A light spring with stiffness coefficient $k$ is attached at one end to a vertical axis around which it can rotate freely, and at the other end to a small weight of mass $m$. The whole system is on a horizontal smooth table, the spring is not stretched. A shock imparts to the weight a velocity $v_{0}$ directed perpendicular to the spring. Find the minimum and maximum distance from the weight to the axis if the velocity of the weight at the maximum distance from the axis is $v_{1}$.
Decision:
At the maximum and minimum distances from the load to the axis, the velocities of the load $\bar{v_{1}}$ and $\bar{v_{2}}$ are perpendicular to the spring. Energy and momentum are conserved as the weight moves:
$\frac{mv^{2}_{0}}{2}=\frac{mv^{2}_{1}}{2}+\frac{k(l_{1}-l_{0})^{2}}{2}=\frac{mv^{2}_{2}}{2}+\frac{k(l_{2}-l_{0})^{2}}{2}$, (1)
$mv_{0}l_{0}=mv_{1}l_{1}=mv_{2}l_{2}$. (2)
Let $l_{0}$ be the length of the unstretched spring; $l_{1},l_{2}$ be the maximum and minimum distances of the weight from the axis.
Let us first show that $l_{2}=l_{0},v_{2}=v_{0}$. Indeed, $l_{2} \leq l_{0}$ -. If $l_{2} < l_{0}$ then according to (2) $v_{2} > v_{0}$ and equality (1) cannot be
be fulfilled. From equations (1), (2) we obtain
$l_{2}=l_{0}=\sqrt{ \frac{v^{2}_{1}m(v_{0}+v_{1})}{k(v_{0}-v_{1})} }, l_{1}=\sqrt{ \frac{v^{2}_{0}m(v_{0}+v_{1})}{k(v_{0}-v_{1})} }$
Decision:
At the maximum and minimum distances from the load to the axis, the velocities of the load $\bar{v_{1}}$ and $\bar{v_{2}}$ are perpendicular to the spring. Energy and momentum are conserved as the weight moves:
$\frac{mv^{2}_{0}}{2}=\frac{mv^{2}_{1}}{2}+\frac{k(l_{1}-l_{0})^{2}}{2}=\frac{mv^{2}_{2}}{2}+\frac{k(l_{2}-l_{0})^{2}}{2}$, (1)
$mv_{0}l_{0}=mv_{1}l_{1}=mv_{2}l_{2}$. (2)
Let $l_{0}$ be the length of the unstretched spring; $l_{1},l_{2}$ be the maximum and minimum distances of the weight from the axis.
Let us first show that $l_{2}=l_{0},v_{2}=v_{0}$. Indeed, $l_{2} \leq l_{0}$ -. If $l_{2} < l_{0}$ then according to (2) $v_{2} > v_{0}$ and equality (1) cannot be
be fulfilled. From equations (1), (2) we obtain
$l_{2}=l_{0}=\sqrt{ \frac{v^{2}_{1}m(v_{0}+v_{1})}{k(v_{0}-v_{1})} }, l_{1}=\sqrt{ \frac{v^{2}_{0}m(v_{0}+v_{1})}{k(v_{0}-v_{1})} }$