how do you find the moment of inertia for wierdly shaped objects?

jjellybean320

http://www.livephysics.com/tables-of...f-inertia.html

the rectangular plate equation on that page is
(1/12)*Mass*(a^2+b^2)
where a and b are the lenght and width

but this equation is if the axis is through the center.

if the axis is at one end like my boomerang, would i just substitute (1/12) with (1/3)?

cosmik debris

It's funny what passes for a boomerang these days :-)

gneill

No. Look up the Parallel Axis Theorem as I suggested. You need to be precise about where the center of rotation is with respect to the center of the plate -- in your case it won't be precisely at the end of the plate, but a bit in from the end. Measure and be sure.

jjellybean320

the center of rotation is about 1 cm from the center of the plate. i looked at the wikipedia page for the parallel axis theorem and it was really confusing. i've only taken ap calculus ab.

gneill

That seems rather close to the center of mass of the plate. From your image it looked like it was more like 1cm from one end (the end where the blades all come together in the center), so about 5.4cm from the center of mass of the plate.

The parallel axis theorem states, in essence, that given a known moment of inertia I about some axis, then if you move the axis of rotation some distance r from that axis, being sure that the new axis is parallel to the old axis, then the moment of inertia about that new axis is given by I_new = I + M*r². M is the mass of the object.

The formula that you found for the moment of inertia for the plate about its center should be 'shifted' by adding M*r² to it, where r is the distance from the center of the plate to the actual axis of rotation.

jjellybean320

thank you. I understand it so much better now. And you were right, it is about 5.4 cm between the center of rotation and the center of the plate. I simply misread what you wrote.