Conservation of linear and angular momentum

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  • #1
dyn
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Hi ; I have a few question regarding the conservation of linear and angular momentum. Would appreciate any help.
1 - When no external forces act are both linear and angular momentum conserved in all 3 directions separately or just the total linear/angular momentum conserved ?
2 - if I approach a problem using conservation of both types of momentum can I use separate origins for linear and angular momentum or can i only use one origin per problem ?
3 - i have been looking at a problem when a cube slides on a frictionless surface and elasticly impacts on the bottom of a rod pivoted at its centre of mass. Using conservation of angular momentum and energy i can find the speed of the cube after the impact. But if i look at the problem using linear momentum i get( linear mom. of cube before) =( linear mom. of cube after impact) because the rod has no linear momentum because it centre of mass is fixed. This means the cube has the same speed after impact as before impact which i know is wrong. What am i doing wrong ?
Thanks
 

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  • #2
FactChecker
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Hi ; I have a few question regarding the conservation of linear and angular momentum. Would appreciate any help.
1 - When no external forces act are both linear and angular momentum conserved in all 3 directions separately or just the total linear/angular momentum conserved ?
You should think of them as vectors. The total momentum vector of a closed system is constant
2 - if I approach a problem using conservation of both types of momentum can I use separate origins for linear and angular momentum or can i only use one origin per problem ?
Each is true, independent of the other, regardless of what origins you pick.
3 - i have been looking at a problem when a cube slides on a frictionless surface and elasticly impacts on the bottom of a rod pivoted at its centre of mass. Using conservation of angular momentum and energy i can find the speed of the cube after the impact. But if i look at the problem using linear momentum i get( linear mom. of cube before) =( linear mom. of cube after impact) because the rod has no linear momentum because it centre of mass is fixed. This means the cube has the same speed after impact as before impact which i know is wrong. What am i doing wrong ?
Thanks
It the cube's center of mass is fixed to the Earth, then the momentum of the Earth is part of the calculation. This is true for both linear and angular momentum. The Earth is part of the entire closed system. It is exerting and reacting to forces.
 
  • #3
A.T.
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1 - When no external forces act are both linear and angular momentum conserved in all 3 directions separately or just the total linear/angular momentum conserved ?
There is no "total linear/angular momentum". You cannot add them due to different units.

2 - if I approach a problem using conservation of both types of momentum can I use separate origins for linear and angular momentum or can i only use one origin per problem ?
The angular momentum has to be conserved around any origin, if you also consider the torques around that origin. For the linear momentum of an extended body you take the center of mass.

3 - i have been looking at a problem when a cube slides on a frictionless surface and elasticly impacts on the bottom of a rod pivoted at its centre of mass. Using conservation of angular momentum and energy...
As @FactChecker notes, this is not a closed system. Also note that for splitting the kinetic energy of a body into linear and rotational parts you have use a consistent reference point, otherwise you can double count the energy.
 
  • #4
dyn
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Hi ; I have a few question regarding the conservation of linear and angular momentum. Would appreciate any help.
1 - When no external forces act are both linear and angular momentum conserved in all 3 directions separately or just the total linear/angular momentum conserved ?

So , I can't say p(before) = p(after) in x-direction and similarly for y and z direction ? I can only state that the sum of the squares of the 3 components of momentum is a constant ?
3 - i have been looking at a problem when a cube slides on a frictionless surface and elasticly impacts on the bottom of a rod pivoted at its centre of mass. Using conservation of angular momentum and energy i can find the speed of the cube after the impact. But if i look at the problem using linear momentum i get( linear mom. of cube before) =( linear mom. of cube after impact) because the rod has no linear momentum because it centre of mass is fixed. This means the cube has the same speed after impact as before impact which i know is wrong. What am i doing wrong ?
Why can this problem be solved using angular momentum but not linear momentum ?
 
  • #5
FactChecker
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For your own sanity, assume that the Earth is fixed and supplies enough force to keep any attached object stationary, in whatever way it is attached (linearly, rotationally, or just specific directions).
CORRECTION: The motion of the Earth (or some motion of part of it) must be included if the object is attached to it and one wants to preserve linear momentum.
So , I can't say p(before) = p(after) in x-direction and similarly for y and z direction ? I can only state that the sum of the squares of the 3 components of momentum is a constant ?
No. the vectors are constant, so each direction is constant.
Why can this problem be solved using angular momentum but not linear momentum ?
They both can be solved. You need to show the details of your work to get specific help.
 
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  • #6
dyn
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I know that it can be solved using angular momentum and conservation of energy. But if I apply conservation of linear momentum ; before impact it is , m x (initial speed of cube) ; after impact it is , m x (final speed of cube) plus linear momentum of rod but its COM does not move so this term is zero. So I get initial speed of cube = its final speed which I know is wrong
 
  • #7
FactChecker
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I know that it can be solved using angular momentum and conservation of energy. But if I apply conservation of linear momentum ; before impact it is , m x (initial speed of cube) ; after impact it is , m x (final speed of cube) plus linear momentum of rod but its COM does not move so this term is zero. So I get initial speed of cube = its final speed which I know is wrong
You are correct if the motion of the Earth is ignored. So I guess that you will have to accept that unless you want to open a real can of worms.
 
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  • #8
Ibix
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But if I apply conservation of linear momentum
...you get the wrong answer because momentum is not conserved in this problem, as noted in #3 and hinted in #2.
 
  • #9
dyn
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So conservation of linear momentum cannot be used because an external force acts at the pivot ? Is the reason conservation of angular momentum can be used because the external force acts at the pivot and so does not contribute to angular momentum about the pivot ?
 
  • #10
Ibix
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As long as you conserve angular momentum about the pivot, so that the force gives zero torque, yes.
 

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