concepts of the field theory of gravitation..." Even when considering tidal forces on planets, it is sufficient to use centers of mass to find the overall motion. In practice, for non-uniform fields, one simply does not speak of a "center of gravity"
So I'm using the wrong term? I should say center of mass rather than gravity?

I see this as a more complicated version of the see-saw experiment.

The moon as a uniform sphere is like the kid standing in the middle of the seesaw. But the moon as we have it today is like the kid has jumped a meter down one beam. The mass of the moon is displaced towards the earth. With this hypothetical change to the CoM and with the known mass of the moon, can we calculate energy made available?
 
Last edited by a moderator:
  • #8
tiny-tim
Science Advisor
Homework Helper
25,832
249
Hi Str1pe! :smile:

(just got up :zzz: …)
The Center of Gravity of the moon is currently located slightly to the near side from the center of mass for the moon.
ah, that's not what your nasa article says …
http://exploration.grc.nasa.gov/education/rocket/moon.html" [Broken] so I figure its CoG is as well.
the article only mentions the centre of mass, not the centre of gravity.

For all practical purposes, the Earth's gravitational field can be regarded as uniform throughout the Moon, so the centre of gravity can be regarded as being in the same place.

(Both, of course, are not in the geometric centre of the Moon.)
So I'm using the wrong term? I should say center of mass rather than gravity?
Yes, "centre of mass" is considered better, since it is defined whether gravity is present or not. :wink:
I see this as a more complicated version of the see-saw experiment.

The moon as a uniform sphere is like the kid standing in the middle of the seesaw. But the moon as we have it today is like the kid has jumped a meter down one beam. The mass of the moon is displaced towards the earth.
Yes. :smile:
With this hypothetical change to the CoM and with the known mass of the moon, can we calculate energy made available?
Not following you … made available for what? … compared with what? :confused:
 
Last edited by a moderator:
  • #9
12
0
ah, that's not what your nasa article says … the article only mentions the centre of mass, not the centre of gravity.
OK, I'll fix my terminology .. actually, I'll describe a hypothetical .. it might help us more (see-saw comparison in brackets).

Imagine a planetary mass that is the only matter in existence. Its own gravity is the only gravity it ever experiences. It is a perfect sphere and its mass is evenly distributed so the CoG is co-located with the CoM at its geometric center (the see-saw is balanced with the kid in the middle).

God comes along and takes a large chunk out of one side of the planet and places it on the opposite side of the planet (the kid jumps a meter down one beam). Now the mass of the planet (and the see-saw/kid system) is unevenly distributed. Its Centers of Mass and Gravity are changed.

Gravity prefers that the planet be a sphere, so gravity will act toward this end (gravity will drag downwards the side of the see-saw with more kid on it).

MY QUESTION IS:
Can we use the change in CoG - or CoM - to calculate the energy made available (like you said we could use the change in CoG from the see-saw experiment to calculate energy made available)?

Hope you slept well. :)
 
  • #10
tiny-tim
Science Advisor
Homework Helper
25,832
249
Imagine a planetary mass that is the only matter in existence

Gravity prefers that the planet be a sphere, so gravity will act toward this end (gravity will drag downwards the side of the see-saw with more kid on it).
What gravity? You said there's not other matter in existence.

If the seesaw was floating in space, putting the kid near one end would make no difference … in the absence of gravity, the seesaw will stay as it is (or will continue rotating uniformly).
MY QUESTION IS:
Can we use the change in CoG - or CoM - to calculate the energy made available (like you said we could use the change in CoG from the see-saw experiment to calculate energy made available)?
No, the energy was calculated from the distance the centre of mass of the kid moved before the seesaw hit the ground.

Walking the kid along the seesaw only makes energy available because it makes movement of the kid available (in the middle of the seesaw he couldn't move)
 
  • #11
12
0
What gravity?
The gravity of the planet. The gravity of a planet is what makes it round.

If the seesaw was floating in space, putting the kid near one end would make no difference … in the absence of gravity, the seesaw will stay as it is (or will continue rotating uniformly).
Agreed, but the kid on the see-saw is not in space. He's on earth. :)

No, the energy was calculated from the distance the centre of mass of the kid moved before the seesaw hit the ground. Walking the kid along the seesaw only makes energy available because it makes movement of the kid available (in the middle of the seesaw he couldn't move)
OK.

When mass is moved from one side of a spherical planet to the other, that mass (actually - all the mass of the planet) is now available to move. The non-spherical planet prefers to return to the spherical because of its own gravity.
 
  • #12
Q_Goest
Science Advisor
Homework Helper
Gold Member
2,974
39
If the CG of a planet is shifted, it takes energy to do that. Putting energy in to move the CG means it can be taken back out to some degree. However, taking the energy back out by shifting the CG a second time will never be a reversible process. Some energy is inevitably converted to heat.
 
  • #13
12
0
If the CG of a planet is shifted, it takes energy to do that. Putting energy in to move the CG means it can be taken back out to some degree. However, taking the energy back out by shifting the CG a second time will never be a reversible process. Some energy is inevitably converted to heat.
OK .. let's ignore energy required to move the CG .. how much energy will be made available?

What sort of maths do I need to do to to figure that out?
 
  • #14
Q_Goest
Science Advisor
Homework Helper
Gold Member
2,974
39
I'm not going to do the math for you. Perhaps someone else would like to try. But to do it there has to be some kind of model that one can apply the math to. I'll give you that much.

