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Advanced S.H.M. Problem

  1. Oct 29, 2010 #1
    1. The problem statement, all variables and given/known data
    A uniform cylindrical pulley of mass M and radius R can freely rotate about the horizontal axis O. The free end of a thread tightly wound on the pulley carries a dead weight A. At a certain angle [tex]\alpha[/tex] it counter balances a point mass m fixed at the rim of the pulley. Find the angular frequency of small oscillations of the arrangement (refer figure).


    3. The attempt at a solution

    Let the mass of the dead weight be m1
    Considering rotational equilibrium about O, we have
    m1gR = mgRsin[tex]\alpha[/tex]
    i.e. m1 = msin[tex]\alpha[/tex]

    Let the pulley be displaced through a small angle [tex]\theta[/tex] in the clockwise direction.

    The Total Energy of the system after rotation through the small angle is-
    E = mgRcos([tex]\alpha + \theta[/tex]) - m1gR([tex]\alpha+\theta[/tex]) + Iω²/2 + m1ω²R²/2

    As the system is conservative, the time derivative of energy is 0
    dE/dt = 0

    on solving the above equation,
    I got dω/dt = mgRsin[tex]\alpha[/tex] + m1gR/(I + m1R2)

    Now m1 = msin[tex]\alpha[/tex] and I = MR²/2 + mR² ( I did not include m1R² as I already took its kinetic energy as m1ω²R²/2)

    on solving further,

    d[tex]\omega[/tex]/dt = 2mgsin[tex]\alpha[/tex]/[(1+sin[tex]\alpha[/tex])mR +MR/2]

    which is not proportional to [tex]\theta[/tex]!!!
    and hence not a S.H.M.

    The answer given is ω² = 2mgcos[tex]\alpha[/tex]/[MR + 2mR(1+sinα)]

    Please help!!
     

    Attached Files:

    Last edited: Oct 29, 2010
  2. jcsd
  3. Oct 29, 2010 #2
    a figure will help
     
  4. Oct 29, 2010 #3
    I attached it now.
     
  5. Oct 29, 2010 #4
    Did it help?
     
  6. Oct 29, 2010 #5

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Five comments:
    1. Check your signs. You have a sign error here.
    2. The mass A is undergoing pure translational motion. The cylinder is undergoing pure rotational motion. Combining these two to form one value, Iω²/2, is not a good idea.
    3. You have made a notational error here that will hurt you later. You are obviously using ω=dθ/dt here. That's fine, but later on you are using ω to denote the harmonic oscillator frequency. That frequency is not equal to dθ/dt.
    4. Purely stylistic comment: Don't mix latex math with HTML math. The results are strikingly ugly and hard to read. Stick with one or the other.
    5. Here's an HTML alpha: α

    This does not follow from your earlier result. Hint: What happened to θ?
     
  7. Oct 30, 2010 #6
    Thanks Sir! I got my answer.

    I am new to Latex, sorry for inconvenience.
    Thanks for your advise.
     
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