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Angular momentum and acceleration

  1. May 11, 2013 #1
    1. The problem statement, all variables and given/known data

    m1 and m2 are two blocks tied with a rope with a pulley inbetween, like those in this picture
    http://labella.altervista.org/images/mechanicsoftwopoint_2.png
    there are no frictions.
    find: the linear acceleration of the blocks and the tension of the rope on both m1 and m2.


    3. The attempt at a solution
    I'd say it is:
    ## m_2 \cdot g = I \alpha + m_1 \cdot a + m_2 \cdot a## with ##a= \alpha \cdot R_0## being ##R_0## the radius of the pulley
    ## m_2 \cdot g= m_1 \cdot a + I \frac{a}{R_0}+m_2 \cdot a## so
    ##a=\frac{m_2 \cdot g}{mb + \frac{I}{R_0}}##

    and then the tension of the rope on m1 is simply
    T=a*m1
    while the tension on m2 is
    T=m2*g+m2*a

    Is it correct? If not, why?
    Thank you for checking :)
     
    Last edited: May 11, 2013
  2. jcsd
  3. May 11, 2013 #2

    Doc Al

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    Does the pulley have mass?
     
  4. May 11, 2013 #3
    it isn't written, but I don't think so.
     
  5. May 11, 2013 #4

    Doc Al

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    Then you can forget about I for the pulley.

    Apply Newton's 2nd law to each mass.
     
  6. May 11, 2013 #5
    okay, thank you. :)
    But if the pulley had mass, would it be right?
     
  7. May 11, 2013 #6

    Doc Al

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    No, it's not correct.

    If you want to know why it's wrong, show how you got that result.
     
  8. May 11, 2013 #7
    i thought ##m_2 \cdot g## is the force which moves the system, so it equals
    ##m_1 \cdot a## with an unknown acceleration, ##+m_2 \cdot a## because the block goes down with the same acceleration, and it also makes the pulley spin with ##I \cdot \alpha##
    then I've read the formula ## a= \alpha \cdot R_0## on a book, and thought of applying it
     
  9. May 11, 2013 #8

    Doc Al

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    Rather than try to do it all at once, and risk making an error, why not divide and conquer. Apply Newton's 2nd law to the two blocks and to the pulley.
     
  10. May 11, 2013 #9
    like:
    ##F_g= m_2 \cdot g##
    ##F_1= m_1 \cdot a##
    ##F_p= I \cdot \alpha##
    ##F_2= F_g-F_1-F_p##
    ?
    i keep getting the same result
     
  11. May 11, 2013 #10

    Doc Al

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    This one I understand. Where ##F_1## is the tension in string 1.

    These I don't understand.

    What forces act on ##m_2##?
    What forces act on the pulley? What torque do they exert?
     
  12. May 12, 2013 #11
    on m2 we have the gravitational force:
    ##F_p=m_2 \cdot g -T= m_2 \cdot a## where T is the tension of the string
    ##T= m_1 \cdot a##
    by subtracting the two:
    ##m_2 \cdot g - m_1 \cdot a = m_2 \cdot a##
    and ##I \cdot \alpha= m_2 \cdot g - T ##
    ##I \cdot \alpha= m_2 \cdot g - m_1 \cdot a##
    but i really don't know what I am doing :(
     
  13. May 12, 2013 #12
    If the pulley had mass, the tension at its two ends shouldn't be same !

    In fact,

    (T1-T2)*r= Ia/r

    T1 is that downward tension, and T2 leftwards.
    You need to get two more equations by applying newtons second law on each mass for translational motion.

    Approach 2 : Use conservation of mechanical energy on the two blocks system.
     
  14. May 14, 2013 #13
    Thank you :)!
    but I don't understand how you found the left part of this equation: (T1-T2)*r= Ia/r
    i mean, why is the difference between the two tensions multiplied by r?
     
  15. May 14, 2013 #14
    Torque is a vector. Determine the directions of torques produced by tensions in the rope.
     
  16. May 14, 2013 #15

    Doc Al

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    When applied to rotation, Newton's 2nd law becomes:
    Ʃ Torque = Iα

    Again, start by answering the questions I had asked earlier. What forces act on the pulley? What torques do they exert? Then you can apply Newton's 2nd law to the rotation of the pulley.
     
  17. May 14, 2013 #16
    i'd say ##T_2=m_2 \cdot g## and ##T_1=M_1 \cdot a## are the only forces acting on the pulley
    and so ##\Sigma \tau = (T_1- T_2)x \vec{r}## where r is the motion (which equals the radius?). which gives
    ##(m_2 \cdot g - m_2 \cdot a) x r = I \cdot \frac{a}{r}=##
    ##a=\frac{m_2 \cdot g}{\frac{I}{r^2} + m_1}## right?
     
  18. May 14, 2013 #17

    Doc Al

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    Yes, the two tensions are the only relevant forces acting on the pulley. But your first equation, for block 2, is incorrect. (If ##T_2=m_2 \cdot g## then the net force on block 2 would be zero.)

    Do it systematically. For the pulley:

    ##\Sigma \tau = (T_1 - T_2) \cdot R = I \cdot \frac{a}{R}##

    That's one equation. Now write equations for each of the blocks. (The equation you wrote for block 1 is correct; fix the one for block 2.)
     
    Last edited: May 14, 2013
  19. May 15, 2013 #18
    is block 2:
    ##F= -m_2 \cdot g + m_2 \cdot a##?
     
  20. May 15, 2013 #19

    Doc Al

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    Please use the same notation you used earlier. What's F supposed to be?

    What forces act on block 2? What are their directions?
     
  21. May 15, 2013 #20
    oops, sorry
    then, on block two there should be the gravity force:
    ##-m_2 \cdot g## and the tension of the string: ##T_1## which have opposite directions. then the block moves downwards with an acceleration = "a"
    so:
    ##T_2-m_2 \cdot g =- m_2 \cdot a##
     
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