How Does a Cylindrical Pulley System Exhibit Simple Harmonic Motion?

[zorro]

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 \alpha 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\alpha
i.e. m1 = msin\alpha

Let the pulley be displaced through a small angle \theta in the clockwise direction.

The Total Energy of the system after rotation through the small angle is-
E = mgRcos(\alpha + \theta) - m1gR(\alpha+\theta) + 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\alpha + m1gR/(I + m1R²)

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

on solving further,

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

which is not proportional to \theta!
and hence not a S.H.M.

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

Please help!

 

[issacnewton]

a figure will help

 

[zorro]

I attached it now.

 

[zorro]

Did it help?

 

[D H]

E = mgRcos(\alpha + \theta) - m1gR(\alpha+\theta) + 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\alpha + m1gR/(I + m1R²)

This does not follow from your earlier result. Hint: What happened to θ?

 

[zorro]

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.