1. Oct 29, 2010

### zorro

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 $$\alpha$$ 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).

3. 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)]

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Last edited: Oct 29, 2010
2. Oct 29, 2010

### issacnewton

a figure will help

3. Oct 29, 2010

### zorro

I attached it now.

4. Oct 29, 2010

### zorro

Did it help?

5. Oct 29, 2010

### D H

Staff Emeritus
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?

6. Oct 30, 2010

### zorro

Thanks Sir! I got my answer.

I am new to Latex, sorry for inconvenience.