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Internal forces and centre of mass

  1. May 10, 2005 #1
    heres another conceptual one .
    there is a standard yet very elegant problem about a rocket bllowing up mid air and finding the final postions of the parts of the projectile using the principle of centre of mass remaining the same since the forces a re internal when the rocket blows up . now what happens if i drop a glass cup from a height will the cordinates of the centre of mass remain the same ...will the force still be internal ... can i change the system i am working with to make the force internal ?
  2. jcsd
  3. May 10, 2005 #2

    Doc Al

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    Right. Ignoring air resistance, the center of mass of the rocket pieces will have the same trajectory it would have had if the rocket had not exploded.
    Not sure what forces you are talking about. Drop the glass and it falls, due to gravity, an external force. When the glass breaks, additional external forces act on the glass. You can always enlarge the system to make the forces internal; in this case by including the earth, gravity becomes an internal force: the center of mass of the earth-glass system maintains its original trajectory.
  4. May 11, 2005 #3


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    In the case of the rocket, the only significance of the forces being internal is that the center of mass has not been acted upon by some outside force. Therefore, the center of mass will continue to behave inertially, following a ballistic trajectory until it reaches the ground. The important thing is that, if the force of the explosion is both internal and symmetrical, the center of mass will behave as though no force had acted upon it.

    The glass, when dropped, will also follow a ballistic trajectory until it reaches the ground. It's center of mass will behave as though no force had acted upon it for the simple reason that none has. So the point at which the ballistic path of the original object's center of mass reaches the ground will be the center of mass for the debris field. In the case of the glass, the debris field will be more dense because debris did not begin spreading out from the center of mass until it reached the ground.

    I would wager that in real life, predicting the trajectory of the center of mass from an exploding racket is slightly more complex. If the explosion is asymmetrical (which I think it would have to be, to some degree) then the vaporized explosive escaping from one side of the rocket would constitute a reaction mass, altering the path of the solid debris. Although I suppose if one could include the cloud of gases from the explosion as part of the "debris", the center of mass for the entire system would still be the same whether the explosion were symmetrical or asymmetrical.
  5. May 11, 2005 #4
    lurch ,
    if you consider the gases from the rocket to be light which i suppose they are their masses can be neglected ,then the coordinates of the projectile can be found with some accuracy .
    regarding the glass ,suppose the cup broke into 4 pieces 3 of which are at the corners of a square whose centre is where the cup landed before breaking then would the 4th piece would be at the 4th corner of the sqr.?
    u say that no external forces act on the cup what about gravity and interaction with the ground or are u ignoring them since they act along the vertial axis ...
    Last edited: May 11, 2005
  6. May 12, 2005 #5


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    I ignore gravity because it is the same for both objects. Once the rocket's engine stops providing thrust, its center of mass will follow a balistic trajectory. This is true whether the rocket has blown up, or the engine has simlpy shut off. The glass's center of mass will also follow a balistic trajectory once it is released. So the two cases are the same. (I say "no forces" because gravity isn't a true "fore").

    I was disregarding impact with the ground because the it marks the end of the perameters for the prediction (in either case). Any effect this impact will have on the debris field is going to happen after the impact, when the center of mass has become, to all practical purposes, stationary. This of course does not take into account the "bounce" of the various materials after hitting.
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