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Acelleration of a pulley

  1. Nov 7, 2009 #1
    1. MAD_ia_6.jpg

    Find magnitude of angular accelleration alpha, express in terms of radius r and gravity




    2. torque = rF, F=ma



    3. T = mg
    torque = rF = -Tr = I*alpha
    -mgr = I*alpha
    alpha = -mgr/I

    with I = 0.5 mr^2 I get:

    alpha = -2g/r, which is incorrect.
     
  2. jcsd
  3. Nov 7, 2009 #2
    Why t=ma.It is not stationary.it is moving.so mg-T=ma
     
  4. Nov 7, 2009 #3
    The body is moving with an acceleration of a which is alpha*r
     
  5. Nov 7, 2009 #4
    I'm still not getting the solution.

    Using ma = mg - T
    I*alpha = -Tr
    a = r*alpha

    I obtained:

    alpha = (2a - 2g)/r

    Using ma - mg = I*alpha/r I obtained:

    a = 0.5*r*alpha + g

    Putting a into the equation for alpha:

    2[(0.5*r*alpha + g) - g]/r
    a = alpha
     
  6. Nov 7, 2009 #5
    I don't really get what you did.You don't have to put - at the equation Tr=I*alpha.Also alpha=a/r.try this
     
  7. Nov 7, 2009 #6
    I*alpha = -Tr because T exerts torque in the negative (clockwise) direction.
     
  8. Nov 7, 2009 #7
    Nevertheless if you put numbers you would found the same.The idea though is the one i wrore you.i hope i helped.if you have more questions just ask
     
  9. Nov 7, 2009 #8
    Sorry, this is not helping. Using, alpha = Tr/I and I = 0.5mr^2, I obtained:

    alpha = [m(g-a)]/(0.5mr^2)
    Solving after quite a bit of algebra:

    alpha = g/(r^2 + r) is incorrect
     
  10. Nov 7, 2009 #9
    Which is the correct answer?Try to solve as to T.T=-I*alpha/r
    and also write alpha as a/r.then put T at the equation mg-T=ma
     
  11. Nov 7, 2009 #10
    is the correct answer 2g/3r?
     
  12. Nov 7, 2009 #11
    ............................... I don't know the correct answer obviously.

    I have done all that already. With T = -I*alpha/r, alpha = a/r, and T = mg - ma, I obtain:

    alpha = [rm(g-a)]/-I = (2a - 2g)/r
     
  13. Nov 7, 2009 #12
    Yes, how did you calculate this?
     
  14. Nov 7, 2009 #13
    How you obtain all this?-Tr=0.5mr^2*a/r so T=-0.5ma
    Then mg -T=ma so mg-0.5ma=ma so 3/2*a=g so a=2g/3
     
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