Thread: Centre of Mass by integration

1. 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...

2. 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 $\left(0, \frac{4}{3\pi}r \right)$

3. 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

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

4. 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?

where does the number 4 come from?????????

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

It comes from the integral in the numerator:

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

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

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

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

7. 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?

8. 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.

9. 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

10. 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:

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

11. Thanks for that
I understand now

12. 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?

13. 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 :

this is pie (something to eat) :

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 $x = r\sin{\theta}$

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

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

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

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

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

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

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

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

$r^2\left[\frac{\pi}{2} - 0 \right] = \frac{\pi r^2}{2}$

14. 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............

15. 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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