Center of Mass for 8 kg Stone & 2.5 kg Stick

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SUMMARY

The center of mass for a system consisting of an 8 kg stone and a 2.5 kg stick, where the stick is 98 cm long, can be calculated using the formula Xcm = (xh*mh + L*ms)/(mh + ms). In this equation, xh represents the center of mass of the stick, L is the length of the stick, ms is the mass of the stone, and mh is the mass of the stick. The center of mass of the uniform stick is located at its midpoint, which is 49 cm from the angle end, while the stone is positioned at the end of the stick.

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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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