Center of Mass of a Hemiwasher (Y Value)

[Seda]

I have two circles centered at the origin, one with radius A and the other with radius B. Only looking at the hemiwasher above the X axis, what values of A and B will locate the Y value of the center of Mass within the region of the washer itself (not the open space.)

I have calculated this (density is constant) as the Y value of the center of mass in terms of A and B, but I do I know which values of A and B will locate the center of mass between B and A? (I have B as the smaller Radius.)

Y= (2(A^3-B^3))/(3pi(A^2-B^2))

 

[Seda]

Change that to Y= (4(A^3-B^3))/(3pi(A^2-B^2))

 

[Seda]

Simplfied my equation a little by factoring A-B out of my quiotient

:Y= (4(A^2+AB+B^2))/(3pi(A+B))

Please Help!

 

[soroban]

Hello, Seda!

\(\displaystyle Y\:=\:\frac{4(A^2\,+\,AB\,+\,B^2}{3\pi(A\,+\,B)}\)

We want this to be greater than \(\displaystyle B.\)

. . \(\displaystyle \frac{4(A^2\,+\,AB\,+\,B^2}{3\pi(A\,+\,B)} \;>\;B\)

. . \(\displaystyle 4(A^2\,+\,AB\,+\,B^2} \;> \:3\pi B(A\,+\,B)\)

. . \(\displaystyle 4A^2\,+\,4AB\,+\,4B^2\;>\;3\pi AB \,+\,3\pi B^2\)

. . \(\displaystyle 4A^2\,+\,4AB\,-\,3\pi AB \,+\,4B^2\,-\,3\pi B^2\;>\;0\)

. . \(\displaystyle 4A^2\,-\,(3\pi\,-\,4)AB\,-\,(3\pi\,-\,4)B^2\;>\;0\)

. . \(\displaystyle (3\pi\,-\,4)B^2\,+\,(3\pi\,-\,4)AB\,-\,4A^2 \;< \;0\)


Change the inequality to an equation and apply the Quadratic Formula:

. . \(\displaystyle B \;=\;\frac{-(3\pi\,-\,4)A\,\pm\,\sqrt{(3\pi\,-\,4)^2A^2 \,-\,4(3\pi\,-\,4)(-4A^2)}}{2(3\pi\,-\,4)}\)

And we have: \(\displaystyle \:B \;=\;\left[\frac{-(3\pi\,-\,4)\,\pm\,\sqrt{(3\pi\,-\,4)(3\pi\,+\,12)}}{2(3\pi\,-\,4)}\right]\cdot A\)


For positive \(\displaystyle B\), we have: \(\displaystyle \:B\;<\;\frac{-(3\pi\,-\,4)\,+\,\sqrt{(3\pi\,-\,4)(3\pi\,+\,12)}}{2(3\pi\,-\,4)}\cdot A\)

. . Therefore: \(\displaystyle \:B \;<\;(0.493658577)A\)