A uniform ring of mass `m`, with the outside radius `r_2`, is fitted tightly on a shaft of radius `r_1`. The shaft is rotated about its axis with a constant angular acceleration `beta`. Find the moment of elastic forces in the ring as a function of the distance r from the rotation axis.

Explanation

Consider an elementary ring of width `dr` at a distant `r` from the axis. The part outside exerts a couple `N+(dN)/(dr)dr` on this ring while the part inside exerts a couple N in the opposite direction. We have for equilibrium
`(dN)/(dr)dr=-dIbeta`
where `dI` is the moment of inertia of the elementary ring, `beta` is the angular acceleration and minus sign is needed because the couple `N(r)` decreases, with distance vanshing at the outer radius, `N(r_2)=0`. Now
`dI=(m)/(pi(r_2^2-r_1^2))2pirdrr^2`
Thus `dN=(2mbeta)/((r_1^2-r_1^2))r^3dr`
or, `N=1/2(mbeta)/((r_2^2-r_1^2))(r_2^4-r^4)`, on integration