A smooth uniform rod AB of mass M and length l rotates freely with an angular velocity `omega_0` in a horizontal plane about a stationary vertical axis passing through its end A. A small sleeve of mass m starts sliding along the rod from the point A. Find the velocity `v^'` of the sleeve relative to the rod at the moment it reaches its other and B.

Explanation

In the process of motion of the given system he kinetic energy and the angular momentum relative to rotation axis do not vary. Hence, it follows that
`1/2(Ml^2)/(3)omega_0^2=1/2m(omega^2l^2+v^('^2))+1/2(Ml^2)/(3)omega^2`
(`omega` is the final angular velocity of the rod)
and `(Ml^2)/(3)omega_0=(Ml^2)/(3)omega+ml^2omega`
From these equations we obtain
`omega=omega_0//(1+(3M)/(M))` and
`v^'=omega_0l//sqrt(1+3m//M)`