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Angular momentum

  1. Oct 27, 2007 #1
    A uniforum stick AB of mass M and length L is moving on a smooth plane with speed v. The motion is normal to the stick. The stick collides with a fixed nail on the plane and impact is at the middle of AC, where C is the C.M. of the stick.

    Q: What is the instantaneous speed of the C.M. of the stick?




    The attempt at a solution:
    Let F be the force between the stick and the nail
    vf be the instantaneous speed of the C.M. of the stick
    ω be the angular velocity of the stick after impact

    △P = F△t
    M(vf - v)= F△t

    Take moment @ C
    ∴ I=1/12 ML^2
    △L (angular momentum) = τ(torque)△t
    ∵τ= rXF
    ∴I(ω-0)= (L/4)F△t
    (1/12 ML^2)ω = (L/4) M(vf - v)
    ...
    (vf - v) = 1/3 Lω

    How can I let ω interms of v or vf ?
    Or..Is there anything I have missed?
     
  2. jcsd
  3. Oct 28, 2007 #2

    rl.bhat

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    Whether the motion of the stick is only tanslatory or both transalatory and rotaional? And again after the collision what is the nature of the motion? In both the cases use conservation of energy and angular momentum.
     
  4. Oct 28, 2007 #3
    Thx~
    I have not try to use conservation of energy becox I'm afraid that there will be energy loss during the collision

    If there are energy loss, how can I solve this problem?
     
  5. Oct 28, 2007 #4

    Shooting Star

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    Suppose that instead of a rod, it was a small sphere given in the problem. Then could you have found the instantaneous speed of the sphere without assuming conservation of KE?

    You have to assume that the KE of the rod is conserved.
     
  6. Oct 28, 2007 #5
    By the use of conservation of energy and the motion of the stick is transalatory and rotational:

    (1/2)Iω^2 + (1/2) mvf^2 = (1/2) mv^2

    ω^2 = 12/L (v^2 - vf^2)
    (vf - v) = (1/3) Lω
    ∴ (vf - v)^2 = (1/9) L^2 ω^2 = (1/9) L2 (12/L) (v^2 - vf^2)

    (7/3) vf^2 + (-2v) vf - (1/3) v^2 = 0

    vf = v or (-1/7) v

    Then i take vf = (-1/7) v becox the final velocity must be smaller the initail one ( as there are rotation, RKE exist)
    Am I right?
     
    Last edited: Oct 28, 2007
  7. Oct 28, 2007 #6

    Shooting Star

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    (Initial KE= KE of CM + RKE about CM. That's right.)

    Let’s call the mid pt of AC as D. Then, instantaneously, D is at rest. Calculate the RKE only taking D to be fixed. Both the segments AD and BD have the same omega, and vf of CM is given by vf=(length of CD)*omega.

    So, mv^2/2=(1/2)I1w^2 +(1/2)I2w^2, where I1 and I2 is about D, and vf=(L/4)*w.

    I get the new speed as sqrt(3/7)v. Cross check it.
     
  8. Oct 28, 2007 #7

    rl.bhat

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    The rod hits the nail normally. So it must return back in the same direction. The rod is not rotating about D. Hence Vf = r*w, where r is the radius of the rod.
     
  9. Oct 28, 2007 #8
    I'm agree that the rod is not rotating about D, but why Vf = r*w?
    r*w is the tangential velocity of the end pt, not the final velocity of the CM
    Am I correct?
     
  10. Oct 28, 2007 #9

    Doc Al

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    Maybe I haven't had enough coffee yet, but I don't get the point of this problem. Instantaneous speed when? Realize that by symmetry the interaction exerts no torque about the c.m. of the stick, so no rotation occurs.
     
  11. Oct 28, 2007 #10

    Shooting Star

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    The rod hits the nail off-centre. So, why will it not rotate?

    The resulting motion will be translation and rotation.

    The OP is not very clear about the kind of impact. We have to assume that the KE is conserved.
     
  12. Oct 28, 2007 #11

    Doc Al

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    Ah... I misread the problem. Impact is at at the middle of AC, not AB. (D'oh!) ...<reaches for the coffee pot>...
     
  13. Oct 28, 2007 #12

    Doc Al

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    The way I read the problem (reread it, actually), what they are looking for is the speed of the C.M. at the "instant" of the collision, which I take to be that instant where the midpoint of AC is momentarily at rest. At that instant the rod can be viewed as in pure rotation about the impact point.
     
  14. Oct 28, 2007 #13
    Someone solve this Q without using conservation of energy:
    Let P = linear impulse acted on the stick by the nail
    Let I = MI of stick about C
    Let V1 = speed of C after impact

    By conservation of linear momentum,
    MV - P = MV1...............................(1)

    By conservation of angular monentum about C,
    P(L/4) = wI................................(2)

    By kinematics,
    V1 = (L/4)w................................(3)

    I don't know why (3) is correct
     
  15. Oct 28, 2007 #14

    Shooting Star

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    Have you read my earlier post? Looks as if we are thinking on the same lines?
     
  16. Oct 28, 2007 #15

    vf=(L/4)*w <--- then we have the relationship between w and vf
    we don't need to use covservaton of energy
     
  17. Oct 28, 2007 #16

    Shooting Star

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    How do we find the relationship between vf and v? We don't know omega.
     
  18. Oct 28, 2007 #17
    Let P = linear impulse acted on the stick by the nail
    Let I = MI of stick about C
    Let Vf = speed of C after impact

    Change in linear momentum,
    MV - P = MVf...............................(1)

    Change in angular monentum about C,
    P(L/4) = wI................................(2)

    By kinematics,
    Vf = (L/4)w................................(3)

    (1) and (2) give the relationship between w, v, vf
    (3) give the relationship between w, vf
    then vf can be interms of v
    and the ans is vf=3/7 v
     
    Last edited: Oct 28, 2007
  19. Oct 28, 2007 #18

    Doc Al

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    Just to be clear, angular momentum of the rod about C is not conserved. (The impact imparts an angular impulse about C.) Instead, consider angular momentum about the point of impact.

    It's correct if you presume that the rod is momentarily in pure rotation about the impact point.
     
  20. Oct 28, 2007 #19

    Shooting Star

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    What's going on? Why would the ang mom be "conserved" about C? P(L/4) is not ang mom, more like a change in ang mom, like impulse.
     
  21. Oct 28, 2007 #20
    Actually..I think the word 'By conservation of angular monentum about C' is not correct
    It should be 'change in ang mom abt C'

    sorry abt that
     
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