1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Angular momentum of a solid cylinder rotating around axis

  1. Apr 17, 2014 #1
    1. The problem statement, all variables and given/known data

    I'll provide picture for clearer understanding. The solid cylinder of mass ##m## and radius ##r## revolves about its geometric axis at an angular rate ##p## rad/s. Simultaneously, the bracket and shaft revolve about x-axis at the rate ##\omega## rad/s. Determine the angular momentum about O.

    WXS1ivg.jpg

    3. The attempt at a solution

    Due to symmetry it follows that ##I_{xy}=I_{xz}=I_{yz}=0## so the angular momentum reduces to ##\boldsymbol{L_{O}}=I_{xx}\omega \boldsymbol{\hat{i}}+I_{yy}p\boldsymbol{\hat{j}}##.

    Further thoughts: If the ##mh^2## is removed, does not that calculate the rotation of the cylinder IF the geometrical axis coincide with the ##y##-axis ??

    I have the calculated ##I_{xx}## correctly but ##I_{yy}## is incorrect.

    ##I_{yy}=\overline{I}_{yy}+md^2=\frac{1}{2}mr^2+mh^2## but there shound NOT be an ##mh^2## ....and I want to understand why
     
  2. jcsd
  3. Apr 17, 2014 #2

    BiGyElLoWhAt

    User Avatar
    Gold Member

    say what? why shouldn't there be? It's a parallel axis shift What I see is 1/2 mb^2 + mh^2 for your moment in the y plane
     
  4. Apr 17, 2014 #3
    Yes, exactly. It could be wrong in the key, because I got ##I_{xx}## correct. I'll double check.
     
  5. Apr 17, 2014 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I don't see why there would be an mh2p term. The rotation p is not about an axis through O, so the parallel axis theorem does not apply.
     
  6. Jun 2, 2015 #5
    Why is Iyz = 0?

    When looking at the YZ plane the geometry is not symmetrical about either axis?
     
  7. Jun 2, 2015 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    It's hard to tell because the exact expression for the ##\omega## term was never posted, but I suspect the OP is a bit careless about whether ##I_{xx}## refers to moment about the mass centre or moment about O.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Angular momentum of a solid cylinder rotating around axis
Loading...