Assume a hollow spherical shell of mass M1 and inertia I1. Inside this spherical shell, imagine a straight bar running from one inside surface, through the center, to the opposite inside surface. Assume this bar is massless to make it easier to do the math. Now put an object with mass M2 and inertia I2 on the center of this bar (center of spherical shell). Now assume the contraption is spinning with some initial rotational speed W1.

If we start out with M2 in the center of the spherical shell, and start to move away from that center, then the rotation of the system causes the masses to be pushed apart. If you watched this happen from the outside, you'd see the spherical shell start to wobble. That wobble would increase in amplitude as the center of the two masses moved away from each other.

Now imagine a string between the center of M1 and M2. That string will be taught because of the centrifugal force pushing the two apart. You could conceivably "collect" this energy. To calculate that you'd need to integrate the force over the distance that the two masses moved. But it would be very difficult to calculate that force because it also is a function of the rotational inertia (I1 and I2) of the two masses.

To return the system to the un-wobbling state, you'd need to put that energy back in so that the two masses came back to the same center. Note that if you took energy out of this system, when you brought the geometry of the system back to the original location, the object wouldn't be spinning as fast. The energy you removed would have slowed down the rotational velocity of the system.
 
  • #15
12
0
I'm not going to do the math for you. Perhaps someone else would like to try. But to do it there has to be some kind of model that one can apply the math to. I'll give you that much.

Assume a hollow spherical shell of mass M1 and inertia I1. Inside this spherical shell, imagine a straight bar running from one inside surface, through the center, to the opposite inside surface. Assume this bar is massless to make it easier to do the math. Now put an object with mass M2 and inertia I2 on the center of this bar (center of spherical shell). Now assume the contraption is spinning with some initial rotational speed W1.

If we start out with M2 in the center of the spherical shell, and start to move away from that center, then the rotation of the system causes the masses to be pushed apart. If you watched this happen from the outside, you'd see the spherical shell start to wobble. That wobble would increase in amplitude as the center of the two masses moved away from each other.

Now imagine a string between the center of M1 and M2. That string will be taught because of the centrifugal force pushing the two apart. You could conceivably "collect" this energy. To calculate that you'd need to integrate the force over the distance that the two masses moved. But it would be very difficult to calculate that force because it also is a function of the rotational inertia (I1 and I2) of the two masses.

To return the system to the un-wobbling state, you'd need to put that energy back in so that the two masses came back to the same center. Note that if you took energy out of this system, when you brought the geometry of the system back to the original location, the object wouldn't be spinning as fast. The energy you removed would have slowed down the rotational velocity of the system.
Hmmm .. OK.

Thanks very much for the input. I will think this over.
 
  • #16
12
0
I'm not going to do the math for you. Perhaps someone else would like to try. But to do it there has to be some kind of model that one can apply the math to. I'll give you that much.

Assume a hollow spherical shell of mass M1 and inertia I1. Inside this spherical shell, imagine a straight bar running from one inside surface, through the center, to the opposite inside surface. Assume this bar is massless to make it easier to do the math. Now put an object with mass M2 and inertia I2 on the center of this bar (center of spherical shell). Now assume the contraption is spinning with some initial rotational speed W1.

If we start out with M2 in the center of the spherical shell, and start to move away from that center, then the rotation of the system causes the masses to be pushed apart. If you watched this happen from the outside, you'd see the spherical shell start to wobble. That wobble would increase in amplitude as the center of the two masses moved away from each other.

Now imagine a string between the center of M1 and M2. That string will be taught because of the centrifugal force pushing the two apart. You could conceivably "collect" this energy. To calculate that you'd need to integrate the force over the distance that the two masses moved. But it would be very difficult to calculate that force because it also is a function of the rotational inertia (I1 and I2) of the two masses.

To return the system to the un-wobbling state, you'd need to put that energy back in so that the two masses came back to the same center. Note that if you took energy out of this system, when you brought the geometry of the system back to the original location, the object wouldn't be spinning as fast. The energy you removed would have slowed down the rotational velocity of the system.
Will this model work for my question? Is it a simple matter to convert the rotational energy into the gravity equivalent? Are there other models that do this for other situations?

The situation I described does not require any rotation of the body, but I think I understand there might be the need to describe the process in terms of rotation.
 
  • #17
Q_Goest
Science Advisor
Homework Helper
Gold Member
2,974
39
The model doesn't need rotation. Increasing rotational rate increases the centrifugal force throwing the two masses apart. That has to be overcome by the gravitational force. The most simple model would be to disregard rotation, then gravitational force pulls the two masses toward each other all the time and you can calculate the potential energy between two masses from:

ea282fdef5680909de300b443993ad9e.png


See: http://en.wikipedia.org/wiki/Potential_energy#Gravitational_potential_energy
 
  • #18
12
0
The model doesn't need rotation. Increasing rotational rate increases the centrifugal force throwing the two masses apart. That has to be overcome by the gravitational force. The most simple model would be to disregard rotation, then gravitational force pulls the two masses toward each other all the time and you can calculate the potential energy between two masses from:

ea282fdef5680909de300b443993ad9e.png


See: http://en.wikipedia.org/wiki/Potential_energy#Gravitational_potential_energy
Oh. Then that's pretty darn simple then, ain't it?
 

Related Threads for: Energy made available by Change in CoG

  • Last Post
Replies
1
Views
2K
Replies
4
Views
2K
  • Last Post
Replies
13
Views
2K
  • Last Post
2
Replies
43
Views
7K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
21
Views
8K
Top