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From: JEMebius on 2 Apr 2010 19:54
JEMebius wrote:
> JEMebius wrote:
>> Existential Angst wrote:
>>> "Existential Angst" <UNfitcat(a)UNoptonline.net> wrote in message
>>> news:4bb3f677$0$31267$607ed4bc(a)cv.net...
>>>> Awl --

(.....)

> About the force exerted on the stick at the pivot: this consists of...
> (A)
> a force equal to the weight gM of the stick pointing vertically upward,
> and...
> (B)
> the centripetal force "needed to keep the stick at bay at its pivot",
> equal to MLw^2/2 and ==pointing from the pivot to the other end of the stick==.

This is of course the other way around: a centripetal force by definition points from the
mass centre to the centre of rotation.

>
> Now for the specific values for a stick of mass M = 1.00 kg and a length
> L = 2.00 m
> in upright position and being infinitesimally gently pushed aside:
> its final angular speed is sqrt (3g/L) = sqrt (3 * 9.81 / 2) = 3.84 rad/s;
> the final linear speed of the free end = sqrt (3 * 9.81 * 2) = 7.67 m/s;
>
> the orthogonal components of the force exerted upon the stick at the pivot:
> the vertical component 9.81 newton pointing upwards;
> the horizontal component is in general MLw^2/2 = ML.(3g/L)/2 = 3gM/2;
> in our case 14.71 newton.
>
> Ciao: Johan E. Mebius
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