axis through center. I dont know what to do.
I=p(integral r^2 dm)
What do you do about dm (change in mass)
The diameter is lenth a.
2007-01-22 16:37:01 · 3 answers · asked by paul 2 in Science & Mathematics ➔ Physics
The moment of inertia of a particle is calculated by mr^2, right?
If the axis is perpendicular through the center then since for a thin hoop all the little particles have the radius that is the same, you can just sum the whole mass and multiply it by r^2 to find the moment of inertia.
Now ... if the axis is through the center and not parallel with the plane of the hoop you need to integrate from 0 to pi radians and multiply by 2 for each little dr.
2007-01-22 17:29:19 · answer #1 · answered by themountainviewguy 4 · 0⤊ 0⤋
It is simply mr^2 if the axis is perpendicular to the plane of the loop. If the axis is along the diameter, it is 1/2 mr^2.
2007-01-23 00:50:33 · answer #2 · answered by Vijay Krishnan 1 · 1⤊ 0⤋
Since all the mass is in the hoop which has no thickness then the intertia is (r^2)(m) where r is a/2 since the diameter is a
2007-01-23 00:46:07 · answer #3 · answered by rscanner 6 · 0⤊ 0⤋