Finding the center of mass of a system

  • #1
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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


XCM = (m1x1+m2x2)/m1+m2[/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 (m1(0)+m1(80))/2m1
The m1's cancel, leaving me with the CM at point (40,0). The m1'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:
(m2(80)+m2(98))/2m2
This gave me the center of mass of the stone to be at point (89,0).

Combining these two equations, I get:
(m1(0)+m1(80)+m2(80)+m298)/2m1+2m2

So m1 = 2.9
and m2 = 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 m1 in terms of m2 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.
 

Answers and Replies

  • #2
haruspex
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the center of mass of the wooden stick part would be at (m1(0)+m1(80))/2m1
The stick is 98cm long. Why subtract the length of the stone?
 
  • #3
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Haruspex has pointed out the problem that you are having.

You also seem to be over complicating the equation. You already calculated the COM of the two individual components, so you can just use the following with the equation you originally wrote:

m1 = 2.9
m2 = 8.3
x1 = ??
x2 = 89

Then plug into:
XCM = (m1x1+m2x2)/m1+m2
 
  • #4
Simon Bridge
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I have an additional concern:
The m1'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.
... or you could just look at the stick and use your understanding of what "center of mass" means.
i.e. if you had to pick it up (just the stick) where should you grab it?

These exercises are supposed to reward understanding the subject - so if you just apply equations and formulae you will be at a disadvantage.
 
  • #5
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Ooooh. That's an embarrassing mistake. I knew where the center of mass would be, because the stick is uniform. But I thought you were supposed to subtract the overlapping length. Thanks all.
 
  • #6
Simon Bridge
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If you subtract the overlapping length, then you have to include that mass with the mass of the head or something.
Don't worry, everyone makes these mistakes sometime.
 

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