A horizontal disc rotates with a constant angular velocity `omega=6.0rad//s` about a vertical axis passing through its centre. A diameter of the disc with a velocity `v^'=50cm//s` which is constant relative to the disc. Find the force that the disc exerts on the body at the moment when it is located at the distance `r=30cm` from the rotation axis.

Explanation

The disc exerts three forces which are mutually perpendicular. They are the reaction of the weight, `mg`, vertically upward, the Coriolis force `2mv^'omega` perpendicular to the plane of the vertical and along the diameter, and `momega^2r` outward along the diameter. The resultant force is,
`F=msqrt(g^2+omega^4r^2+(2v^'omega)^2)`