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Mass of chain in shape of parabola

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Homework Statement

A chain in the shape of [tex]y = x^{2}[/tex] between x = -1 and x = 1, has density of |x|. Find M, and CM.



Homework Equations





The Attempt at a Solution



[tex]\int^{1}_{-1}|x|dx = \int^{0}_{-1}-xdx + \int^{1}_{0}xdx = 1[/tex]

I got this far and realized that I did nothing with the function itself. :redface:

The formulas I've used before deal with masses bounded by curves, not the mass of the curves themselves.
 

Answers and Replies

  • #2
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Does this help? are you ok with finding the length, THEN using the weight of chain per foot, finding total weight?

http://www.descarta2d.com/BookHTML/Chapters/arclen.html#section_63

Good Luck,
LarryR : )


Homework Statement

A chain in the shape of [tex]y = x^{2}[/tex] between x = -1 and x = 1, has density of |x|. Find M, and CM.



Homework Equations





The Attempt at a Solution



[tex]\int^{1}_{-1}|x|dx = \int^{0}_{-1}-xdx + \int^{1}_{0}xdx = 1[/tex]

I got this far and realized that I did nothing with the function itself. :redface:

The formulas I've used before deal with masses bounded by curves, not the mass of the curves themselves.
 
  • #3
4,239
1
I got this far and realized that I did nothing with the function itself. :redface:
No, no. You're on the right tack. Notice that the density of the chain is independent of y, so you don't need to know the function to get the mass.
 
  • #4
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So was I right the first time? How do I find the center of mass?

I redid the problem using

[tex]\int^{1}_{-1} \rho\sqrt{1+(y')^{2}} dx[/tex]

I then threw an x in there to find the moment and the integral became really nasty.

Intuition tells me that it would be at 0,0 because the function and the density function are both symmetric across the y axis so both sides should weigh the same. Right?
 
  • #5
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Yes, you got it right the first time; the mass of the chain is 1.

Use the symmetry of the problem to help solve for the center of mass. By symmetry, the center of mass in the x direction is at y-axis. The center of mass for each leg, in the y direction will be the same for each leg so you only need concern yourself with one leg of the parabola.

You are given rho as a function of x, what is rho as a function of y for the first quadrant? Integrate the mass times displacement from y=0 to y=1.
 
  • #6
gabbagabbahey
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Yes, you got it right the first time; the mass of the chain is 1.
No, the infinitesimal length element of the parabola is not just [itex]dx[/itex]. To get the mass, you need to integrate the density over the length of the parabola.
 
Last edited:
  • #7
gabbagabbahey
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I redid the problem using

[tex]\int^{1}_{-1} \rho\sqrt{1+(y')^{2}} dx[/tex]

I then threw an x in there to find the moment and the integral became really nasty.
This is the correct method. Break the moment integral into two intervals and use the substitution [itex]x\to-x[/itex] on the negative interval.

Intuition tells me that it would be at 0,0 because the function and the density function are both symmetric across the y axis so both sides should weigh the same. Right?
You are half correct, the x-coordinate of the CoM will be zero, because the density at the chain are symmetric about the y-axis. The y-coordinatee need not be zero, and your intuition should tell you that it will be positive since most of the chain is at y>0 and its density is always positive. You can verify this by evaluating both the x and y moment intervals.
 
  • #8
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No, the infinitesimal length element of the parabola is not just [itex]dx[/itex]. To get the mass, you need to integrate the density over the length of the parabola.
reread the problem statement
 
  • #9
gabbagabbahey
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reread the problem statement
I did, and my comment still stands. What makes you think otherwise?
 
  • #10
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I did, and my comment still stands. What makes you think otherwise?
I took my own advise and reread the problem statement. I been more than just unhelpful, spreading confusion. I'm going to go attend to my depleted ego now.
 
  • #11
gabbagabbahey
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I took my own advise and reread the problem statement. I been more than just unhelpful, spreading confusion. I'm going to go attend to my depleted ego now.
Lol, we all make mistakes. I know I've made my fair share.
 

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