Re: Angular momentum

Didn't have to. The moment given was for the man and the chair only, and an
initial angular speed. From that, combined with the radi, I was able to
calculate the tangencial velocity in both states, which in turn, I converted
back to angular velocity for part A of the question. For part B where it
asked to give the Ke of both states (weights at 1m and then at 0.3m) I
simply did this:

    Ke=Sum(fi)=1/2(I Va^2)+m Vt^2 -> where Va is angular velocity and Vt is
tangencial.


Note the 1/2 is not included in the second term because there were two
weights.  Was this wrong?

Thanks



"Gregory L. Hansen" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> In article <[EMAIL PROTECTED]>,
> Timtro <[EMAIL PROTECTED]> wrote:
> >I'm sorry to bother you all with this question. I was just playing around
> >with a problem I found in a book. It is a simple problem where in a man
sits
> >on a stoll free to ratate without friction and is spun around while he
holds
> >two 3kg weights. we are given the moment of inertia of the man+stool,
> >initial speed and the fact that the weights are 1m from the axis of
> >rotation. The man then contracts his arms to 0.3m from the axis of
rotation.
> >Obviously the man will go faster because of this, but when I calculated
the
> >initial and final kinetic energies, they were VERY differant. Where did
all
> >that extra energy come from? Or did I calculate the energy wrong?
>
> Did you recalculate the moment of inertia when the man changed the
> distribution of his masses?
>
> -- 
> "Let us learn to dream, gentlemen, then perhaps we shall find the
> truth... But let us beware of publishing our dreams before they have been
> put to the proof by the waking understanding." -- Friedrich August Kekulé