Center of Mass for 8 kg Stone & 2.5 kg Stick

AI Thread Summary
To find the center of mass of an 8 kg stone attached to a 2.5 kg stick, the formula Xcm = (sum) x*m / m is applied. The stick is uniform and symmetrical, meaning its center of mass is at its midpoint, which is 0.49 m from the handle end. The stone's position is at the end of the stick, 0.98 m from the handle. By substituting the masses and distances into the formula, the overall center of mass can be calculated. This approach effectively combines the contributions of both the stone and the stick to determine the center of mass location.
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Homework Statement


alley club-ax consists of a symmetrical 8 kg stone attached to the end of a uniform 2.5 kg stick that is 98 cm long. How far is the center of mass from the angle end of the club-ax?


Homework Equations


Xcm = (sum) x*m / m


The Attempt at a Solution



I don't know how to approach this.. do i put the ax on a coordinate plane?
and it says that it is it is symmetrical and of uniform weight, does that give extra information that i should be using?
 
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Not sure which is the angle end, but say it is the handle. You can reverse it if not.

The handle is .98m and uniform density. You could take the integral of that to find the center of mass of just the handle, but I'm sure you know already it's in the middle for the handle part. To complete the sum of the moments for the whole shillelagh, then add the moment of the stone at the end.

So with ms the mass of the stone, mh the mass of the handle, and xh the center of mass of the handle,

Xcm = (xh*mh + L*ms)/(mh + ms)
 

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