How to calculate the center of gravity of a cylinder segment?
I have a 6" thick segment of a cylinder, with about a 15' radius to the outer face. It's roughly 10 feet tall, and I'm trying to work out the center of gravity so I can successfully hang it from two points. The inner and outer face weigh the same, and the internal construction is uniform, so any slice through it has a CoG 3" from the face.
It's been a while since I've used calculus and while I'm sure there's a way to do this with pure math, so hoping someone has the math laying around.
(The geometric solution I'm going to try is to find the point where the CoG line (curve) has the same length available on either side of a line drawn through.)
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4 Answers
My geometric method described above is incorrect... *sigh* (The "tails" stick out further from the COG line drawn, so they cause a backwards tilt...)
I may be easier to just lie it down on two rods and move the rod that is easier to move closer until the other gets easier to more. I think of the classic Mr. Wizard experiment where he had a kid put a yard stick on two finger then try to move their fingers together. Since the finger that is farther from the center of gravity has less weight the friction between the finger and the ruler is less. Therefore that finger moves closer to the center of gravity until the other finger is farther from the center or gravity...... process repeat. The fingers end up at the center of the ruler, which is the center of gravity.
Otherwise, brake out the integral calculus.
i'm intrigued. why are you hanging a chunk of cylinder?
Not only integral calculus, you’ll need double integrals for this one.
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