System of particles, impulse and conservation of angular momentum

  • #1
Spector989
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Homework Statement:
A uniform rod of length 6 a and mass 8 m lies on a smooth horizontal table. Two particles of masses m and 2 m moving in the same horizontal plane with speed 2 v and v respectively, strike the rod and stick to the rod after collision as shown in the figure. The velocity of centre of mass and angular velocity about centre of mass just after collision are respectively. ( Diagram is provided)
Relevant Equations:
Integration(F.rdt)= impulse
I1.W1 = I2W2 [ I = Moment of inertia , W = angular speed ]
M1V1 = M2V2
So i was able to solve the angular velocity part but i don't know how to find the velocity of centre of mass . For the first part i simply conserved momentum about COM because if i consider the particles as a part of the same system as rod the collision are internal forces . I am mainly confused as to how to solve it mathematically and properly . Thanks
 

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  • #2
malawi_glenn
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i don't know how to find the velocity of centre of mass

Is the linear momentum conserved?
If it is, what can you say about the motion of the center of mass for the entire system?
 
  • #3
haruspex
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For the first part i simply conserved momentum about COM
?
Do you mean for the second part you used conservation of angular momentum about CoM?
What did you get?
 
  • #4
Spector989
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?
Do you mean for the second part you used conservation of angular momentum about CoM?
What did you get?
Mb , for the second part i conserved angular momentum , i got W = v/5a
 
  • #5
Spector989
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Is the linear momentum conserved?
If it is, what can you say about the motion of the center of mass for the entire system?
Well external force=0 so velocity of COM = 0 ?
 
  • #6
Shreya
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Well external force=0 so velocity of COM = 0 ?
Conservation of linear momentum says that if external force is 0, the momentum is conserved (not that its neccessarily 0). Why dont you try to write the conservation of linear momentum equation & see what you come up with?
 
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  • #7
Spector989
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Conservation of linear momentum says that if external force is 0, the momentum is conserved (not that its neccessarily 0). Why dont you try to write the conservation of linear momentum equation & see what you come up with?
What i meant to say is that as the net external force is zero, the velocity of Centre of mass will be conserved which was zero initially so the centre of mass will stay at rest
 
  • #8
Spector989
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What i meant to say is that as the net external force is zero, the velocity of Centre of mass will be conserved which was zero initially so the centre of mass will stay at rest, or... maybe not , now that i think about it . Imma try something else and check back
 
  • #9
Shreya
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Your idea the CM will stay at rest is correct, but only writing the equation will help you understand why. And dont forget to include the other 2 (moving) particles in your system.
 
  • #10
Spector989
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Your idea the CM will stay at rest is correct, but only writing the equation will help you understand why. And dont forget to include the other 2 (moving) particles in your system.
Yeah that is what i am trying , thanks for clearing it :)
 
  • #11
Spector989
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Is it alright if i post a similar question in this thread or do i create a new one , the question is pretty same but instead of colliding body simply impulse is imparted to an end point
 
  • #12
Shreya
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this thread will do ig
 
  • #13
Spector989
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So i tried to conserve angular momemtum about the point where impulse was imparted , how do i conserve angular momentum when axis of rotation itself is moving
 

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  • #14
Shreya
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Conservation of angular momentum can only be applied if external torque is 0. Do you see why here it is not so?
 
  • #15
Spector989
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Conservation of angular momentum can only be applied if external torque is 0. Do you see why here it is not so?
But external torque about the point of application of impulse will be zero , right ?
 
  • #16
Spector989
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We can converse angular momentum if external torque about any axis is zero , about that axis . ( by i mean:- axis ~ axis of rotation )
 
  • #17
PeroK
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But external torque about the point of application of impulse will be zero , right ?
There are two points where an impulse is applied. You are asked for the AM about the centre of mass, so in any case you should be looking at AM about the centre of mass.
 
  • #18
Shreya
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hmm, alright...Impulse here means a change in angular momentum. If you look at the right ball from your axis, it experiences an impulse, aka a change in angular momentum. Try calculating that.
Dont worry about linear momentum.
 
  • #19
Spector989
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There are two points where an impulse is applied. You are asked for the AM about the centre of mass, so in any case you should be looking at AM about the centre of mass.
Is it a given that it is about the com when it simply says find Angular velocity of system
 
  • #20
Shreya
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There are two points where an impulse is applied. You are asked for the AM about the centre of mass, so in any case you should be looking at AM about the centre of mass.
wouldnt it be the same for both axes ?
 
  • #21
Spector989
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wouldnt it be the same for both axes ?
Yeah mb Angular momentum will be different but angular velocity will be same
 
  • #22
Spector989
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hmm, alright...Impulse here means a change in angular momentum. If you look at the right ball from your axis, it experiences an impulse, aka a change in angular momentum. Try calculating that.
Dont worry about linear momentum.
Solved it , i got confused for no reason. Well thanks a lot for all the help :)
 
  • #23
PeroK
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Is it a given that it is about the com when it simply says find Angular velocity of system
I missed that we've moved on to a new question. The angular velocity is the same about any point on the body. In this case, there's no advantage in considering the AM about the CoM.
 
  • #24
haruspex
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Solved it
So what did you get?
there's no advantage in considering the AM about the CoM.
I suggest that considering angular momentum about the CoM is a little easier here because it avoids having to write down a linear momentum equation.
 

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