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Conservation of Angular momentum problem

  1. Apr 15, 2016 #1
    1. The problem statement, all variables and given/known data
    In the figure, a small 0.235 kg block slides down a frictionless surface through height h = 0.471 m and then sticks to a uniform vertical rod of mass M = 0.470 kg and length d = 2.36 m. The rod pivots about point Othrough angle θ before momentarily stopping. Find θ.

    2. Relevant equations
    L=I*w
    P=mv
    v=wr
    Ki=Kf -----> 0.5mv^2=mgh , and this gives expression for 'v'
    w^2=w0^2 + 2*alpha*(delta ø) , for when alpha(angular acceleration) is constant
    3. The attempt at a solution
    in effort to find ø, alpha and w are needed. I established that m*v=I*w (not sure if this is correct)

    by substituting relevant equations, i got w0=[sqrt(2gh)]/d^2
    and w, is according to the question, 0.

    now i can't seem to find an expression for alpha.
    could u point out what i missed out?
     

    Attached Files:

  2. jcsd
  3. Apr 15, 2016 #2
    Find out angular momentum of the system just after impact. Just before the impact the angular momentum is non-zero only for the hitting block given by mvr*L about the pivot, which moment of momentum. Assume some instantaneous velocity V for the hit tip of the rod, which will be the same for the block too just after it sticks.. Your equation is wrong even dimension-ally.
     
  4. Apr 15, 2016 #3

    haruspex

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    But it won't be.
     
  5. Apr 15, 2016 #4
    Because alpha is not constant you cannot apply that formula. Apply Total L conservation just before and after impact and for rest segments apply energy conservation.
     
  6. Apr 16, 2016 #5
    I've reworked an expression for w, w=sqrt(2gh)/3d
    by applying I*w = I*w
    but then how do i find the angle?
     
  7. Apr 16, 2016 #6

    haruspex

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    I don't understand how you get that. Please post your working.
     
  8. Apr 16, 2016 #7
    why isn't alpha constant?
     
  9. Apr 16, 2016 #8

    haruspex

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    What force is responsible for the angular acceleration? Is its moment about the centre of the arc constant?
     
  10. Apr 16, 2016 #9
    Oh I get it thanks a lot.
    Is it because gravity causes the tangential force to change when the rod pivots which causes the angular acceleration to change?
     
  11. Apr 16, 2016 #10

    haruspex

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    Yes, as the angle changes the tangential component of gravity changes.
     
  12. Apr 16, 2016 #11
    Because Torque acting on the bar and the attached piece is not constant.
     
  13. Apr 16, 2016 #12
    Right before the cube collied with the rod, it's angular moment is I*w=1/3md^2 * (v/d), where v=(2gh)^0.5
    right after the collision, the new I*w=1/3(m+M)d^2 * w

    now that I*w=I*w, we have new w=(mv)/(m+M)d=(1/3)(v/d)=sqrt(2gh)/3d
     
  14. Apr 16, 2016 #13

    haruspex

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    It is described as small, so consider it a point particle. Where do you get the 1/3 from?
     
  15. Apr 16, 2016 #14
    m/(m+M)=0.235/(0.235+0.47)=1/3
     
  16. Apr 16, 2016 #15

    haruspex

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    I still don't get it. You wrote
    so we're talking about the angular momentum of the small block. It has not hit the rod yet, so M cannot play a role.
     
  17. Apr 16, 2016 #16
    I assumed it could as suggested by the other guy
    If this does not hold, which equation shall I establish?
     
  18. Apr 16, 2016 #17

    haruspex

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    Do you mean this:
    If so, you are misreading it. The expression for the angular momentum of the block just before impact is clearly stated there. But the first part, "Find out angular momentum of the system just after impact." may be slightly misleading. Let'sthink means the angular momentum of block+rod just after impact (and this does involve M) is equal to that of the block just before impact. So you need two expressions for angular momentum, one just before impact and one just after, which you then set equal to obtain an equation.
    You do not need to worry about forces and accelerations. Just use conservation laws as appropriate. After impact, what is conserved?
     
  19. Apr 16, 2016 #18
    So the angular momentum of the small block would be I*w of the slide-block system, I=(1/3)mh^2, w=v/r
     
  20. Apr 16, 2016 #19

    haruspex

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    Forget the rod for the moment. We have a small block which has reached a speed of √(2gh) and is travelling horizontally at a distance d below point O. What is the angular momentum of the block about point O? If you think there is a factor 1/3 in the answer, please explain how you arrive at that.
     
  21. Apr 16, 2016 #20
    Oh right, so L=r*p where p=m*v, thus L=r*m*v=h*sqrt(mgh)*m
     
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