# Reaction forces due to imbalance of a shaft

1. Apr 19, 2014

### zeralda21

1. The problem statement, all variables and given/known data

The shaft carries two particles of mass m, and rotates about the z-axis with the constant angular velocity $\omega$. Determine the x- and y-component of the bearing due to imbalance.

3. The attempt at a solution

Rotation only around the $z-$axis so $\omega _{x}=\omega _{y}=0$ and $\boldsymbol{\omega }=\omega \boldsymbol{\hat{k}}$. Since $\dot{\omega}=0$ we have:
$\left\{\begin{matrix} \sum M_{x}=I_{yz}\omega^2\\ \sum M_{y}=-I_{xz}\omega^2 \\ \end{matrix}\right.$

I have trouble now. Is it correct that for $M_{y}$, only forces acting perpendicular to y-axis that should be accounted for? We also see that the intersection of points of the $xz-$plane and the two sphere's are none so that means that $I_{xz}=0$(Is this even correct??). And $I_{yz}=\overline{I}_{yz}+md_{y}d_{z}=0+mR\frac{L}{3}-mR\frac{2L}{3}$

2. Apr 19, 2014

### SammyS

Staff Emeritus
I recommend starting with a Free Body Diagram of the shaft, considering the two particle masses to be external to the shaft.

3. Apr 19, 2014

### zeralda21

Alright. Did you read my attempt? Would like to know if I am all wrong or somehow correct.

The FBD contains the two particles, each with a gravitational force $mg$ acting in negative $x$. We have a parallel-plane motion with constant rotation in $z-$axis, hence the moment equations:

$\left\{\begin{matrix} \sum M_{x}=I_{yz}\omega^2\\ \sum M_{y}=-I_{xz}\omega^2 \\ \end{matrix}\right.$

For the moment equation about$y-$axis we need to determine $I_{yz}$.

$I_{yz}=\int yz {\mathrm{d} m}$ and from here I don't know how to proceed and establish the integral.

Since I don't know how to set up the integral I could bypass that by using

$I_{yz}=\overline{I}_{yz}+md_{y}d_{z}$ for each particle(or is it only for rigid bodies?), hence

$I_{yz}=0+mR\frac{L}{3}$ for particle 1 and $I_{yz}=0+m(-R)\frac{2L}{3}$ for the second.

Last edited: Apr 19, 2014