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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. |
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Answer» Solution :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 
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