Advanced S.H.M. Problem

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Homework Statement


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).


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!!
 

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Answers and Replies

  • #2
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a figure will help
 
  • #3
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I attached it now.
 
  • #4
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Did it help?
 
  • #5
D H
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E = mgRcos([tex]\alpha + \theta[/tex]) - m1gR([tex]\alpha+\theta[/tex]) + Iω²/2 + m1ω²R²/2
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

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)
This does not follow from your earlier result. Hint: What happened to θ?
 
  • #6
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Check your signs. You have a sign error here.

Thanks Sir! I got my answer.

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.

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

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