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

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Homework Help Overview

The problem involves calculating the center of mass (COM) of a club-axe, which consists of a stone and a stick of specified weights and lengths. The original poster attempts to determine the distance from the handle end to the center of mass using a formula for COM.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the correct interpretation of the variables in the COM equation, particularly the positions of the center of mass for both the stick and the stone. There is a focus on the importance of using the correct coordinates for these masses.

Discussion Status

Some participants have provided clarifications regarding the setup of the problem, emphasizing the need to accurately define the positions of the center of mass for the components involved. There is an ongoing exploration of the implications of these definitions on the calculations.

Contextual Notes

There are indications of confusion regarding the lengths and positions used in the calculations, particularly with the handle's length being noted as unusually long. This may affect the interpretation of the problem setup.

michaeltozer13
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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??
 
Last edited by a moderator:
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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 !
 
michaeltozer13 said:
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.
 
michaeltozer13 said:
The length of the handle is L1=91.0m
That's a very long handle!
 

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