A small disk of mass m lies on the highest point of a sphere of radius R. A slight push makes the disk start sliding down. Find the force of pressure of the disk on the sphere as a function of the angle its radius vector makes with the vertical. Where does the disk lose contact with the sphere? Friction is to be neglected.

Answer

`mg(3 cos alpha -2); a_("ret") = alpha_("rel") = a r cos ""2/3=48^@`

Explanation

According to Newton.s third law, the pressure of the washer against the sphere is equal in magnitude to the reaction. The forces acting on the object in the direction normal to the velocity vector are the reaction N and the component of the force of gravity `F_2 = mg cos alpha` (see Fig). According to Newton.s second law
`F_(2) - N =mv^(2)//R`
To find the velocity apply the law of conservation of energy mgh `= mv^2//2` Since h = (1 — cos ), we obtain after some simple transformations
`N = mg (3 cos alpha - 2) `
When the washwer leaves the sphere it ceases to press against it and the reaction becomes zero. The condition for the loss of contact is `cos alpha = 2//3 , alpha = 48^@ h= R//3`