Consider a rigid rod-plus-disc system
Find the moment of inertia of the pendulum as it freely rotates about the point P. The rod has a length 0.62 m, and the disc has a radius half that. The pivot at P is a fourth of the way from the end of the rod. The rod has a mass of 1.94 kg and the disc is twice as massive with uniform mass distribution.
Answer in units of kg · m2.
Irod = (1/12)*m*R2
Idisc = (1/2)*2m*(R/2)2
Ix-axis parallel = m*[(R - (¼*R))/2]² + m*R --> Are these correct?
The Attempt at a Solution
Itotal = Irod + Idisc + Ix-axis parallel
Irod = (1/12)*1.94kg*0.62m2
Idisc = (1/2)*(2*1.94kg)*(0.62m/2)2
Ix-axis parallel = 1.94kg*[(0.62m - (¼*0.62m))/2]² + 1.94kg*0.62m
Not sure if I've correctly accounted for Pivot Point P.