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!!