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Instantaneous centres of rotation

  1. Oct 11, 2006 #1


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    Anyone know a good web site, or two, that will tell me about working out the instantaneous centre of rotation; for links only I imagine.

  2. jcsd
  3. Oct 14, 2006 #2


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    The first thing that came up on google [:biggrin:]: http://www.fsid.cvut.cz/en/U2052/node27.html. Not much, depends on how deep you need to get into it.

    By the way, if I recalled it correctly, if you know the speed [tex]\vec{v}_{A}[/tex] of a point A, and the angular speed [tex]\vec{\omega}[/tex], then you can find the speed of any point B with [tex]\vec{v}_{B}=\vec{v}_{A}+\vec{\omega}\times \vec{r}_{BA}[/tex], where [tex]\vec{r}_{BA}[/tex] is the vector from A to B. The condition on the centre of velocity is [tex]\vec{v}_{B} = \vec{v}_{c} = \vec{0}[/tex], so you can find its position.
    Last edited: Oct 14, 2006
  4. Oct 14, 2006 #3


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    Many thanks.
    I found a site (eventually - it took me ages, even on google) that told me what I was supposed to do. I used Kennedy's theorem to find the IC's and worked things out from there.
    It was faster the first way I did it though. I just resolved velocities at the link ends and worked my way through the mechanism. Instantaneous centres gave the same answer, but took longer :frown:
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