A smooth light horizontal rod AB can rotate about a vertical axis passing through its end A. The rod is fitted with a small slieeve of mass m attached to the end A by a weightless spring of length `l_0` and stiffness `x`. What work must be performed to slowly get this system going and reaching the angular velocity `omega`?

Explanation

Let the deformation in the spring be `Deltal`, when the rod AB has attained the angular velocity `omega`.
From the second law of motion in projection form `F_n=mw_n`.
`kDeltal=momega^2(l_0+Deltal)` or, `Deltal=(momega^2l_0)/(k-momega^2)`
From the energy equation, `A_(ext)=1/2mv^2+1/2kDeltal^2`
`=1/2momega^2(l_0+Deltal)^2+1/2kDeltal^2`
`=1/2momega^2(l_0+(momega^2l_0)/(k-momega^2))^2+1/2k((momega^2l_0^2)/(k-momega^2))^2`
On solving `A_(ext)=k/2(l_0^2eta(1+eta))/((1-eta)^2)`, where `eta=(momega^2)/(k)`