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Linear momentum and Angular momentum

  1. Jan 30, 2016 #1
    1. The problem statement, all variables and given/known data
    Two objects with mass of m are connected with a rod with length 2l and with no mass. The center of the rod is pinned so that it can spin. Object with mass M comes with speed v and sticks to m. There is no friction.

    1) What is the angular speed w after collision?
    2) FInd the value and the direction of Impulse that was sent through the rod during collision
    3) Right after collision, find the value and the direction of the force acting on the rod by the center of the rod
    As an alternate version of this question, suppose that the collision was perfect elastic collision and that the velocity after collision was v1 for M and v2 for m (the direction of the velocity didn't change). M>2m
    4) Find the value of v1
    5) Find the value of impulse sent through the rod
    2. Relevant equations

    Conservation of momentum and angular momentum

    3. The attempt at a solution

    1) I don't know whether linear and angular momentum are exchangeable. So I used only linear momentum. Mv= (M+2m)v1 and solved for v1.

    2) The collision happens at the left side of the rod but the right side of the rod moves as well, and it does at velocity v1. The momentum for the object on the right side changes from 0 to mv1 so the impulse should be mv1 to direction -y

    3) I didn't really get this question. But I though that perhaps it has got to do with the center of the rod not being the same as the center of mass.

    4) Mv=Mv1+2mv2 and (v2-v1)/(v-0)=1. So v1=(M-2m)v/(M+2m)
    5) Thinking the same way as I did in 2) I answered mv2

    I tried to answer all the questions but I'm not sure if they are correct. If you could check these I would be thankful
     

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  3. Jan 30, 2016 #2

    Andrew Mason

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    You cannot use conservation of linear momentum because the system is constrained by the pin i.e. the system cannot have translational motion. The system is free to rotate, however. Can you use the principle of conservation of angular momentum here? What is the angular momentum before the collision? What is it after?
    The system receives an impulse of Mv. Since the system is constrained and can have no linear momentum, what is the impulse that the the pin must exert on the system?
    This is a rather tricky question. This has to do with the fact that the pin is not located at the centre of mass of the system.

    Let's see where that gets you. You may find that the second scenario solution is clearer.

    AM
     
  4. Jan 30, 2016 #3
    The angular momentum before is 0 and after is Mv/(M+2m)l right? I'm not too sure about v1 so I can't be sure if this is correct.

    Well, impulse is the same as difference in momentum, but can we use angular momentum here?

    Perhaps the force in 3) is r*F where r is the distance from the center of mass and the center of the rod and F is difference between the two centripetal force( one of the bigger mass and one of the small mass)
     
  5. Jan 30, 2016 #4

    Andrew Mason

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    No. You are just taking into account the angular momentum of the 2m and rod before the collision. M has angular momentum before the collision.

    The angular momentum of each mass is mass x angular speed x radius2. Angular speed and radius is the same for all three masses AFTER, but not before, the collision. You have to include the angular momentum of M immediately prior to the collision, relative to the point of rotation. The initial angular momentum is just:

    ##L_i = Mvl + 0 = M\omega_i l^2##

    Work out the angular momentum after the collision in terms of the only unknown: ##\omega_f## so you can solve for it.

    The total final linear momentum of the system is 0 after the collision. The initial linear momentum was Mv, just before the collision. The impulse imparted by M to the system (Mv) + the impulse (J) imparted by the pin to the system, which now includes M, results in the final linear momentum (i.e. 0). So work out what J must be.

    Think of the rotating system as the centre of mass of the whole system rotating about the pivot. What is the force that the pivot must exert on the centre of mass?

    AM
     
    Last edited: Jan 30, 2016
  6. Jan 31, 2016 #5
    Then wf=Mwi/(M+2m)

    The impulse would be opposite in direction with M and have the same value as Mv.

    The distance between the two points is Ml/(M+2m) and the mass is M+2m so F=(M+2m)w2f(Ml/M+2m)
     
  7. Jan 31, 2016 #6

    Andrew Mason

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    ## = \frac{Mv}{(M+2m)l}##
    Correct.
    which can be reduced to.....

    AM
     
  8. Jan 31, 2016 #7
    w2fMl !
     
  9. Jan 31, 2016 #8

    Andrew Mason

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    So the force on the pivot of the rotating system is the same as .... (describe what Mωf2l represents).

    AM
     
    Last edited: Jan 31, 2016
  10. Feb 1, 2016 #9
    The centripetal force of M after collision!
     
  11. Feb 1, 2016 #10
    So before the collision there was no need for the pin to exert force because the center of mass and the center of the rod were the same, but since M broke this equilibrium the pin needs to exert force to sustain the new visitor!
     
  12. Feb 1, 2016 #11
    So before the collision there was no need for the pin to exert force because the center of mass and the center of the rod were the same, but since M broke this equilibrium the pin needs to exert force to sustain the new visitor!
     
  13. Feb 2, 2016 #12

    Andrew Mason

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    Ok so far! Now try 4 and 5. Don't worry about how the collision takes place. It says that it is elastic and no change in direction of M, so just assume that to be the case. In this case,what quantities(y) are(is) conserved?
    AM
     
  14. Feb 4, 2016 #13
    Do we use
    ##Mvl=Mv_1l+2mv_2l##?
     
  15. Feb 4, 2016 #14

    Andrew Mason

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    That is part of the solution. What conservation laws apply? You have 2 unknowns so you need another equation.

    AM
     
  16. Feb 4, 2016 #15
    The collision is elastic so the energy of the whole system is conserved.
    Before it was: ##\frac{1}{2}Mv^2##
    After it is ##\frac{1}{2}Mv_1^2+ml^2w_2^2## where ##w_2=v_2l##

    The ##v_1## that I got was

    ##v_1=\frac{l^4M-2m}{l^4M+2m}v##

    and ##v_2## was

    ##v_2=\frac{2M}{l^4M+2m}v##
     
  17. Feb 4, 2016 #16

    Andrew Mason

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    Please show how you got that. Start with the two equations and show how you solved it.

    Check this. It is dimensionally incorrect (ie. there is an extra distance ^-4). It should have dimensions of distance/time. Also, please show your work.

    AM
     
  18. Feb 5, 2016 #17
    Yeah I made a mistake I used ##w_2=v_2l## instead of the correct ##w_2=\frac{v_2}{l}##
    The value of ##v_1## was calculated by

    ##Mv=Mv_1+2mv_2## (taken by dividing both sides of the equation above by ##l##)
    and
    ##Mv^2=Mv_1^2+2mv_2^2##

    from the first equation
    ##v_2=\frac{M(v-v_1)}{2m}##
    substituting this value into the second equation I got

    ##\frac{M^2(v-v_1)^2}{2m}=M(v^2-v_1^2)##
    dividing both sides by ##M(v-v_1)## we get

    ##\frac{M(v-v_1)}{2m}=v+v_1##
    solving for ##v_1##

    ##v_1=\frac{M-2m}{M+2m}v##
     
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