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I'm Having a similar Problem for a conic

Assume constant density, then mass = volume

I'm looking for the center of gravity

So I assume this means point of equal volume

Let Cone lie on side, let theta = @ ; then Tan@ = radius/height

CONE: V = 1/3 x R^2 x H ; A = pi x R^2 ; R = H x Tan@

Centroid of cone?

cone v = 1/3 x r^2 x h ; A = pi x r^2 ; r = h x Tan@

Doesn't

2 x v = V ???

Then 2/3 x pi x (Tan@)^2 x h^3 = 1/3 x pi x (Tan@)^2 x H^3

therefore 2 x h^3 = H^3 h = 2^(1/3) x H

One/cuberoot of 3 = 0.7937... does not equal 3/4 x H

What is wrong with my logic?????

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migrated from physics.stackexchange.com Sep 4 '13 at 19:14

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Then 2/3 x pi x (Tan@)^2 x h^3 = 1/3 x pi x (Tan@)^2 x H^3 therefore 2 x h^3 = H^3 h = 2^(1/3) x H One/cuberoot of 3 = 0.7937... does not equal 3/4 x H What is wrong with my logic????? –  Mark Sep 4 '13 at 18:41
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Welcome to Physics.SE! One hint: you can use LaTeX markup in your posts by using dollar signs. For example, $\tan(\theta)$ renders as $\tan(\theta)$. –  episanty Sep 4 '13 at 18:42
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Possible duplicate: Determining the center of mass of a cone. –  episanty Sep 4 '13 at 18:48
    
Then....... 2/3 x pi x (Tan@)^2 x h^3 = 1/3 x pi x (Tan@)^2 x H^3...... therefore...... 2 x h^3 = H^3....... h = 2^(1/3) x H....... One/cuberoot of 2 = 0.7937....... does not equal 3/4 x H...... What is wrong with my logic????? –  Mark Sep 4 '13 at 18:50
    
Got it, leverage CofG not equal to equal volume –  Mark Sep 4 '13 at 19:14

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