Homework Help: 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