Force in Center of Mass frame equaling zero

[CGandC]

Homework Statement

Suppose I have particles with masses m1 and m2 , both are in freefall:

Then , looking at the center of mass coordinates , I know that the sum of momentums of masses m1 and m2 in the center of mass coordinates is equal to zero : P1c + P2c = 0

so now I know that the total force acting on the masses in the center of mass frame is :
Fcmf = d(P1c + P2c )/dt = 0

From here I get : Fcmf = 0

But , If I sum the forces ( only weights ) of the two bodies, I don't get zero : Fcmf= -M2g -M1g ≠ 0
(I get contradiction , because I know as a fact that Fcmf must be equal to zero , Fcmf = 0 )
Why do I get a contradiction?

Homework Equations

F = ma

The Attempt at a Solution

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[kuruman]

P1c + P2c = 0

This is true only if the CM does not accelerate, i.e. if there are no external forces acting on it. Here there is gravity.

 

[CGandC]

This is true only if the CM does not accelerate, i.e. if there are no external forces acting on it. Here there is gravity.

so what will be the sum of momentums P1c + P2c if there are external forces acting on the system? I always get that P1c + P2c = 0 , regardless of whether there are external forces acting on the system or not.

 

[conscience]

This is true only if the CM does not accelerate, i.e. if there are no external forces acting on it. Here there is gravity.

This isn't correct .

OP is right that the net momentum of the two masses in CM frame is zero .

Total momentum of the system is always zero in CM frame irrespective of whether CM is accelerating or not .

 

[conscience]

so what will be the sum of momentums P1c + P2c if there are external forces acting on the system? I always get that P1c + P2c = 0 , regardless of whether there are external forces acting on the system or not.

You are right

the total force acting on the masses in the center of mass frame is :
Fcmf = d(P1c + P2c )/dt = 0

F = dP/dt is valid only in inertial (non accelerating) frames .

Since CM is accelerating , it is not an inertial frame .

You are right when you say that ΣP = 0 in CM frame . But you are wrong if you apply F=dP/dt in CM frame .

Note that F=dP/dt is nothing but Newton's 2^nd Law which is valid only in inertial frames .

Hope that helps .

 

[kuruman]

This isn't correct .

You are correct. I meant to say what you said much more eloquently in post #5.

 

[CGandC]

You are right
F = dP/dt is valid only in inertial (non accelerating) frames .

Since CM is accelerating , it is not an inertial frame .

You are right when you say that ΣP = 0 in CM frame . But you are wrong if you apply F=dP/dt in CM frame .

Note that F=dP/dt is nothing but Newton's 2^nd Law which is valid only in inertial frames .

Hope that helps .

Just to make sure,
If I were to use Newton's second law for the CM accelerating frame , I'd have to fix it like this ( written below ) ? :
F+Ffict=dP/dt , where Ffict is a pseudo-force / fictional force
and This pseudo force is equal to : Ffict = +M1g +M2g , so that F+Fict = 0 , Is this correct?

 

[conscience]

Just to make sure,
If I were to use Newton's second law for the CM accelerating frame , I'd have to fix it like this ( written below ) ? :
F+Ffict=dP/dt , where Ffict is a pseudo-force / fictional force
and This pseudo force is equal to : Ffict = +M1g +M2g , so that F+Fict = 0 , Is this correct?

I am not sure if you are properly applying the fictitious force .

This is what is happening in your example .

Fictitious force acts on all the particles whose motion we are interested in as seen from the accelerated frame .

In your example , fictitious force acts on both $M_1$ and $M_2$ in the direction opposite to the acceleration of the frame (CM frame in your example) .

So , considering downward positive , as seen from the CM frame , fictitious force $-M_1a_{cm}$ acts on $M_1$ apart from $M_1g$ .Similarly fictitious force $-M_2a_{cm}$ acts on $M_2$ apart from $M_2g$ .

$a_{cm}=g$

The net force acting on $M_1$and $M_2$ as seen from the CM frame = $-M_1a_{cm} - M_2a_{cm} +M_1g+M_2g$ . This evaluates to 0 .