A top of mass `m=1.0kg` and moment of inertia relative to its own axis `I=4.0g*m^2` spins with an angular velocity `omega=310rad//s`. Its point of rest is located on a block which is shifted in a horizontal direction with a constant acceleration `w=1.0m//s^2`. The distance between the point of rest and the centre of inertia of the top equals `l=10cm`. Find the magnitude and direction of the angular velocity of precession `omega^'`.

Explanation

The effective `g` is `sqrt(g^2+w^2)` inclined at angle `"tan"^-1w/g` with the vertical. Then with reference to the new "vertical" we proceed as in problem. Thus
`omega^'=(mlsqrt(g^2+w^2))/(Iomega)=0*8rad//s`.
The vector `overset(rarr')omega` forms an angle `theta="tan"^-1(w)/(g)=6^@` with the normal vertical.