# Centre of Mass by integration

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• October 22nd 2007, 09:40 PM
superphysics
Centre of Mass by integration
I've come across this question (which looks more like physics than maths, to be honest), and I really need some help...

Q. Find the centre of mass of a homogenous semi-circular plate. Given that $x_{cm}=\frac{1}{M}\int{x}\ dm$ and
$y_{cm}=\frac{1}{M}\int{y}\ dm$

where $x_{cm}$ and $y_{cm}$ refer to the centre of mass along the respective axis, $M$ is the mass of the plate, and $dm$ is the small change in mass.

Oh, by the way, if there's a type of solution apart from calculus (geometric or otherwise), I'd be glad to know of it...
• October 23rd 2007, 06:43 AM
earboth
Quote:

Originally Posted by superphysics
I've come across this question (which looks more like physics than maths, to be honest), and I really need some help...

Q. Find the centre of mass of a homogenous semi-circular plate. Given that $x_{cm}=\frac{1}{M}\int{x}\ dm$ and
$y_{cm}=\frac{1}{M}\int{y}\ dm$

where $x_{cm}$ and $y_{cm}$ refer to the centre of mass along the respective axis, $M$ is the mass of the plate, and $dm$ is the small change in mass.

Oh, by the way, if there's a type of solution apart from calculus (geometric or otherwise), I'd be glad to know of it...

Hello,

I don't know how to get the formula but I can give you possible way to get a valid solution:

The "centre of mass" of a plane area of homogenous material which is bounded by the x-axis, the curve of y = f(x), the straight lines $x = x_1$ and $x = x_2$ is the centroid of this area. The coordinates of the centroid are calculated by:

$x_c=\frac{\int_{x_1}^{x_2}x \cdot y dx}{\int_{x_1}^{x_2}y dx}$ and $y_c=\frac{\int_{x_1}^{x_2}y^2 dx}{2 \int_{x_1}^{x_2}ydx}$

$y = \sqrt{r^2-x^2}$ , that menas the center of the semi-circle is (0, 0), $x_1 = -r$, $x_2 = r$

Obviously $\int_{-r}^{r}ydx = \frac12 \cdot \pi \cdot r^2$

Plug in all values you know:

$x_c=\frac{\int_{-r}^{r}x \cdot \sqrt{r^2-x^2} dx}{\frac12 \cdot \pi \cdot r^2} = 0$ Use integration by substitution

$y_c=\frac{\int_{-r}^{r}(r^2-x^2)dx}{\pi \cdot r^2}=\frac{\frac43 r^3}{\pi \cdot r^2}=\frac{4}{3\pi}r$

Therefore the centre of mass has the coordinates $\left(0, \frac{4}{3\pi}r \right)$
• August 19th 2009, 07:52 AM
Karl Harder
Centre of mass by integration semi-circle with thickness
Supposing that you wanted to find the centre of mass of a semi-circle plate with radius A and this semi circle had a smaller semi-circle plate with radius a removed then would the centre of mass be equal to

$y_c=\frac{\int_{-A}^{A}(A^2-x^2)dx}{\pi \cdot A^2}-\frac{\int_{-a}^{a}(a^2-x^2)dx}{\pi \cdot a^2}=\frac{\frac43 a^3}{\pi \cdot a^2}=\frac{4}{3\pi}(A-a)$
• February 20th 2010, 02:51 PM
wolfhound
Quote:

Originally Posted by earboth
Hello,

I don't know how to get the formula but I can give you possible way to get a valid solution:

The "centre of mass" of a plane area of homogenous material which is bounded by the x-axis, the curve of y = f(x), the straight lines $x = x_1$ and $x = x_2$ is the centroid of this area. The coordinates of the centroid are calculated by:

$x_c=\frac{\int_{x_1}^{x_2}x \cdot y dx}{\int_{x_1}^{x_2}y dx}$ and $y_c=\frac{\int_{x_1}^{x_2}y^2 dx}{2 \int_{x_1}^{x_2}ydx}$

$y = \sqrt{r^2-x^2}$ , that menas the center of the semi-circle is (0, 0), $x_1 = -r$, $x_2 = r$

Obviously $\int_{-r}^{r}ydx = \frac12 \cdot \pi \cdot r^2$

Plug in all values you know:

$x_c=\frac{\int_{-r}^{r}x \cdot \sqrt{r^2-x^2} dx}{\frac12 \cdot \pi \cdot r^2} = 0$ Use integration by substitution

$y_c=\frac{\int_{-r}^{r}(r^2-x^2)dx}{\pi \cdot r^2}=\frac{\frac43 r^3}{\pi \cdot r^2}=\frac{4}{3\pi}r$

Therefore the centre of mass has the coordinates $\left(0, \frac{4}{3\pi}r \right)$

Hello ,I understand that the r^2 cancels with the r^3 to give r on the top, but where does the number 4 come from please?
• February 20th 2010, 03:11 PM
wolfhound
where does the number 4 come from?????????
• February 20th 2010, 03:40 PM
Aryth
Quote:

Originally Posted by wolfhound
where does the number 4 come from?????????

