Center of mass of club-axe with stone and stick

[Zhalfirin88]

Homework Statement

A club-axe consists of a symmetrical 16.0 kg stone attached to the end of a uniform 1.6 kg stick, as shown in the figure. The length of the handle is L1 = 71.0 cm and the length of the stone is L2 = 16.0 cm. How far is the center of mass from the handle end of the club-axe?

The Attempt at a Solution

$$\frac{(1.6*71)+(16*16)}{(16+1.6)}$$

= 21 cm

Now this means that the center of mass is at 21 cm, so to answer the question the center of mass away from the club-axe would be 50 cm?

 

[Gear300]

I don't see the diagram, but to make sure of things, it might be a good idea to differentiate between the center of mass along the number of dimensions you're taking, in which for your case, its probably 2. For symmetrical objects, the center of mass can be taken at the geometric center...so what you could do is take the two separate center of masses (rod and stone) and take the center of mass for the two-point system.

If what you have is right (along one dimension), then it is 50cm away from the handle.