mass centre of a half sinus?

[Kedas]

Hi,

 

Does someone know where the mass centre of a half sinus period is located?

I know pi/2 on the x-axes but on the y-axes ?

 

to give an idea for a half circle it's located at R*4/(3*pi) that is 0.42 for R=1

[bloodhound]

whats a half sinus period?

[Kedas]

whats a half sinus period?

 

I mean 0 to pi , only the first half of one full sinus.

 

consider a plate that has that shape, at which point would I have to hang it so it would stay flat. (centre of mass)

[bloodhound]

hmm... i still dont know what a sinus is..

 

the centre of mass will lie on the line of symmetry. if u know one condition on the position of centre of mass , and u know the line of symmetry , then u should know the point itself.

[Kedas]

hmm... i still dont know what a sinus is..

 

the centre of mass will lie on the line of symmetry. if u know one condition on the position of centre of mass ' date=' and u know the line of symmetry , then u should know the point itself.[/quote']

 

we clearly have a communication problem y=sin(x) <--a sinus

http://www.jazzalmanach.de/images/musikalswissenschaft/sinus.jpg

 

(I want to calculate te potential energy of a water wave.)

[bloodhound]

well i can't remember the formula..

 

just find the formula for the centre of mass for a lamina.

[bloodhound]

http://mathworld.wolfram.com/GeometricCentroid.html

 

just take the density fucntion to be constant.

 

i remember doing this is my mechanics a level. and i remember not having anything to do with double integrals... but have a go anyway

[Kedas]

http://mathworld.wolfram.com/GeometricCentroid.html

 

just take the density fucntion to be constant.

 

i remember doing this is my mechanics a level. and i remember not having anything to do with double integrals... but have a go anyway

 

I know the formula but I was hoping someone already did the double integrals and simplefied it. (I expect something simple)

 

btw I got integral ((y+cos(Z)) * Z * tg(Z)).dZ (Z from pi/2 to 0, with y the constant I'm looking for)

but I serioulsy doubt it's correct, since I made it

[bloodhound]

well i just computed it.. i got the [math]\bar{y}=\frac{\pi}{8}[/math]

[Kedas]

well i just computed it.. i got the [math']\bar{y}=\frac{\pi}{8}[/math]

 

This sure can be correct

and which program did you use for it?

[bloodhound]

if the density function is constant then the formula can be simplified to

 

[math]\bar{y}=\frac{\frac{1}{2}\int_{0}^{\pi}y^2}{M}[/math]

 

where M is just the area under the curve

[bloodhound]

This sure can be correct

and which program did you use for it?

 

but of course i did it by hand...

 

 

 

in the above simplified formula the the bounds in the integral only works for this example!! dont go using it for other laminas as well. the general formula is the same except you have to sort out the limits.

[Kedas]

M the surface is 2 but I fail to see the simplifying to y² = sin²(x)

 

anyway it looks OK so thanks for your help.

my math is failing me.

[bloodhound]

i forgot the add that the integral is with respect to x....

 

anyway this is how you do it

 

you find M easily giving its value 2.

 

then u have to find [math]\int \int_{A}y dA[/math] where A is the area bounded by the sin curve between 0 and pi and the horizontal axis.

 

writing that as interated integral you get

[math]\int_{0}^{\pi} \int_{0}^{\sin(x)}y dy dx[/math]

[math]=\int_{0}^{\pi}[\frac{y^2}{2}]_{0}^{\sin(x)} dx[/math]

[math]=\frac{1}{2}\int_{0}^{\pi}\sin^2(x) dx[/math]

[math]=\frac{1}{2}\frac{\pi}{2}[/math]

 

dividing by M u get . [math]\bar{y}=\frac{\pi}{8}[/math]

 

actually there is the constant density function in both the M integral and the above one... but since they are constants, you can take them out the integral in both of them and they will just cancel out.

[Kedas]

Thanks for the full explanation I'm sure that I can figure it out for myself next time