- #1

- 18

- 0

## Homework Statement

[/B]

I don't understand what I'm doing wrong with this problem:

An old club-ax consists of a symmetrical 8.3 kg stone attached to the end of a uniform 2.9 kg stick. The stick is 98 cm long, and the 18 cm long stone is drilled through its center and mounted on one end of the stick. How far is the center of mass of the club-ax from the handle end of the club-ax?

## Homework Equations

X

_{CM}= (m

_{1}x

_{1}+m

_{2}x

_{2})/m

_{1}+m

_{2}[/B]

This is the only equation I need to use, I believe.

## The Attempt at a Solution

So what I have tried is as follows:

I set up a coordinate axis so that the origin is at the handle end of the club.

So by using the equation and the x-axis, the center of mass of the wooden stick part would be at (m

_{1}(0)+m

_{1}(80))/2m

_{1}

The m

_{1}'s cancel, leaving me with the CM at point (40,0). The m

_{1}'s are just points I imagined that lie on the tips of the stick, with equal masses because the stick is uniform. Then I just used the equation.

For the stone, I did a similar thing, except I changed the x-coordinates:

(m

_{2}(80)+m

_{2}(98))/2m

_{2}

This gave me the center of mass of the stone to be at point (89,0).

Combining these two equations, I get:

(m

_{1}(0)+m

_{1}(80)+m

_{2}(80)+m

_{2}98)/2m

_{1}+2m

_{2}

So m

_{1}= 2.9

and m

_{2}= 8.3

Plugging in and solving, I get 76.3125 as the x-coordinate of the center of mass of the system. Since my coordinate origin is at the tip of the handle, that means the center of mass is 76.3125 cm from the tip, right?

But the online homework says this value it is wrong. I also tried to put m

_{1}in terms of m

_{2}and vice versa, but those values I get are also wrong. (I have to round numbers when I do this though.)

What am I doing wrong? Any comments would be helpful.