Moment of inertia for physical pendulum

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SUMMARY

The discussion focuses on calculating the moment of inertia (I) for a physical pendulum consisting of two 55kg masses positioned at different points on a rigid, massless strut of length 11m. The correct formula for point masses is I = ΣmD², where D is the distance from the axis of rotation. After correcting the initial miscalculations, the total moment of inertia for the pendulum is determined to be 8318.75 kg-m².

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  • Familiarity with the integral calculus for continuous objects
  • Knowledge of point mass calculations in physics
  • Basic principles of physical pendulums
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Physics students, mechanical engineers, and anyone involved in dynamics or rotational motion analysis will benefit from this discussion.

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I have a rigid, massless strut of length 11m. A mass of 55kg is located on the free end, and another mass of 55kg is located at the midpoint. I'm having some trouble calculating the moment of inertia I for this physical pendulum.

I = \int r^{2} dm (for continuous objects)

I = \sum m D for all particles/objects composing the system.


I'm a little confused. Should I calculate I for each 55kg mass and add them together?
 
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BoogieBot said:
Should I calculate I for each 55kg mass and add them together?
Exactly.
 
so:
55*5.5 = 302.5
55*11 = 605

so I for the whole pendulum is 907.5 kg-m^2, correct?
 
BoogieBot said:
so:
55*5.5 = 302.5
55*11 = 605

so I for the whole pendulum is 907.5 kg-m^2, correct?
No, your formula for I of a point mass is incorrect. (I didn't notice that earlier.)

BoogieBot said:
I = \int r^{2} dm (for continuous objects)

I = \sum m D for all particles/objects composing the system.
For both continuous objects and point masses, the distance must be squared. (Imagine integrating the first formula to get the second. All of the mass is at the same distance from the axis, so the integral is trivial.)
 
Ah, so then:

55*5.5^2 = 1663.75
55*11^2 = 6655

1663.75 + 6655 = 8318.75 kg-m^2
 
You got it.
 
awesome! thanks!
 

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