Centre of mass integrals

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  • #1
T7
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Hi,

I appreciate that we can calculate the x-coordinate of the centre of mass of a system of bodies like this:

[tex]x = (m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + ... + m_{n}x_{n}) / M[/tex]

where M is the total mass.

This could be rewritten as

[tex] x = \frac {1} {M} \sum_{i=1}^n m_{i}x_{i} [/tex]

My next step would be to say that

[tex] x = \frac {1} {M} \int_{a}^{b} m dx[/tex]

(I am imagining a scanner sweeping horizontally through an object X and recording its mass at very small intervals).

However, looking in the textbook, it is expressed as

[tex] x = \frac {1} {M} \int x dm[/tex]

I'm a bit of a beginner as far as calculus goes, but it seems to me that[tex] \int x dm[/tex] is not the same as [tex]\int m dx[/tex].

Looking at the problem geometrically, if you plot m against x, then [tex]\int_{a}^{b} m dx[/tex] will give you an area between the graph of m and the x-axis, whereas [tex]\int_{a}^{b} x dm[/tex] will give you an area between the graph of m and the y-axis.
 
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Answers and Replies

  • #2
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If you imagine a "scanner," as you put it, recording the center of mass (a position!) of each slice, and you add those up, then you get an integral of positions with respect to the mass of the arbitrarily small slice, not the other way around.


If you're imagining a bunch of thin slices, and recording the mass at each point, then the mass of each slice is [itex]dm[/itex], not m (or [itex]dx[/itex]). A solid object can be treated as a continuous distribution of particles, but each is so small that it has the differential mass dm. So, the sums ([itex]m_ax_a+m_b+x_b+...+m_nx_n[/itex]) become [itex]x_adm_a+x_bdm_b+...+x_ndm_n[/itex], or the integral [tex]\int xdm[/tex]
 
  • #3
T7
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Cheers scholzie.
 
  • #4
HallsofIvy
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T7 said:
Hi,

I appreciate that we can calculate the x-coordinate of the centre of mass of a system of bodies like this:

[tex]x = (m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + ... + m_{n}x_{n}) / M[/tex]

where M is the total mass.

This could be rewritten as

[tex] x = \frac {1} {M} \sum_{i=1}^n m_{i}x_{i} [/tex]

My next step would be to say that

[tex] x = \frac {1} {M} \int_{a}^{b} m dx[/tex]

(I am imagining a scanner sweeping horizontally through an object X and recording its mass at very small intervals).
But xi is a coordinate- it is not a "small piece" of something. mi is the mass of a small section of solid centered at xi. Perhaps it would be better to write it [itex]\Delta m_i[/itex]. That would make it clearer that you are adding small pieces of mass, each multiplied by the coordinate of its center and so
However, looking in the textbook, it is expressed as

[tex] x = \frac {1} {M} \int x dm[/tex]
is the correct formula.
 
  • #5
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centre of mass

Sketch the region which is bounded by the graps y =x^2+1 of , x = 0, x = 1 and y-0 . Find the coordinates of its centre of mass.

pls help on how to find the centre of mass
 
  • #6
arildno
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Ideas, please?
 
  • #7
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Teng, did you sketch the region ?

marlon
 
  • #8
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what is the formula of finding the centre of mass??is it same as finding centroid of a region??
 
  • #9
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yaya.now sketching
 
  • #10
arildno
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teng125 said:
what is the formula of finding the centre of mass??is it same as finding centroid of a region??
That is correct, unless you have been given a non-constant density function.
 
  • #11
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okok.....thanx
 

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