Question

In an old-fashioned rolling mill, grain is ground by a disk-shaped millstone that rolls in a circle on a flat surface, driven by a vertical shaft. Because of the stone's angular momentum, the contact force with the surface is greater than the weight of the wheel.

Assume that the millstone is a uniform disk of mass M, radius b, and width w, and that it rolls without slipping in a circle of radius R with angular velocity w. Find the ratio of the contact force with respect to the surface to the weigth of the stone.

Answer 1

Given uniform sisc of mass= M

Radius of the disc = B

Width of the disk = w

ANGULAR VELOCITY = w

CIRCLE OF THE RADIUS = r

Given that the contact force with the surface is greater than the weight of the wheel.

The stone moves around the vertical shaft with angular velocity w.

Moment of inertia

By Newton's thirdlaw, the contact force

exerted by the millstone upon the flat surface is equal to and opposite to normal force

So the ratio of contact force to weight of th

So the ratio of contact force to weight of th