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## Homework Statement

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 · m

^{2}.

## Homework Equations

I

_{rod}= (1/12)*m*R

^{2}

I

_{disc}= (1/2)*2m*(R/2)

^{2}

I

_{x-axis parallel}= m*[(R - (¼*R))/2]² + m*R --> Are these correct?

## The Attempt at a Solution

I

_{total}= I

_{rod}+ I

_{disc}+ I

_{x-axis parallel}

I

_{rod}= (1/12)*1.94kg*0.62m

^{2}

I

_{disc}= (1/2)*(2*1.94kg)*(0.62m/2)

^{2}

I

_{x-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.

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