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Calculate velocity

  1. Oct 23, 2015 #1
    1. The problem statement, all variables and given/known data
    A wheel drive in the horizontal plane with angular velocity 13 rad / s. Two bodies mass mA = 1 kg and mB = 1.4 kg slide without friction along a rail fixed to the disk and passing through its center. The two bodies are connected by a bar inextensible (negligible mass) and 0.3m long.

    Knowing that the body A is initially 15 cm from the center, determine the initial radial velocity that should print the body so that his movement tends to break even.


    2. Relevant equations

    aA = xA w^2 - (mB l w^2) / (mA+mB)
    xA(eq) = (mBl/(mA+mB))

    3. The attempt at a solution
    I determine the equations above, but I don't how to use them to calculate what I need to...
    Te correct answer is 0,325 m/s
     
  2. jcsd
  3. Oct 23, 2015 #2

    haruspex

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    You need to write your first equation as a differential equation and solve it. Then use the information that x is bounded as time tends to infinity.
     
  4. Oct 24, 2015 #3
    Hi!
    So I did it and I found out that

    rA = C1 e^-wt + C2 e^wt
    rA (t=0) = C1 + C2
    vA (t=0) = -C1 w + C2 w

    we need C2=0 so I told that v (t=0) was also -wC1 -wC2

    Then I wrote

    rA = C1 e^-wt
    vA = -wC1 e^-wt

    rA(t=0) = r0 = C1

    So

    rA = r0 e^-wt
    vA = -r0 w e^-wt

    and then vA (t=0) - 1,95 m/s

    But theres something I'm missing because the result is wrong... There's something about rAeq that I'm not guessing.... Can you please help me?
     
  5. Oct 24, 2015 #4

    haruspex

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    How are you defining rA?
     
  6. Oct 24, 2015 #5
    What do you mean?
    rA = r0 e^-wt
     
  7. Oct 24, 2015 #6

    haruspex

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    That's an equation, not a definition. What does you variable rA represent? If it represents the distance of mass A from the axis then your differential equation solution is clearly wrong. It would imply that the mass reaches the axis at infinite time.
     
  8. Oct 24, 2015 #7
    Oh yes now I'm understanding I guess... Thanks! So How can I fix the solution? Can you explain?
     
  9. Oct 24, 2015 #8

    haruspex

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    Please post the differential equation you obtained and the steps up to the solution you posted.
     
  10. Oct 25, 2015 #9
    rA = C1 e^-wt + C2 e^wt
    rA (t=0) = C1 + C2
    vA (t=0) = -C1 w + C2 w

    we need C2=0 so I told that v (t=0) was also -wC1 -wC2

    -C1w +C2w = -wC1 - wC2
    C2 = 0

    Then I wrote

    rA = C1 e^-wt
    vA = -wC1 e^-wt
    vA(t=0)= -C1w - C2w
    rA(t=0) = r0 = C1

    So

    rA = r0 e^-wt
    vA = -r0 w e^-wt

    vA (t=0) = - r0 w e^-wt

    and then vA (t=0) - 1,95 m/s
     
  11. Oct 25, 2015 #10

    haruspex

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    No, I asked for the differential equation. Your very first line there is an incorrect solution to the equation.
     
  12. Oct 25, 2015 #11
    Ah

    aA - w^2 rA = - w^2 (req - X)

    if X = rAeq

    aA - w^2rA = 0
     
  13. Oct 25, 2015 #12

    haruspex

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    We're not looking for the equilibrium position. We need the general equation of motion of this system. Consider forces.
     
  14. Oct 25, 2015 #13
    I'm really not understanding what i need to do
     
  15. Oct 25, 2015 #14

    haruspex

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    Consider the forces acting on the system and write the equation for the resulting acceleration along the rod.
    E.g. let x be the distance of mass A from the axis. Find an equation relating ##\ddot x## to x, ##\omega##, length of rod L, and the two masses.

    Edit: it's the first of your two 'relevant equations'
     
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