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Finding Moment of Inertia - Rigid Rod Plus Disc System

  • Thread starter MEJJ
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  • #1
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Homework Statement



Consider a rigid rod-plus-disc system
2qu4wv7.jpg


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



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.
 
Last edited:

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi MEJJ! Welcome to PF! :smile:
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.

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

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.
I don't follow your Ix-axis parallel equation :confused:

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 :wink:)
 
  • #3
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Hi & Thanks, Tiny-Tim!

Yes, by "Ix-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?
 
  • #4
tiny-tim
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Hi MEJJ! :smile:
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. :redface:
 
  • #5
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So IParallel 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?
 
  • #6
tiny-tim
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So IParallel 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? :confused:

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:​
 
  • #7
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Ehh, I dunno what I'm doing. Time to visit the professor. Thanks for trying to help.
 

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