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Two objects on a pulley

  1. Feb 19, 2006 #1
    i willl leave out the numbers to make this go more smoothly


    [​IMG]

    find the acceleration of m1

    heres what i thought:

    (m2/(m1+m2))g= a

    but this is wrong. can you guys please help me out. thanks
     
  2. jcsd
  3. Feb 19, 2006 #2
    friction is neglected
     
  4. Feb 19, 2006 #3

    arildno

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    Note that the rope length increases on BOTH sides of m2 equally.
    Thus, in order for the TOTAL length of ropes to remain constant, the speed of m2 must be HALF of the speed of m1
     
  5. Feb 19, 2006 #4
    thanks for helping me, but im still getting the wrong answer. after what you said, i realized that the tension of m(m2), too, had to be 1/2 of the tension of M (m1)

    so heres how i did it this time:

    a= (2(m)g)/(M-2m)

    this is also wrong
     
  6. Feb 19, 2006 #5

    arildno

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    Yes, that's wrong.
    Here's a correct way.
    Let m1's velocity (and acceleration) be positive, m2's velocity (and acceleration) be negative.
    The SPEED of m2 is half that m1's, that is the velocities are related v2=-v1/2

    There is conservation of mechanical energy, so we have:
    [tex]\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}+m_{2}gz=C[/tex]
    Differentiating, we get:
    [tex]m_{1}v_{1}a_{1}+m_{2}v_{2}a_{2}+m_{2}gv_{2}=0[/tex]
    Note that our choice of v2 negative implies the correct loss of potential energy.

    By using the relation between the velocities (and accelerations), we get, by dividing with v1/2:
    [tex](2m_{1}+\frac{m_{2}}{2})a_{1}=m_{2}g[/tex]
    that is:
    [tex]a_{1}=\frac{2m_{2}g}{4m_{1}+m_{2}}[/tex]
     
  7. Feb 19, 2006 #6
    thanks arildno :!!)

    :cool:
     
  8. Feb 19, 2006 #7

    arildno

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    OOPS!
    What you say about the tension is totally wrong!
    Since the rope is assumed massless, and there is no friction, the tension must be the same THROUGHOUT the rope!

    Here's how you should to this from F=ma:
    For m1, we have:
    [tex]T=m_{1}a_{1}[/tex]
    For m2, consider the system of the mass plus the half-circle arc of massless rope around it, with vertical ends of rope.
    Since the rope is massless, we get the equation for the system's C.M:
    [tex]T+T-m_{2}g=m_{2}a_{2}+0*a_{rope}\to{2T}-m_{2}g=-m_{2}\frac{a_{1}}{2}[/tex]
    Eliminating the tension in the last equation by using the first equation reproduces the result in the previous post.
     
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