A top of mass `m=0.50kg`, whose axis is tilted by an angle `theta=30^@` to the vertical, precesses due to gravity. The moment of inertia of the top relative to its symmetry axis is equal to `I=2.0g*m^2`, the angular velocity of rotation about that axis is equal to `omega=350rad//s`, the distance from the point of rest to the centre of inertia of the top is `l=10cm`. Find:
(a) the angular velocity of the top's precession,
(b) the magnitude and direction of the horizontal component of the reaction force acting on the top at the point of rest.

Explanation

Here `M=Iomega` is along the symmetry axis. It has two componets, the part `Iomegacostheta` is constant and the part `M_(_|_)=Iomegasintheta` presseses, then
`|(dvecM)/(dt)|=Iomegasinthetaomega^'=mglsintheta`
or, `omega^'`=precession frequency `=(mgl)/(Iomega)=0*7rad//s`
(b) This force is the centripetal force due to precession. It acts inward and has the magnitude
`|vecF|=|summ_iomega^('^2)vecrho_i|=momega^('^2)lsintheta=12mN`.
`vecrho_i` is the distance of the `ith` element from the axis. This is the force that the table will exert on the top. See the diagram in the answer sheet