Finding the moment of inertia of a compound object

AI Thread Summary
The discussion focuses on calculating the moment of inertia for a compound object consisting of a uniform bar and two small point masses attached to its end. The moment of inertia for the bar is calculated using the formula I=1/3ML^2, resulting in 5.33 kgm². The contribution from the two point masses is calculated using I=Σmr², yielding an additional 1.0 kgm². The total moment of inertia is then found to be 6.33 kgm². The conversation concludes with clarification on the correct formulas and a resolution of confusion regarding the calculations.
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Homework Statement


A uniform bas has two small balls glued to its end, the bar is 2.00 m long, and it has a mass 4.00 Kg, while the balls, have mass 0.500Kg and can be treated as point masses. Find the moment of inertia of this combination about each of the following axes. a) an axis perpendicular to the bar through its center.


Homework Equations


I=1/3ML^2
I=\Sigmamr^2

The Attempt at a Solution


I=1/3ML^2
I=1/3(4.00Kg)(2.00m)^2
I=5.33 Kgm^2

I=\Sigmamr^2
I=(0.500Kg)(2.00m)^2(2)
I=1.0Kgm^2

I tot =6.33 Kgm^2
I am supposed to add both results?, how is the result of hit sexercise 2.33 Kgm^2?
thanks.
 
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I of bar is 1/12*M*L^2.
 
I believe that's the equation when the axis of rotation goes through one of the bar's end. when the axis crosses the center of the bar I=1/3ML^2 according to my textbook.
 
you were right, I am so sorry. thanks. I finally got it.!
 
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