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Center of Mass Rotational Motion

  1. Aug 3, 2016 #1
    1. The problem statement, all variables and given/known data
    The system shown in (Figure 1) consists of two balls Aand B connected by a thin rod of negligible mass. Ball Ahas three times the inertia of ball B and the distance between the two balls is ℓ. The system has a translational velocity of v in the x direction and is spinning counterclockwise at an angular speed of ω=2v/ℓ.

    Mazur1e.ch12.p12.jpg
    2. Relevant equations
    mArA = mBrB
    rA + rB = ℓ
    ω = 2v/ℓ
    vA = v - (2v/ℓ)rA
    vB = v + (2v/ℓ)rB

    3. The attempt at a solution

    mArA = mBrB ; mA = 3mB

    3mBrA = mBrB
    3rA = rB

    rA + rB = ℓ
    rA + 3rA = ℓ
    4rA = ℓ
    rA/ℓ =1/4

    ω = 2v/ℓ

    vA = v(1 - 2rA/ℓ)
    = v(1 - 2/4)
    = (1/2)v

    vB = v(1 + 2rB/ℓ)
    = v(1 + 2(3rA)/ℓ)
    = v(1 +6rA)/ℓ)
    = v(1 + 6/4)
    = (3/2)v

    vA/vB = [(1/2)v]/[(3/2)v] = (1/2)/(3/2) = (1/2)x(2/3) = 1/3

    1/3 is not the correct answer, where am I going wrong?
     
  2. jcsd
  3. Aug 3, 2016 #2

    TSny

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    The problem statement doesn't state what you are to find.

    EDIT:
    Do you have the correct signs here?
     
    Last edited: Aug 3, 2016
  4. Aug 3, 2016 #3
    Oh, I'm sorry. I'm supposed to find the ration of Va/Vb. Ans for the second portion, should it be 10/4v then for vB? So 1/5 is the ratio ?
     
  5. Aug 3, 2016 #4
    Also, yes I believe I have the correct signs.
     
  6. Aug 3, 2016 #5

    TSny

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    Does VA represent the speed of A relative to the "earth"?

    In what direction is the velocity of A due to the translational motion?

    In what direction is the velocity of A due to the rotational motion?

    How would these combine (add or subtract)?
     
  7. Aug 3, 2016 #6
    Ohhhhh, thank you! So the answer should be -3, or 3 since its a ratio.
     
  8. Aug 3, 2016 #7

    haruspex

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    That's not what I get. Please post your most recent working.
     
  9. Aug 3, 2016 #8
    vA = v(1 - 2rA/ℓ)
    = v(1 + 2/4)
    = (3/2)v

    vB = v(1 + 2rB/ℓ)
    = v(1 + 2(3rA)/ℓ)
    = v(1 +6rA)/ℓ)
    = v(1 - 6/4)
    = (-1/2)v

    vA/vB = [(3/2)v]/[(-1/2)v] = (3/2)/(-1/2) = (3/2)x(-2/1) = -3 but its a ratio so just 3. It said it was correct ?
     
  10. Aug 3, 2016 #9

    haruspex

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    As TSny indicated, the sign above looks wrong, but maybe you are defining rA in such anway that its value will be negative.
    You seem to have substituted rA=-l/4. How did you get that?
     
  11. Aug 3, 2016 #10
    mArA = mBrB ; mA = 3mB

    3mBrA = mBrB
    3rA = rB

    rA + rB = ℓ
    rA + 3rA = ℓ
    4rA = ℓ
    rA/ℓ =1/4


    I got the answer of 3 which was correct. Thank you
     
  12. Aug 3, 2016 #11

    haruspex

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    Sure, but that's not -l/4.

    Anyway, I made a mistake. I agree with your 3:1, but your sign errors were confusing.
     
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