Moment of Inertia: 3 Point Mass, Massless Rods, Rotational Motion

[Celestiela]

Simple question:

I have 3 point masses of 110 kg apiece. All are connected by 3 massless rods to a massless point in the middle. The system will rotate about that point in the middle. The rods are 3.9 m long. They are 120 degrees apart from each other like:

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/ \

Kinda. Anyways, when I'm calculating the moment of interia, am I correct in saying saying (110 kg)(3.9 m)(3.9 m)+(110 kg)(3.9 m)(3.9 m)+(110 kg)(3.9 m)(3.9 m)=5019.3 kg m^2? Do I need to worry about the way the weight is distributed? Or is everything ok because the center of mass is at that point? What if it wasn't? I think I'm confusing the two...

 

[OlderDan]

Simple question:

I have 3 point masses of 110 kg apiece. All are connected by 3 massless rods to a massless point in the middle. The system will rotate about that point in the middle. The rods are 3.9 m long. They are 120 degrees apart from each other like:

.|
/ \

Kinda. Anyways, when I'm calculating the moment of interia, am I correct in saying saying (110 kg)(3.9 m)(3.9 m)+(110 kg)(3.9 m)(3.9 m)+(110 kg)(3.9 m)(3.9 m)=5019.3 kg m^2? Do I need to worry about the way the weight is distributed? Or is everything ok because the center of mass is at that point? What if it wasn't? I think I'm confusing the two...

Your calculation is correct for point masses, which is what you were given. If each of those masses were distributed in some way (like disks or spheres) a correct calculation would involve the parallel axis theorem. Each term in your sum would then have an additional term that depended on the shape of the mass.

 

[Celestiela]

sweet! Thanks!