Center of Mass Calculation for a Club-Axe: How Far from the Handle is the COM?

[michaeltozer13]

Homework Statement

A club-axe consists of a symmetrical 23.0 kg stone attached to the end of a uniform 2.8 kg stick. The length of the handle is L1=91.0m and the length of the stone is L2=13.0cm. How far is the center of mass from the handle end of the club?

Homework Equations

Center of mass equation: COM=x1m1+x2m2/(m1+m2)

The Attempt at a Solution

so for my attempt i kept the lengths in cm, and my equation looks like this COM=(91.0x2.8)+(104x23.0)/(2.8+23.0).

The answer I am getting is 101.59 cm, which is wrong. The correct answer is 91.9cm. Any help??

 

[BvU]

x₁ isn't the length of the handle: it's the position of the center of gravity of the handle !
Likewise, x₂ isn't the end of the stone but the x coordinate of the center of mass of the stone.

I do hope the glue at the end of the handle is strong enough :)

welcome to the world of PF !

 

[haruspex]

COM=(91.0x2.8)+(104x23.0)/(2.8+23.0).

I assume you mean ((91.0x2.8)+(104x23.0))/(2.8+23.0) (parentheses matter!)
That would be right if all of the mass of the stick were at 91cm from the end, all the mass of the stone at 104cm from the end. But they're not.

 

[Mark44]

The length of the handle is L1=91.0m

That's a very long handle!