Calculating coordinates of extra balancing mass (https://imgur.com/a/vvW5Hkc)
Hello, I have unbalanced mechanism which center of rotation is not in line with center of its mass and I'm trying to calculate coordinates $x_1$ and $y_1$ of mass that would balance it.
I've tried to use forces and torques - there is centrifugal force and the Torque by it ($M \omega^2 \delta \sin a$, $\sin a = 0$) and Angular Force( $M \varepsilon \delta$ and non zero Torque - ($ M \varepsilon \delta$)
Also, $ \varepsilon $ is Angular Acceleration
So from my calculation:
$$ x_1 = 0 $$
$$ y_1 = \frac{M}{m} \delta $$
But i think the answer is not right and in this case balancing is not that easy.
Please help!
 A: The pseudo force components are:
$$\vec{F}_c=\left[ \begin {array}{c} -m \left( -{\omega}^{2}x-\epsilon\,y-2\,
\omega\,{\dot y} \right) \\ -m \left( -{\omega}^{2}y
+\epsilon\,x+2\,\omega\,{\dot x} \right) \\ 0
\end {array} \right] 
$$
where $\omega=\int \dot\epsilon\,dt+\omega_0$
pseudo force mass $M$ with $x=0\,,y=-\delta$
$$\vec F_M=\left[ \begin {array}{c} -M\epsilon\,\delta\\ M{
\omega}^{2}\delta\\ 0\end {array} \right]
$$
pseudo force mass $m$ with $x=x_1\,,y=y_1$
$$\vec F_m=\left[ \begin {array}{c} -m \left( {\omega}^{2}x_{{1}}-\epsilon\,y_{{
1}}+2\,\omega\,{\dot x}_{{1}} \right) \\ -m \left( {
\omega}^{2}y_{{1}}+\epsilon\,x_{{1}}+2\,\omega\,{\dot y}_{{1}}
 \right) \\ 0\end {array} \right] 
$$
$$ \begin{bmatrix}
   \ddot x_1 \\
   \ddot y_1 \\
   0 \\
 \end{bmatrix}=\vec F_m+\vec F_M\tag 1$$
assuming $\dot x_1=0\,,\dot y_1=0\,,\ddot x_1=0\,,\ddot y_1=0~$ and solve eq. (1)
for $x_1\,,y_1~$ you obtain your solution
$$x_1=0\,,~,y_1=\frac{M\,\delta}{m}$$
general case you must solve the ODE's eq. (1),to obtain the solutions $~x(t)\,,y(t)$
A: Using the parallel axis theorem,
$$ I_\mathrm{disk} = I_{CM} + M d^2 $$
the moment of inertia of the disk about the axis, assuming the mass is distributed uniformly in the disk, is
$$ I_{\mathrm{disk}} = \frac{1}{2} M r^2 + M \delta^2 $$
Let's treat the balancing mass as a point mass.
$$ I_m = m r_1^2 $$
where $r_1$ will follow the following relation:
$$ r_1^2 = x_1^2 + y_1^2 $$
For simplicity, pick $x_1 = 0$. (This doesn't hold if the mass of the disk is not uniformly distributed. Then you'll have to pick $x_1 = - x_\mathrm{CM, ~ disk}$.)
Now, for the mechanism to be balanced, so that the mechanism rotates at constant angular momentum $\omega$, the center of mass of the composite (disk + mass) system has to be located at the axis, that is, we want $r_\mathrm{CM, system} = (0, 0)$. The center of mass of the uniform disk is located at $(0, -\delta)$. Then we can find $y_1$ using the formula for center of mass:
$$ \begin{align}
y_\mathrm{CM, ~ system} &= \frac{ M y_\mathrm{CM, ~ disk} + m y_1}{ M + m } \\
(0) &= \frac{ M (-\delta) + m y_1 }{ M + m } \\
y_1 &= \frac{M}{m} \delta
\end{align} $$
So, OP's answer is already correct.
