Question

A device (shown in figure above) consists of a smooth L-shaped rod located in a horizontal plane and a sleeve $A$ of mass $m$ attached by a weightless spring to a point $B$. The spring stiffness is equal to $x$. The whole system rotates with a constant angular velocity $\omega$ about a vertical axis passing through the point $O$. Find the elongation of the spring. How is the result affected by the rotation direction?

Answer 1

At first draw the free body diagram of the device as shown. The forces, acting on the sleeve are its weight, acting vertically downward, spring force, along the length of the spring and normal reaction by the rod, perpendicular to its length.
Let $F$ be the spring force, and $\Delta l$ be the elongation.
From $F_n=m{w}_n$
$N\sin \theta +F\cos \theta =m{\omega }^{2}r$ (1)
where $r\cos \theta =(l_0+\Delta l)$.
Similary from $F_t=m{w}_t$
$N\cos \theta -F\sin \theta =0$ or, $N=\frac{F\sin \theta }{\cos \theta }$ (2)
From (1) and (2)
$F\left(\frac{\sin \theta }{\cos \theta }\right)\cdot \sin \theta +F\cos \theta =m{\omega }^{2}r$
$=m{\omega }^2\frac{(l_0+\Delta l)}{\cos \theta }$
On putting $F=\kappa \Delta l$,
$\kappa \Delta l{\sin }^2\theta +\kappa \Delta l{\cos }^2\theta =m{\omega }^2(l_0+\Delta l)$
on solving, we get,
$\Delta l=m{\omega }^2\frac{l_0}{\kappa -m{\omega }^2}=\frac{l_0}{(\frac{\kappa }{m}{\omega }^2-1)}$
and it is independent of the direction of rotation.