# Internal forces and centre of mass

• govinda
In summary: But again, that would be the same in either case. So I'm not ignoring any forces; I'm just stating that they are equal in either case.In summary, the conversation discusses the concept of finding the final positions of the parts of a projectile, such as a rocket or a glass cup, after it has been destroyed or dropped. The principle of center of mass remaining the same is used, as the forces are considered to be internal. However, the force of gravity, which is an external force, does affect the trajectory of the objects and can be included in the system to make the forces internal. The conversation also mentions that predicting the trajectory of the center of mass for an exploding rocket may be more complex due to asymmetrical
govinda
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 ?
thanks
govinda

govinda said:
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 .
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.
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 ?
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.

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.

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 ...

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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.

## 1. What are internal forces?

Internal forces are forces that act within a system or object, such as the forces between molecules in a solid or the forces between parts of a structure. These forces do not change the overall motion of the object, but rather they cause internal stresses and deformations.

## 2. How do internal forces affect the center of mass?

The center of mass is the point at which the mass of an object is concentrated and is affected by all of the forces acting on the object, including internal forces. Internal forces can cause the center of mass to shift or move, depending on their direction and magnitude.

## 3. What is the difference between internal forces and external forces?

External forces act on an object from outside of the system, while internal forces act within the system. External forces can change the overall motion of an object, while internal forces only cause internal changes within the object.

## 4. How do internal forces contribute to the stability of structures?

Internal forces play a crucial role in maintaining the stability of structures. They balance out external forces and prevent the structure from collapsing or deforming. By distributing the forces evenly throughout the structure, internal forces help keep it in equilibrium and maintain its shape.

## 5. Can internal forces be ignored in calculations and analysis?

No, internal forces should not be ignored in calculations and analysis. While they may not affect the overall motion of an object, they can significantly impact the internal stresses and deformations within the object. Neglecting internal forces can lead to inaccurate results and potentially dangerous consequences in structural design and engineering.

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