Question

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

Answer 1

Let the deformation in the spring be $\Delta l$, when the rod $AB$ has attained the angular velocity $\omega$. From the second law of motion in projection form $F_n=m{w}_n$.
$\kappa \Delta l=m{\omega }^2(l_0+\Delta l)$ or, $\Delta l=\frac{m{\omega }^2{l}_0}{\kappa -m{\omega }^2}$
From the energy equation, $A_{ext}=\frac{1}{2}m{v}^2+\frac{1}{2}\kappa \Delta l^2$
$=\frac{1}{2}m{\omega }^2{(l_0+\Delta l)}^2+\frac{1}{2}\kappa \Delta l^2$
$=\frac{1}{2}m{\omega }^2{(l_0+\frac{m{\omega }^2{l}_0}{\kappa -m{\omega }^2})}^2+\frac{1}{2}\kappa {\left(\frac{m{\omega }^2{l}_0}{\kappa -m{\omega }^2}\right)}^2$
On solving $A_{ext}=\frac{\kappa }{2}\frac{l_0^2\eta (1+\eta )}{{(1-\eta )}^2}$, where $\eta =\frac{m{\omega }^2}{\kappa }$