Calculate Moment of Inertia of Rods | Solution Included

AI Thread Summary
The discussion focuses on calculating the moment of inertia for two rods about their center of mass. The moment of inertia for one rod is given as L^2m/12, and the center of mass is identified at L/4 in both x and y directions. To find the total rotational inertia, the parallel axis theorem is applied, leading to the formula L^2m/12 multiplied by 2, plus L^2m/8 multiplied by 2. The conversation also touches on how the calculation would differ if the configuration involved a square of rods, noting that the distances to the center of mass would vary. Understanding these principles is crucial for accurately determining moment of inertia in different geometrical arrangements.
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[SOLVED] moment of inertia

Homework Statement



what is the moment of interia of these 2 rods (picture attached) about their center of mass
each rod mass is m with length L

Homework Equations


one rod i know is \frac{L^2m}{12}


The Attempt at a Solution

 

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First things first: Find the center of mass.
 
good question

i thought the answer is \frac{L^2m}{12}*2+\frac{L^2m}{4}*2
obviously it isnt

i think center of mass is at \frac{L}{4}\hat{x}-\frac{L}{4}\hat{y}
 
Last edited:
supercali said:
i think center of mass is at \frac{L}{4}\hat{x}-\frac{L}{4}\hat{y}
Good--it's right in the middle between the two rod centers. So what's the distance from the center of each rod to that center of mass? To find the rotational inertia of the rods about that center of mass, use the parallel axis theorem.
 
yeah i got it it is \frac{L^2m}{12}*2+\frac{L^2m}{8}*2 and one more thing if it was a square or rods i would have to multiply by 4?
 
supercali said:
yeah i got it it is \frac{L^2m}{12}*2+\frac{L^2m}{8}*2
Good.
and one more thing if it was a square or rods i would have to multiply by 4?
If it were a square of rods, the distance from each rod to the center of mass would be different. (But there would be a factor of 4 involved.)
 
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