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Moment of inertia / torque question

  1. Dec 11, 2005 #1
    Deleted Deleted
    Last edited: Dec 12, 2005
  2. jcsd
  3. Dec 12, 2005 #2
    ummm. i think you mean to say it is attached to a solid 'brick'
    [tex] \vec{\tau} = \vec{r} \times \vec{F} [/tex]
    where [itex] \vec{r} [/itex] is the position vector from the axis of rotation to where the force is acting, and [itex] \vec{F} [/itex] is the force.
    the direction of the vector produced by the cross product is given by the right hand rule.
  4. Dec 12, 2005 #3
    oops. i meant disk.
    Last edited: Dec 12, 2005
  5. Dec 12, 2005 #4
    bump bump bump
  6. Dec 12, 2005 #5
    anyone? anyone?
  7. Dec 12, 2005 #6
    Last edited by a moderator: Apr 21, 2017
  8. Dec 12, 2005 #7
    One qn.. How did u actually arrive at this?
    Last edited: Dec 12, 2005
  9. Dec 12, 2005 #8
    added the moment of inertia of the hoop (mr^2) witht he moment of inertia of the disk (.5mr^2).. good chance its wrong, but itdoesnt really matter at this point, i just turned the assignment in a few minutes ago
  10. Dec 12, 2005 #9
    Moment of inertia shld be the same as the formula for torque=F*d=mgd.
    I'm not sure bout it either. But are u sure mr^2 is the right formula for moment of a force?
  11. Dec 12, 2005 #10


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    Science Advisor

    You deleted the original question. How are we going to help if we don't know what you are asking? However, if it is a homework problem, it belongs in the homework section.
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