Center of mass and momentum- A question about systems

[palaphys]

Homework Statement

A triangular wedge of mass M has a smaller mass m on its top most point. Initially the system is at rest. Considering the fact that the ground is frictionless, analyze the motion of the center of mass of the wedge block system

Relevant Equations

P=mv, center of mass= sum of mass * coordinates / sum of all masses, conservation of momentum

consider this situation. The diagram I have drawn is just after the smaller mass m begins sliding.

So first thing I wanted to do is define my system. Here I have defined my system as "Small mass m+ Larger wedge M".
DOUBT 1: I had a rather silly idea- is it right to consider the "floor" as part of the system? Is that possible in the first place?

okay, now, I drew a free body diagram of the masses to predict how they would behave:-
|

here N_g represents the normal reaction from the ground, and N_1 is the internal normal reaction between the two blocks. I have divided that into two components. So from here, it is clear that the larger mass M is going to move towards the left, with some acceleration.

DOUBT 2: FBD of the "system"
so in addition to these, I am going to draw the FBD of the system as defined earlier:

the forces acting on the system are (M+m)g and the normal force from the ground, N_g.
POINT 1: as there are no external forces operating in the x- direction,
$$p_{ix}= p_{fx} $$
As far as I have learnt, I have learnt that "if there are no external forces acting on a system, the acceleration of the center of mass is 0, i.e the velocity of the center of mass is constant." This implies that the position of the center of mass in the x direction (here) does not change at all.
DOUBT 3:- My main query with this question other than the ones above, is regarding the displacement/ behaviour of the y- coordinates of the center of mass. As per the FBD on the right, the vectors $$ (M+m)g , N_g $$ seem to cancel out each other.
Does this imply that momentum is conserved in the y direction as well? is the CM of the system in rest throughout the motion?
I came to this assumption that it is in rest by considering the equation
$$ F_{ext,y} = (M+m)a_{cm,y} $$, External force here is PROBABLY 0, and under that assumption, a_cm,y becomes zero.
However, when I went through this problem again
$$ Y_cm = (My_1 + my_2)/ (M+m) $$
Differentiating twice,
$$ a_{cm,y} = (Ma_{y1} + ma_{y2} )/ (M+m) $$ (1)

Now going back to the FBD of the the individual blocks,

where $$ a_x1$$ is the acceleration of the larger mass and $$ a_x2 $$ is the acceleration of the smaller mass.
So clearly, $$ a_{cm,x} = 0,$$ using an equation similar to (1) above.

now, it seems to me that $$a_{y1}$$ is zero, as I do not see the larger mass M accelerating along the y direction. However, it is clear that the smaller mass has some acceleration in the y direction.

from this, substituting values in equation (1), the y- component of the acceleration turns out to be non zero. However this contradicts the statement made above.

Where have I gone wrong in this analysis? is the CM stationary along the y plane?

DOUBT 4- I remember reading somewhere that only objects with the same acceleration can be considered as part of a valid system. Is this a myth or is this true?

I would greatly appreciate all help, please do point out mistakes if there are any in this question!

 

[erobz]

The center of mass in the vertical direction is changing position as the little block is descending. The little mass is accelerating down the wedge ergo the center of mass of the system (block+wedge) in vertical direction is also accelerating.

 

[Orodruin]

Homework Statement: A triangular wedge of mass M has a smaller mass m on its top most point. Initially the system is at rest. Considering the fact that the ground is frictionless, analyze the motion of the center of mass.

The center of mass of what? The problem should specify this. Without that information the problem is arbitrary. Presumably of the block-wedge system, but that’s not stated clearly.

So first thing I wanted to do is define my system. Here I have defined my system as "Small mass m+ Larger wedge M".
DOUBT 1: I had a rather silly idea- is it right to consider the "floor" as part of the system? Is that possible in the first place?

You can define a system any way you want. The only question is whether or not it is helpful to do so. If what you are after is the center of mass of the block-wedge system, including the floor does not seem helpful.

okay, now, I drew a free body diagram of the masses to predict how they would behave:-
|View attachment 357556
here N_g represents the normal reaction from the ground, and N_1 is the internal normal reaction between the two blocks. I have divided that into two components. So from here, it is clear that the larger mass M is going to move towards the left, with some acceleration.

DOUBT 2: FBD of the "system"
so in addition to these, I am going to draw the FBD of the system as defined earlier:View attachment 357558
the forces acting on the system are (M+m)g and the normal force from the ground, N_g.
POINT 1: as there are no external forces operating in the x- direction,
$$p_{ix}= p_{fx} $$
As far as I have learnt, I have learnt that "if there are no external forces acting on a system, the acceleration of the center of mass is 0, i.e the velocity of the center of mass is constant." This implies that the position of the center of mass in the x direction (here) does not change at all.

Indeed. The com x-position is constant.

DOUBT 3:- My main query with this question other than the ones above, is regarding the displacement/ behaviour of the y- coordinates of the center of mass. As per the FBD on the right, the vectors $$ (M+m)g , N_g $$ seem to cancel out each other.

