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Centre of Mass

  • Thread starter ritwik06
  • Start date
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1. Homework Statement

Consider the following situation:
http://img206.imageshack.us/img206/9739/fbdow8.th.jpg [Broken]
The height of the incline is 'h'.
a) Find the net acceleration of the centre of mass of the combined system (of the incline and the block) relative to ground.
b) Find the speed of the incline when the block slides down to the bottom of the incline.
[tex]g=10 m s^{-2}[/tex]
And all the surfaces are assumed to be frictionless.




3. The Attempt at a Solution

First I would list all the external forces acting upon the system.
1. Mg on the incline due to earth's pull.
2. mg on the block due to earth's pull.
3. The normal provided by the ground to the Incline (M+m)g
As these forces cancel out to zero, the accleeration of th centre of mass of the mentioned system is zero as there is no net external force on the system.

As the block falls down the incline the incline will also move in order to conserve the momentum (which is zero).
The velocity of the block as it reaches the bottom is = [tex]\sqrt{2gh}[/tex] relative to the incline and along the incline. The velocity can be written as [tex]\sqrt{2gh} cos \theta \hat{i}+\sqrt{2gh} sin \theta \hat{j}[/tex]
relative to the incline.
On writing the velocity relative to ground, I get=[tex]\sqrt{2gh} cos \theta - V \hat{i}+\sqrt{2gh} sin \theta \hat{j}[/tex]
Let V be the velocity of the incline in X direction.
Conserving momentum
[tex]m (\sqrt{2gh} cos \theta - V \hat{i}+\sqrt{2gh} sin \theta \hat{j})+MV\hat{i}=0[/tex]
I am confuse now, where does the j component come from. How will I conserve momentum????
 
Last edited by a moderator:

LowlyPion

Homework Helper
3,079
4
1. Homework Statement

Consider the following situation:

a) Find the net acceleration of the centre of mass of the combined system (of the incline and the block) relative to ground.
b) Find the speed of the incline when the block slides down to the bottom of the incline.
[tex]g=10 m s^{-2}[/tex]
And all the surfaces are assumed to be frictionless.


3. The Attempt at a Solution

First I would list all the external forces acting upon the system.
1. Mg on the incline due to earth's pull.
2. mg on the block due to earth's pull.
3. The normal provided by the ground to the Incline (M+m)g
As these forces cancel out to zero, the accleeration of th centre of mass of the mentioned system is zero as there is no net external force on the system.

As the block falls down the incline the incline will also move in order to conserve the momentum (which is zero).
The velocity of the block as it reaches the bottom is = [tex]\sqrt{2gh}[/tex] relative to the incline and along the incline. The velocity can be written as [tex]\sqrt{2gh} cos \theta \hat{i}+\sqrt{2gh} sin \theta \hat{j}[/tex]
relative to the incline.
On writing the velocity relative to ground, I get=[tex]\sqrt{2gh} cos \theta - V \hat{i}+\sqrt{2gh} sin \theta \hat{j}[/tex]
Let V be the velocity of the incline in X direction.
Conserving momentum
[tex]m (\sqrt{2gh} cos \theta - V \hat{i}+\sqrt{2gh} sin \theta \hat{j})+MV\hat{i}=0[/tex]
I am confuse now, where does the j component come from. How will I conserve momentum????
I was wondering where it was you were accounting for the conservation of energy? Won't change in Potential energy = to Total Kinetic energy ? (They started at rest.)
 
42
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The j component comes from the change in the height of the incline. If you setup your axis at the base of the incline so that the bottom of the incline is ending at the x axis, you will clearly see that there is a change in the y direction, thus accounting for your j. Also, look at the relationship between your change in potential energy and kinetic energy. If the block is not moving, all the energy is potential, as the block moves down the incline, the potential energy is converted into kinetic energy. Since the surface is frictionless, no energy is lost via heat or sound due to friction.
 
