How far is the center of mass from the handle end of the club-axe?

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SUMMARY

The center of mass of a club-axe, consisting of an 11.0 kg stone and a 2.9 kg stick, is calculated to be 87.26 cm from the handle end. The calculations involve determining the moments created by the weights of the stone and the stick, with the stone's weight significantly influencing the balance point due to its greater mass. The final formula used was (46.5 x 2.9 + 98 x 11) / (2.9 + 11), leading to the accepted answer of 87.26 cm from the end of the stick.

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bkhofmann
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Homework Statement



A club-axe consists of a symmetrical 11.0 kg stone attached to the end of a uniform 2.9 kg stick, as shown in the figure. The length of the handle is L1 = 93.0 cm and the length of the stone is L2 = 10.0 cm. How far is the center of mass from the handle end of the club-axe?

Homework Equations





3. The Attempt at a Solution [/b
I'm not sure how to do this, I can up with 58.9cm on my own but most likely not the correct way, any help appreciated!
 
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bkhofmann said:

Homework Statement



A club-axe consists of a symmetrical 11.0 kg stone attached to the end of a uniform 2.9 kg stick, as shown in the figure. The length of the handle is L1 = 93.0 cm and the length of the stone is L2 = 10.0 cm. How far is the center of mass from the handle end of the club-axe?

Homework Equations



3. The Attempt at a Solution [/b
I'm not sure how to do this, I can up with 58.9cm on my own but most likely not the correct way, any help appreciated!


How much do they weigh together?
Won't half the weight be on each side at the balance point?

Starting from the end with the stone, how much does the weight of the stone increase the weight to it's side of the balance point on a per cm basis? If the stick passes through the stone, how much does the stick add per cm? Now all you have to do is divide that into half the total weight to give you where it balances.
 
I believe that's what I have done. Do you come up with 58.9 cm?
 
bkhofmann said:
I believe that's what I have done. Do you come up with 58.9 cm?

How is that answer possible?

The balance point must lie within the stone as it is 4 times heavier than the whole stick.
 
Ok the stone weighs 1.1kg/cm and the stick is 1.1 kg/35.279cm. Is this what you are talking about?
 
bkhofmann said:
Ok the stone weighs 1.1kg/cm and the stick is 1.1 kg/35.279cm. Is this what you are talking about?

So no, you should deal with the center of mass of the stone.

Where is the stone? Added to the end? Or is the stick through the middle?
 
Last edited:
I got it (46.5x2.9+98x11)/(2.9+11), the stone is just added on.
 
bkhofmann said:
I got it (46.5x2.9+98x11)/(2.9+11), the stone is just added on.

OK. Sorry, my first post was a little misleading.

What you are balancing then is something on the left (the stone end) that looks like

1.1*(10 - x)/2 the center of mass times the moment arm that equals
1.1*x/2 +2.9*(x+46.5)

So the total from the handle end is 93 + x.
 
the center of mass is at 87.26cm from the end of the stick and lon-capa accepted it.
 
  • #10
bkhofmann said:
the center of mass is at 87.26cm from the end of the stick and lon-capa accepted it.

OK. So the stick goes to the end of the stone. Not the stone at the end of the stick.

Sounds about right then.
 

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