Why would they cancel? Being in the same direction does not mean having the same magnitude.

Does this imply that momentum is conserved in the y direction as well?

Only if you can support the claim that the forces actually cancel (they won’t ).

Where have I gone wrong in this analysis? is the CM stationary along the y plane?

See above.

DOUBT 4- I remember reading somewhere that only objects with the same acceleration can be considered as part of a valid system. Is this a myth or is this true?

It is not the case. As long as you are considering the position/velocity/acceleration of thd center of mass, normal rules apply.

 

[anuttarasammyak]

It seems that the mass m block will pop up to the air after it hits the floor.

 

[palaphys]

The center of mass of what? The problem should specify this. Without that information the problem is arbitrary. Presumably of the block-wedge system, but that’s not stated clearly.



You can define a system any way you want. The only question is whether or not it is helpful to do so. If what you are after is the center of mass of the block-wedge system, including the floor does not seem helpful.


Indeed. The com x-position is constant.


Why would they cancel? Being in the same direction does not mean having the same magnitude.


Only if you can support the claim that the forces actually cancel (they won’t ).


See above.


It is not the case. As long as you are considering the position/velocity/acceleration of thd center of mass, normal rules apply.

center of mass of the wdge block system

 

[palaphys]

Why would they cancel? Being in the same direction does not mean having the same magnitude.

i thought the system was on the floor moving only in x direction right? unless the system is not flying or stuff how does it not cancel? i am confused about that. Intuitively the normal from ground must balance the weight of the system right?

also another question i just got- suppose a system is defined. External forces acting on any component of the system are considered as external forces acting on the system itself right?

 

[kuruman]

i thought the system was on the floor moving only in x direction right?

Wrong. Part of the system, namely the block, is accelerating down. You need to write the vertical equation of motion in a way that includes this acceleration.

also another question i just got- suppose a system is defined. External forces acting on any component of the system are considered as external forces acting on the system itself right?

Right. See above.

 

[Orodruin]

i thought the system was on the floor moving only in x direction right? unless the system is not flying or stuff how does it not cancel? i am confused about that. Intuitively the normal from ground must balance the weight of the system right?

The wedge is at constant vertical position but the block is going down so clearly the center of mass of the system as a whole is moving down. Unlike in the horizontal direction in which, as we have already concluded, the center of mass does not move (both parts move, but the net momentum in the horizontal direction is zero).

also another question i just got- suppose a system is defined. External forces acting on any component of the system are considered as external forces acting on the system itself right?

Yes.

 

[Orodruin]

It seems that the mass m block will pop up to the air after it hits the floor.

I don't think hitting the floor is actually a part of the question. Questions like this one where that is asked for are many times poorly worded as it is unclear how the collision with the floor is to be treated. It can be anything from a gradual transition of the velocity of the block to horizontal (i.e., energy conservation) or the vertical component of velocity simply nullifies if you treat the collision as inelastic. You will of course get different results depending on the assumptions.

 

[jbriggs444]

Intuitively the normal from ground must balance the weight of the system right?

This would be correct only if the center of mass of the system is not accelerating in the vertical direction.

 

[palaphys]

The wedge is at constant vertical position but the block is going down so clearly the center of mass of the system as a whole is moving down. Unlike in the horizontal direction in which, as we have already concluded, the center of mass does not move (both parts move, but the net momentum in the horizontal direction is zero).


Yes.

So the FBD on the left is more apt for the system of m+M? External forces on system are basically external forces on the CM right? so the diagram on the left implies that the CM is not at rest and has some acceleration in the y direction? If I use second law for the CM and find the acceleration, the result does not look the same as obtained from this equation:

are these two results the same?

 

[erobz]

The vertical acceleration of the wedge itself is?

 

[palaphys]

The vertical acceleration of the wedge itself is?

0? the wedge is just moving along a straight line right?

 

[palaphys]

Another question :- What happens when the block leaves the system wedge? Is there some external force now on the wedge? I am not enquiring about the nature of collisions of the block with the floor, but rather about the movement of the wedge.

 

[haruspex]

i thought the system was on the floor moving only in x direction right? unless the system is not flying or stuff how does it not cancel? i am confused about that. Intuitively the normal from ground must balance the weight of the system right?

It’s the wedge that has no vertical acceleration, so the vertical forces on the wedge balance. What are the vertical forces on the wedge?

 

[anuttarasammyak]

Another question :- What happens when the block leaves the system wedge? Is there some external force now on the wedge? I am not enquiring about the nature of collisions of the block with the floor, but rather about the movement of the wedge.

The wedge will do inertial motion with constant x-speed. The block also will keep its x-direction speed if there exists neither friction nor inelasticity between block and floor.
[EDIT] Please disregard the latter which is of my own interest. The floor can provide force of y-direction only so block x-mementum is conserved in spite of possible pop-up or spinning.
[EDIT2] Not only the block but also the edge could topple after they leave depending on the unstable position of the edge COM. But as said above for the bock, its x-momentum will be conserved.

 

[Orodruin]

The wedge will do inertial motion with constant x-speed. The block also will keep its x-direction speed if there exists neither friction nor inelasticity between block and floor.

