Finding Moment of Inertia - Rigid Rod Plus Disc System

[MEJJ]

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 · m².

Homework Equations

I_rod = (1/12)*m*R²
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*0.62m²
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.

 

[tiny-tim]

Welcome to PF!

Hi MEJJ! Welcome to PF!

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.
…
I_{x-axis parallel} = m*[(R - (¼*R))/2]² + m*R --> Are these correct?
…
I_total = I_rod + I_disc + I_{x-axis parallel}

I_rod = (1/12)*1.94kg*0.62m²
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.

I don't follow your I_{x-axis parallel} equation

Is that supposed to be the parallel axis theorem ?

That says that I = Ic.o.m. + md², where d is the distance from the pivot to the centre of mass.

(and btw R is .31, not .62 )

 

[MEJJ]

Hi & Thanks, Tiny-Tim!

Yes, by "I_{x-axis parallel}", I really meant the equation for Parallel Axis Theorem.

You're right: R is 0.31m, not 0.62m.

Even so, I take it my Parallel Axis Theorem is not correct for the situation?

 

[tiny-tim]

Hi MEJJ!

Even so, I take it my Parallel Axis Theorem is not correct for the situation?

It's the right theorem, but you've done it completely wrong.

 

[MEJJ]

So I_{Parallel Axis Theorem} = m*[(R - (¼*R))/2]² + m*R for the rod & disc is not at all correct? I am only correct in needing to use the theorem-formula?

 

[tiny-tim]

So I_{Parallel Axis Theorem} = m*[(R - (¼*R))/2]² + m*R for the rod & disc is not at all correct? I am only correct in needing to use the theorem-formula?

Your (R - (¼*R))/2 is 3R/8 … what is that supposed to be?

And your m*R isn't even the right dimensions.

These should both be md², where d (different for each one) is the distance from the pivot to the centre of mass of the part in question.

And I'm off to bed now :zzz:​

 

[MEJJ]

Ehh, I don't know what I'm doing. Time to visit the professor. Thanks for trying to help.