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)]