# I Motion of the Center of Mass

1. Dec 18, 2016

### Aldnoahz

Hi all,

I've been stuck on thinking about a situation where a mass splits in two in midair. The mass originally undergoes a projectile motion with no air resistance, and it splits in two in midair but one of the piece reaches the ground before the other. In this case, I know that the trajectory of center of mass no longer stays on the original path, but what I want to know is how the trajectory will change after one piece hits ground. Will it make steeper angle to reach the ground or will it make a less steep angle to reach the ground.

I know that the net external force is smaller after one piece hits the ground, but I seem not to understand the effect of this change in force on the path of center of mass.

Thanks

2. Dec 18, 2016

### vanhees71

In such a case you have to include the Earth in your analysis, i.e., a closed system, and then momentum conservation holds true and thus the center of mass of this complete system moves with constant velocity.

3. Dec 18, 2016

### Aldnoahz

Is it possible to only look at the mass as a system and regard gravity as an external force and have an intuitive sense as how the center of mass would travel?

4. Dec 19, 2016

### Cutter Ketch

For a qualitative sense, just imagine what the center of mass would have done if the piece hadn't hit the ground. If it hadn't hit the ground the piece would have continued further down range and lower. Therefore after the piece stops the center of mass is ever more up range and higher than where it would otherwise be. The mass of the piece times the vector between where the piece would be if hadn't hit the ground and the point where it did hit the ground can be subtracted from the undisturbed trajectory to get the new trajectory of the center of mass.

Now my question is why would you want to know? The center of mass trajectory ceases to be a useful concept once a piece hits the ground. What possible meaning or use could this have for you?

5. Dec 19, 2016

### jbriggs444

When the one piece hits the ground, there is an impulsive force that brings that piece to a stop. That force changes the velocity of the center of mass. It may change the speed of the center of mass or may change its direction or may do both. It is possible to come up with scenarios where this makes the angle of trajectory of the center of mass either steeper or shallower. [Careful choice of angles, ejection velocities and relative masses can achieve either result]

Following the collision, the motion of the center of mass is simple. As you say, the net external force has been reduced by the addition of an external supporting force on the detached mass. One way to analyze the trajectory of the center of mass would be to simply apply the equations of motion (they still apply) to a virtual object which is positioned at the center of mass of the combined system, has the mass of the combined system and is subject to a net force equal to the force of gravity on the piece that is still moving.

6. Dec 19, 2016

### albertrichardf

Well, if the original mass is undergoing projectile motion, it is simpler to suppose it splits at the apex of the quadratic curve, such that the mass originally has zero velocity.
In the absence of air resistance and gravity, the split masses would have to move in opposite directions to conserve momentum. Momentum was zero originally since the whole mass was stationary, and it can't change in the absence of a force.

With gravity, the original mass would still be stationary at the apex of the quadratic, but momentum in the downwards direction would not be conserved because of gravity. The momentum in the two other directions would be zero for the centre of mass because that has to be conserved since there is no net force in those directions. While the masses are still moving in opposite directions, their downwards motion is increasing because of gravity. Thus one mass would have to initially move upwards and the other downwards, so that the mass with downwards velocity hits the ground first.

Now, in this scenario, the centre of mass had no horizontal* component of velocity, so its path was completely vertical, at 90º to the ground. The two masses, are however assumed to have a velocity component in horizontal opposite directions due to whatever separated them to begin with, which add up to zero. Upon collision, the centre of mass moves in the same direction as the moving piece of mass, making its velocity a bit more horizontal. This decrease the angle it makes to the horizontal, and thus makes it less steep. The larger the horizontal velocity component, the less steep it will be. If the two masses had no horizontal component, then there is no change in the angle, since it will fall straight down anyway.
If the mass originally had a horizontal velocity as well, the centre of mass would have to have the same velocity by conservation of components of momentum. There are two ways for that to happen: either the pieces move in the same direction, at a lower horizontal velocity than the centre of mass, or they move in opposite directions, with at least one piece having more momentum than the centre of mass, so that the higher momentum is counterbalanced by the momentum of the other piece.

*What I mean by horizontal is any direction that lies in the plane perpendicular to the direction of gravity.

If they both move in the same direction, the centre of mass loses horizontal velocity when one piece strikes the ground, but also loses vertical velocity as well. At this point it becomes impossible to tell how the angle will change qualitatively. The angle of the velocity of the centre of mass after the piece strikes the ground to the horizontal is the same as that of the still-falling piece. The same holds true if they were moving in opposite directions.

The argument still holds for when the mass does not split at the apex, though it is not necessary for each piece of mass to have vertical velocities in opposite directions in that case. Besides, you could always change your frame of reference to an inertial one such that the velocity of the mass when it splits is zero. Also, note that whatever the angle is initially, it will become steeper and steeper since it is accelerating downwards.