580
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I was wondering where it was you were accounting for the conservation of energy? Won't change in Potential energy = to Total Kinetic energy ? (They started at rest.)
Please read my question over again. I am applying conservation of mometum which is always true in absence of external force. I am not bothered about energy terms.
 

Dick

Science Advisor
Homework Helper
26,258
618
i) don't worry about canceling the j component of the momentum. It's canceled by the very very very slight upward motion the incline and earth. ii) You can't use sqrt(2gh) for the velocity of the block. That comes from conservation of energy NOT considering the horizontal motion of the incline. You need to go back to first principles and leave the velocity V of the incline and the velocity v of the block as unknowns. Now write conservation of energy for block and incline and conservation of momentum for the x components of the momentum. That's two equations in two unknowns. Solve them.
 

LowlyPion

Homework Helper
3,079
4
Please read my question over again. I am applying conservation of mometum which is always true in absence of external force. I am not bothered about energy terms.
Just because you're not bothered by them doesn't mean that you can't make use of them. Because they are conserved. They sure look like a pretty handy means of relating the velocities and the masses. Applying them both looks awfully appealing to me anyway.
 
580
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i) don't worry about canceling the j component of the momentum. It's canceled by the very very very slight upward motion the incline and earth. ii) You can't use sqrt(2gh) for the velocity of the block. That comes from conservation of energy NOT considering the horizontal motion of the incline. You need to go back to first principles and leave the velocity V of the incline and the velocity v of the block as unknowns. Now write conservation of energy for block and incline and conservation of momentum for the x components of the momentum. That's two equations in two unknowns. Solve them.
Just because you're not bothered by them doesn't mean that you can't make use of them. Because they are conserved. They sure look like a pretty handy means of relating the velocities and the masses. Applying them both looks awfully appealing to me anyway.
Thanks guys for the help. I have got the correct answer now. As u all say the second equation was necessary.
But the question which still torments me is that, the system has a net force 0 along y direction and yet there is change in momentum of th system in that dirrection. How???
 

Dick

Science Advisor
Homework Helper
26,258
618
The whole system is block+incline+earth. As the block gains downward momentum, incline+earth gains upward momentum. The sum is automatically zero. The reason you don't have to worry about this is that the velocity of this motion is so small it can be ignored in the conservation of energy equation. If you don't believe me, figure out what the kinetic energy of the earth is if it has a momentum of 1kg*m/sec.
 
580
0
The whole system is block+incline+earth. As the block gains downward momentum, incline+earth gains upward momentum. The sum is automatically zero. The reason you don't have to worry about this is that the velocity of this motion is so small it can be ignored in the conservation of energy equation. If you don't believe me, figure out what the kinetic energy of the earth is if it has a momentum of 1kg*m/sec.
I agree to that. But when I say that I take the block and the incline as one system. How do you define creation of momentum in the y direction????? That question torments me.
 

Dick

Science Advisor
Homework Helper
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618
I agree to that. But when I say that I take the block and the incline as one system. How do you define creation of momentum in the y direction????? That question torments me.
There's an unbalanced external force (the component of gravity parallel to the incline) acting on the block. Isn't there? Stop being tormented. No momentum is being created.
 
580
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There's an unbalanced external force (the component of gravity parallel to the incline) acting on the block. Isn't there? Stop being tormented. No momentum is being created.
The force you are talking about is a part of the gravitational pull. Isn't it completely quelled by the normal from the friction less ground?
Listen to this point of view of mine:
as the block begins to gain momentum in the downward, i.e. the -ve y-direction. The incline will gain momentum in th +ve y direction, due to which the normal decreases and then a net force acts on the body (M+m)g -N causing the system to accelerate down.
 

Dick

Science Advisor
Homework Helper
26,258
618
A system that's sitting on 'frictionless ground' cannot be considered as isolated. You are implicitly thinking of the 'ground' as being infinitely massive, since it's FIXED in position. It can soak up any amount of momentum without moving, and it is absorbing your reaction momentum.
 

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