Question

A small bar $A$ resting on a smooth horizontal plane is attached by threads to a point $P$ (figure shown above) and, by means of a weightless pulley, to a weight $B$ possessing the same mass as the bar itself. Besides, the bar is also attached to a point $O$ by means of a light non-deformed spring of length $l_0=50cm$ and stiffness $\kappa =5\frac{mg}{l_0}$, where $m$ is the mass of the bar. The thread $PA$ having been burned, the bar starts moving. Find its velocity at the moment when it is breaking off the plane.

Answer 1

When the thread $PA$ is burnt, obviously the speed of the bars will be equal at any instant of time until it breaks off. Let $v$ be the speed of each block and $\theta$ be the angle, which the elongated spring makes with the vertical at the moment, when the bar $A$ breaks off the plane. At this stage the elongation in the spring,
$\Delta l=l_0\sec \theta -l_0=l_0(\sec \theta -1)$ (1)
Since the problem is concerned with position and there are no forces other than conservative forces, the mechanical energy of the system (both bars $+$ spring) in the field of gravity is conserved, i.e. $\Delta T+\Delta U=0$
So, $2\left(\frac{1}{2}m{v}^2\right)+\frac{1}{2}\kappa l_0^2{(\sec \theta -1)}^2-mg{l}_0\tan \theta =0$ (2)
From Newton's second law in projection form along vertical direction
$mg=N+\kappa l_0(\sec \theta -1)\cos \theta$
But, at the moment of break off, $N=0$.
Hence, $\kappa l_0(\sec \theta -1)\cos \theta =mg$
or, $\cos \theta =\frac{\kappa l_0-mg}{\kappa l_0}$ (3)
Taking $\kappa =\frac{5mg}{l_0}$, simultaneous solution of (2) and (3) yields
$v=\sqrt{\frac{19g{l}_0}{32}}=1.7m/s$