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Reaction forces due to imbalance of a shaft

  1. Apr 19, 2014 #1
    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.


    QLfdH8V.jpg

    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. jcsd
  3. Apr 19, 2014 #2

    SammyS

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    I recommend starting with a Free Body Diagram of the shaft, considering the two particle masses to be external to the shaft.
     
  4. Apr 19, 2014 #3
    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
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