How Do You Calculate the Center of Mass for a Club-Ax?

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SUMMARY

The calculation of the center of mass for an Alley club-ax involves a symmetrical 8 kg stone and a uniform 2.5 kg stick. The correct formula to use is x(cm) = (m1x1 + m2x2) / (m1 + m2), where m1 is the mass of the stone, m2 is the mass of the stick, x1 is the center of mass of the stick, and x2 is the center of mass of the stone. The final result for the center of mass is 77.3 cm from the handle's end, determined by correctly defining the coordinate system and applying the masses and positions accurately.

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  • Knowledge of coordinate systems in physics
  • Ability to perform arithmetic operations with units
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Homework Statement


"Alley club-ax consists of a symmetrical 8 kg
stone that is 18cm long, and is attached to the end of a uniform 2.5 kg
stick that is 98 cm long."

Homework Equations


cm = centre of mass

x(cm) = x1m1 + m2x2 / (m1+m2)


The Attempt at a Solution


m1=8
m2=2.5

x1=9
x2=40

This seems like a very straightforward question, but I can't seem to get it right... the answer is 77.3cm from the handle's end, and I keep on getting 18.52cm which is WAY off!
 
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Formula is good, but the data you have put in it isn't. You say that the solution is 77.3 cm from the handle's end. Handle length is 80 cm and length of a stone is 18.
First, you have to define the coordinate system. In this case, we'll say that coordinate system starts from the handle's end. That means that the handle's CM is located at 40 cm (xh=40cm). CM of the stone is at 89 cm because coordinate system starts from the handle's end, so you have to add handle's length which is 80 cm plus 9 cm from the stone (xs=89cm).

xm=(mhxh+msxs)/(mh+ms)=(2.5*40+8*89)/(2.5+8)=77,3333cm
 

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