It comes from the integral in the numerator:

$\int_{-r}^r (r^2 - x^2)dx = [r^2x]_{-r}^r - \left[\frac{1}{3}x^3\right]_{-r}^r$

$= (r^3 + r^3) - \left(\frac{1}{3}r^3 + \frac{1}{3}r^3\right)$

$= 2r^3 - \frac{2}{3}r^3$

$= \frac{4}{3}r^3$
• February 20th 2010, 03:50 PM
wolfhound
Quote:

Originally Posted by Aryth
It comes from the integral in the numerator:

$\int_{-r}^r (r^2 - x^2)dx = [r^2x]_{-r}^r - \left[\frac{1}{3}x^3\right]_{-r}^r$

$= (r^3 + r^3) - \left(\frac{1}{3}r^3 + \frac{1}{3}r^3\right)$

$= 2r^3 - \frac{2}{3}r^3$

$= \frac{4}{3}r^3$

but why don't you get r^3/3 (on first line) when you integrate it?
• February 20th 2010, 03:51 PM
Aryth
Quote:

Originally Posted by wolfhound
but why don't you get r^3/3 (on first line) when you integrate it?

No, the integral is done with respect to x not respect to r. So r remains a constant in the integral.
• February 20th 2010, 03:56 PM
wolfhound
Quote:

Originally Posted by Aryth
No, the integral is done with respect to x not respect to r. So r remains a constant in the integral.

Ok, could you explain the 2nd line please, when you evaluate r and -r, how did you get this bit on the 2nd line. please
thanks
• February 20th 2010, 04:02 PM
Aryth
Quote:

Originally Posted by wolfhound
Ok, could you explain the 2nd line please, when you evaluate r and -r, how did you get this bit on the 2nd line. please
thanks

Alright.

$[r^2x]_{-r}^r - \left[\frac{1}{3}x^3\right]_{-r}^r$

We know that:

$[f(x)]_a^b = [f(b) - f(a)]$

So that:

$[r^2x]_{-r}^r = [(r^2)(r) - (r^2)(-r)] = (r^3 - (-r^3)) = 2r^3$

And:

$\left[\frac{1}{3}x^3\right]_{-r}^r = \left[\frac{1}{3}r^3 - \frac{1}{3}(-r)^3\right] = \left(\frac{1}{3}r^3 - \frac{1}{3}(-r)^3\right) = \frac{2}{3}r^3$
• February 20th 2010, 04:06 PM
wolfhound
Thanks for that
I understand now
• February 21st 2010, 07:05 AM
wolfhound
one more question
$y = \sqrt{r^2-x^2}$ , that menas the center of the semi-circle is (0, 0), $x_1 = -r$, $x_2 = r$

Obviously $\int_{-r}^{r}ydx = \frac12 \cdot \pi \cdot r^2$

if y = sqroot of r^2 - x^2
when its integrated how does it become 3.14xr^2
where does pie come from?
• February 21st 2010, 07:38 AM
skeeter
Quote:

Originally Posted by wolfhound
$y = \sqrt{r^2-x^2}$ , that menas the center of the semi-circle is (0, 0), $x_1 = -r$, $x_2 = r$

Obviously $\int_{-r}^{r}ydx = \frac12 \cdot \pi \cdot r^2$

if y = sqroot of r^2 - x^2
when its integrated how does it become 3.14xr^2
where does pie come from?

first of all ...

this is the lower case Greek letter pi :

http://dustyloft.files.wordpress.com...pg?w=174&h=184

this is pie (something to eat) :

http://scienceblogs.com/corpuscallos...umpkin_pie.jpg

this is where $\pi$ comes from in calculating the integral of the semicircle ...

$\int_{-r}^r \sqrt{r^2-x^2} \, dx$

$2\int_0^r \sqrt{r^2-x^2} \, dx$

let $x = r\sin{\theta}$

$dx = r\cos{\theta} \, d\theta$

$2\int_0^{\frac{\pi}{2}} \sqrt{r^2 - r^2\sin^2{\theta}} \cdot r\cos{\theta} \, d\theta$

$2r^2\int_0^{\frac{\pi}{2}} \sqrt{1 - \sin^2{\theta}} \cdot \cos{\theta} \, d\theta$

$2r^2\int_0^{\frac{\pi}{2}} \sqrt{\cos^2{\theta}} \cdot \cos{\theta} \, d\theta$

$2r^2\int_0^{\frac{\pi}{2}} \cos^2{\theta} \, d\theta$

$2r^2\int_0^{\frac{\pi}{2}} \frac{1 + \cos(2\theta)}{2} \, d\theta$

$r^2\int_0^{\frac{\pi}{2}} 1 + \cos(2\theta) \, d\theta$

$r^2\left[\theta +\frac{\sin(2\theta)}{2} \right]_0^{\frac{\pi}{2}}$

$r^2\left[\frac{\pi}{2} - 0 \right] = \frac{\pi r^2}{2}$
• February 21st 2010, 07:45 AM
wolfhound
I dont understand this
is this the easiest way?
how can I pluck values of sin and cos out of the sky?
I like apple pie but pi is making me angry............
• February 21st 2010, 08:07 AM
skeeter
Quote:

Originally Posted by wolfhound
I dont understand this
is this the easiest way?
how can I pluck values of sin and cos out of the sky?
I like apple pie but pi is making me angry............

other than recognizing that the area of a semicircle is $\frac{\pi r^2}{2}$ , no.

the method of integration is called trig substitution ... not "plucking" values of sine and cosine out of the sky.

Trigonometric substitution - Wikipedia, the free encyclopedia
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