OP is explicitly stating that they are not concerned with what happens when the block hits the floor ...

Another question :- What happens when the block leaves the system wedge? Is there some external force now on the wedge? I am not enquiring about the nature of collisions of the block with the floor, but rather about the movement of the wedge.

(my emphasis)

 

[kuruman]

The wedge will do inertial motion with constant x-speed. The block also will keep its x-direction speed if there exists neither friction nor inelasticity between block and floor.
[EDIT] Please disregard the latter which is of my own interest. The floor can provide force of y-direction only so block x-mementum is conserved in spite of possible pop-up or spinning.

This is simply not true. Newton's third law says that the block exerts a force to the left on the wedge and the wedge exerts a force to the right on the block. The two masses are initially at rest and they accelerate in opposite directions as long as they are in contact. The center of mass that has constant velocity, in this case zero whether the masses are in contact or not.

There, I "analyzed" the motion of the CM in the horizontal direction.

Edited to fix a misunderstanding, see posts #20 and #21.

 

[kuruman]

Another question :- What happens when the block leaves the system wedge? Is there some external force now on the wedge? I am not enquiring about the nature of collisions of the block with the floor, but rather about the movement of the wedge.

Can you draw a FBD of a wedge of mass $M$ on a horizontal, frictionless plane?

 

[anuttarasammyak]

This is simply not true. Newton's third law says that the block exerts a force to the left on the wedge and the wedge exerts a force to the right on the block. The two masses are initially at rest and they accelerate in opposite directions as long as they are in contact. It is the center of mass that has constant velocity, in this case zero.

You say about when the two are in contact. I agree with what you say. But I said about after they leave as questioned. To be more clear I might have to add that in the moment they leave there is no force between them.

[EDIT]
While they are in contact:
X-momentum conservation
MV=mv_x
Energy conservation
\frac{1}{2}MV^2+\frac{1}{2}m(v_x^2+v_y^2)+mgh=mgh_0
Where
\frac{v_y}{v_x+V}=\tan \theta
With convention that all
V,v_x,v_y,h> 0
are in natural direction.　Solving for V
V=k\sqrt{\frac{2g(h_0-h)}{sec^2\theta-k}}
Thus
v_x=(1-k)\sqrt{\frac{2g(h_0-h)}{sec^2\theta-k}}
v_y=\sqrt{\frac{2g(h_0-h)}{sec^2\theta-k}}\tan\theta
where
k=\frac{m}{M+m}<1

 

[kuruman]

You say about when the two are in contact. I agree with what you say. But I said about after they leave as questioned. To be more clear I might have to add that in the moment they leave there is no force between them.

Sorry, I missed what question you were answering. I stand corrected.

 

[erobz]

0? the wedge is just moving along a straight line right?

So you can simplify your relationship for $a_{cm,y}$.

 

[jbriggs444]

To be more clear I might have to add that in the moment they leave there is no force between them.

There can be an impulsive force between them at the moment of detachment. The magnitude of that impulse can depend on the angle of the slope, the dimensions (height/width ratio) of the block and the coefficient of restitution for a block/wedge impact.

For a 45 degree slope and a block with height equal to its width, there would be no impulse. We would ideally have the block skating forward, maintaining its 45 degree angle, precisely balanced on its bottom corner.

 

[anuttarasammyak]

There can be an impulsive force between them at the moment of detachment.

Like explosives to detach rocket and booster ? I do not find explosives or springs here. But anyway it may not violate x-momentum conservation of the system.

 

[jbriggs444]

Like explosives to detach rocket and booster ? I do not find explosives or springs here. But anyway it may not violate x-momentum conservation of the system.

There is an impulsive vertical force from the collision of block with ground.

If the center of mass of the block is not directly above the point of impact, this means that there is an impulsive torque from ground on block (about its CM).

If the block is forward-leaning then the torque is clockwise and will tend to rotate the bottom of the block into the wedge, possibly resulting in a leftward impulse of block on wedge.

If the block is rearward-leaning then the torque is counter-clockwise and will tend to rotate the top of the block into the wedge, again possibly resulting in a leftward impulse of block on wedge.

A question requiring calculation is whether the separation rate engendered by the vertical impulse from the ground is enough to match the closure rate engendered by the rotation due to the impulsive torque from the ground.

But you are correct that this does not violate x-momentum conservation of the system.

 

[anuttarasammyak]

There is an impulsive vertical force from the collision of block with ground.

I draw sketch to take a look at whether the block will contact with the edge again after its ground touch.

The sketch shows the moment of floor contact. The velocities are calculated in [EDIT] of my post #20. Say the floor would provide perfect elastic y-force but no x-force. The shape and density homogeneity/inhomogeneity of the block is open. My conjecture is no contact again because V and vx are in opposite direction but I am not sure of it for various cases.

By re-contact |x-momentum| of each body increase or decrease the sum of which is zero ? Increase requires conversion of y-energy. Falling part of the block should contact the edge slope.
Decrease seems impossible. How edge slope which is frictionless and perfect elastic can decrease |x-momentum| of the block?