Problem set 7 3C-4 asks us to find the center of mass of a sector of given angle. Can anyone explain why it isn't possible to find the 'polar' centre of mass by simply integrating r*rdrdθ? I get that it won't come out dependent on θ in this case, but I don't understand why we can't find COM in polar coordinates as we can in rectangular... I'm fond of intuitive geometrical explanations but I can't think of one!

Question

phi · Answer

what makes you think you can't?

deltanu · Answer

Because when we naively try to use the equation $$r _{cm}=\int\limits_{0}^{2\alpha}\int\limits_{0}^{a}\delta\ r*rdrd \theta $$
we get a nonsense answer of 2a/3. I'm just wondering why it is we can't do this in the same way that we found xcm...

baru · Answer

the formula for $x_{cm}$ is derived by balancing moments generated by masses about an axis parallel to the y axis

baru · Answer

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baru · Answer

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baru · Answer

we want to find $x_cm$ such that $m_1r_1+m_2r_2=0$
$$m_1(x_{cm}-x_1) +m_2(x_{cm}-x_2)=0$$

baru · Answer

$$(m_1+m_2)x_{cm}=m_1x_1+m_2x_2$$
$$x_{cm}=\frac{m_1x_1+m_2x_2}{(m_1+m_2)}$$

baru · Answer

its the same derivation which can be extended to continuously distributed mass...
if you replace x with r or theta, it just dosent make sense

baru · Answer

because the idea of balancing moments is lost

deltanu · Answer

Great explanation. That really clears things up! Definitely changed the way I think about centre of mass. The concept of moments is key to the whole definition... Thanks for the help

baru · Answer

no prob :)

baru · Answer

just to clarify: you can still switch to polar and find $x_{cm}$ (not $r_{cm}$)
$$x_{cm}=\int\limits\limits\int\limits\limits_{R}\delta\ rcos \theta*rdrd \theta$$