Take the following example: A rod (of length L and mass m) is held horizontally at both ends by supports. One is instantaneously removed.
The specific problem is to prove that the force on the other support drops from mg/2 to mg/4, which I proved by first considering the centre of mass as the instantaneous reference frame, and thus considering a rotation around the support.
Resolving angular forces: (F = Force at pivot, I = Moment of Inertia = m(L^2)/12, ω = angular velocity)
FL/2 = I * dω/dt FL/2 = m(L^2)/12 * dω/dt F = mL/6 * dw/dt (1)
Now taking the instantaneous reference frame around the pivot: (I = M(L^2)/3, ω' = angular velocity, force at CoM = mg)
mg * L/2 = I * dω'/dt mgL/2 = mL2/3 * dw'/dt dw'/dt = 3g/2L (2)
The desired solution can be found by substituting (2) into (1), i.e. by equating dw'/dt and dw/dt. Why is it that this can be done?