8/1     If I want a pendulum of period 1 sec, what is the length I should
        have? I know the gravity and that the pendulum is independent of
        its weight. Thanks!
        \- hello, making some simplifying assumptions, like the moment
           of intertia is about that of a point mass at length l from the
           fulcrum, and angle between the pendulum arm and the vector from
           the funcrum to the center of mass isnt too large then
           T = 2pi sqrt(l/g). Do you need a pendulum with large amplitude?
           Also I assume this is in simple pendulum like a grandfather clock.
           If you have a more complicated object thenthe radius of gyration
           is k = sqrt [moment of intertia(w.r.t. center of mass)/mass] ...
           I assume you know what a moment of inertia is.
           [mnemonic: pi^2 is actually pretty close to the grav acceleration
           at the earth surface ... this is useful if you have to do approx
           on the back of an envelope. so figure a quarter meter. this isnt a
           homewortk problemis it?]
           \_ of course it was a homework problem --!OP
           \_ isn't it easier to just find the
                integration [intergation of (cos Angle)*gravity]*distance  ?
                \- the newtonian approach with some simplifications is not
                   too difficult. a richer approach is the "Highbrow
                   mechanics" which has at it's core the kinetic and potential
                   energy functions rather than the newtonian s''=v'=a func
                   appriach. however you may need a little geometry and some
                   knowledge of elliptical integrals. ok tnx.
                   \_ I found some notes from an old math class with elliptical
                      integrals on it.  I used to know that stuff.  I feel
                      really dumb now.
                      \_ More like your brain has pushed that crap aside for
                         more important things.