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M3 statics of rigid bodies

Show that the centre of mass of a uniform semicircular lamina lies on its axis of symmetry at a distance of 4r/3pi from the bounding diameter.

I have shown this using integration, however:

I am using the standard equations for calculating the coordinates of the centre of mass of this lamina. I have considered the elementary strip of width dx. I have shown that the area of the strip is 2ydx. where y=sqrt(r^2 - x^2). In doing so I have managed to calculate the x-coordinate of the centre of mass. Now due to the symmetry I have located the y-coordinate without using the equation, i.e. y=0

But When I try and use integration to calculate the y-coordinate I donot manage to obtain zero?

Please help

I donot understand why, I cannot use the equation My(bar)=int(r,0)py^2dx

this is the typical equation. I can use something similar to find xbar: the Mx=int(r,0)x(2p)sqrt(r^2-x^2) dx

where y=sqrt(r^2 - x^2)

For all the other questions this reasoning seems to work, however for the hemisphere it does not- why?
Original post by sulexk

I donot understand why, I cannot use the equation My(bar)=int(r,0)py^2dx

this is the typical equation.


I don't know where that equation has come from. If you're trying to find y-bar shouldn't you be taking elementary strips parallel to the x-axis and integratnig wrt y.

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