Solve Easy Center of Mass Homework for Club-Axe

[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

[tex] \frac{(1.6*71)+(16*16)}{(16+1.6)} [/tex]

= 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.

 

[tomcue]

I would like to commend you for your attempt at solving this problem using the formula for calculating center of mass. Your calculation and answer seem to be correct. However, I would like to suggest that you also consider the vertical positions of the two masses in your calculation. The center of mass is not just the average of the distances from the handle end, but also takes into account the distribution of mass along the length of the club-axe. It would be helpful to use the formula for center of mass in one dimension (x-coordinate) which is given by x_cm = (m1x1 + m2x2)/(m1 + m2), where m1 and m2 are the masses and x1 and x2 are their respective distances from a chosen reference point. In this case, the reference point could be the handle end of the club-axe. By considering the vertical positions of the masses, you can obtain a more accurate and precise answer for the center of mass. Keep up the good work!