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## Homework Statement

Find the centre of mass of a semi-circle

## Homework Equations

##y_{cm}=\frac{1}{M} \int y dm ##

## The Attempt at a Solution

So ## y= R cos \theta ## where theta is measured from the vertical, and the base of the semi-circle is along the horizontal

Now apparently from here you can change coordinated to polar coordinates and replace ##dm## with ## \rho r dr d \theta## to obtain the correct answer of ##\frac {4}{3 \pi}## But I'm confused as to why you can't also replace ##dm## with ##2 \rho R sin \theta R d \theta## to split to 'dm' segment into rows perpendicular to the horizontal base?

Many thanks :)