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Finding center of mass

  • Thread starter uriwolln
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  • #1
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Hey all, small question,

a mass is distributed along a rod by the function M(X)= bx - a(x^2). The length of the rod is 3b/4a. What is the center of mass.
Ive tried working it out by using integrals to find the the total mass, and then divide it by 2, and then try to find out the x for that. The equation I used is
l= integral{ sqrt (1+ (dy/dx)^2) dx }
it came out with cosh and things like that, and I just can not find the x.

Help?
 

Answers and Replies

  • #2
gneill
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On a thin rod, the center of mass would be the location where the mass from that point to one end of the rod is the same as the mass from that point to the other end. So set up two integrals to represent the masses of the two rod sections and equate them.
 
  • #3
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yeah, that's the thing. I've tried that, and here what it comes to:

2(b-2ax)cosh(sinh^-1 (b-2ax)) + 2(sinh^-1(b-2ax)) =
(-b/2)(cosh(sinh^-1 (-b/2)) + sinh^-1(-b/2) + bcosh(sinh^-1 (b)) + sinh^-1 (b)

How can I get the x of out there? there is just no way....

help?
 
  • #4
gneill
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Does the given function, M(X)= bx - a(x2), represent a linear density function or the total mass of the rod from the origin (x = 0) up to distance x along the rod?
 
  • #5
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i believe its linear density function
 
  • #6
gneill
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i believe its linear density function
Okay, if that's the case then may I suggest that you declare a new variable, L, to be the total length of the rod. The length of the rod is given as 3b/4a, so L is equal to that.

Next make a variable substitution, replacing b with (4/3)aL. Why? You'll see in a moment!

Next, find the total mass of the rod. That would be an integral from 0 to L of the expression you just found. What do you get?
 
  • #7
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sorry gneil,
I do not understand.
To which equation do I do the integral?
Is it to
(4/3)aL. If I do that, I just get (4/3)aL^2 which I do not see how it fits.
or for the integral{sqrt(1+(dy/dx)^2)dx} ?
 
  • #8
gneill
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If the linear density of the rod is ρ(x) = bx - ax2, then the making the substitution that I suggested yields

ρ(x) = (4/3) a L x - a x2

where L = 3b/4a

Then the mass of the rod is

[tex] m_{rod} = \int_0^L \left(\frac{4}{3} a L x - a x^2 \right) dx [/tex]
 
  • #9
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OK!
so now the total mass is (1/3)aL^2.

So now I use the regular equation for Center of Mass:

1/M integral (0 to L) {x dm dx}

right?
 
  • #10
gneill
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Sure. Or just set

[tex]
\int_0^{x_{cm}} \left(\frac{4}{3} a L x - a x^2 \right) dx = \frac{1}{2} \left(\frac{1}{3} a L^2 \right)
[/tex]
 
  • #11
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Thanks!!
 
  • #12
gneill
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Thanks!!
Don't thank me yet! You're still facing a cubic to solve! :smile:
 
  • #13
SammyS
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[tex]x_{CM}=\frac{\int_0^Lx\,\lambda(x)\,dx}{\int_0^L\lambda(x)\,dx}\,,[/tex] where λ(x) is the linear mass density function.
 
  • #14
SammyS
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On a thin rod, the center of mass would be the location where the mass from that point to one end of the rod is the same as the mass from that point to the other end.
I'm quite sure that this is incorrect.
 
  • #15
gneill
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Ha! That makes it easy indeed! Nice call.
 
  • #16
gneill
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I'm quite sure that this is incorrect.
Hi Sammy. You're absolutely right. I don't know what I was thinking! I guess I've grown used to always seeing uniform rods. No excuse, though. Thanks for catching that.

To uriwolln, I apologize for the slip and (clearly) wasting your time down an incorrect path. Sammy's equation will get you to a solution quickly and cleanly.

-g
 
  • #17
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gneil,
The equations in my point of view are much the same, though I did need the extra push for the replacement part of b.

Though now, you two got me confused a bit. Why, on a rod, the center of mass will not be at the point where, theortically, on the left side there is the same mass on the right side?
 
  • #18
gneill
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gneil,
The equations in my point of view are much the same, though I did need the extra push for the replacement part of b.

Though now, you two got me confused a bit. Why, on a rod, the center of mass will not be at the point where, theortically, on the left side there is the same mass on the right side?
It will work that way for a uniform rod, where the mass is evenly distributed.

Think of the case where the "rod" consists of two discrete masses, M1 and M2, separated by some distance D. Definitely a non-uniform mass distribution!. Now, no matter where you place yourself on the line between M1 and M2 you can't "balance" the amount of mass from one side to the other because no mass "changes sides" as you move.

However, by defining the center of mass in terms of the geometrical distribution of mass, as the formula the SammyS does, a consistent picture emerges. It's the average of their mass-weighted positions.

Again, sorry for the earlier fumble.
 
  • #19
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I see what you mean. When there are two masses connected by a weightless rod kinda thing?
But in that case, its just theoretical right?
I mean, I do get that the correct way to think of it as the average mass-weighted positions.

Thanks again
 
  • #20
gneill
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I see what you mean. When there are two masses connected by a weightless rod kinda thing?
But in that case, its just theoretical right?
That was just an example (one which is pretty good for, say, a moon orbiting a planet!), but it is true in general. One could concoct all sorts of mass distributions where the same issue would arise.

I mean, I do get that the correct way to think of it as the average mass-weighted positions.

Thanks again
Yes. The formula that SammyS provided is way to go. It's basically the definition of center of mass.
 
  • #21
SammyS
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Hi Sammy. You're absolutely right. I don't know what I was thinking! I guess I've grown used to always seeing uniform rods. No excuse, though. Thanks for catching that.

-g
g,

I've made more than a few mis-statements myself. I've read lots of your very helpful replies on this site and very much admire your patience and respect your physics knowledge.

You don't know how many times I went over this before finally making my second post. I was sure you would catch it yourself or that I was overlooking something.

Thanks for the gracious response.

Sam
 

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