Lead and aluminum balls of equal radius $r$ are connected by a weightless non-stretchable thread, the length of which is much greater than the size of the balls. The balls were lowered in a vessel with glycerol. They then came to motion with zero initial velocity. The drag force on the balls is proportional to their velocity, and the coefficient of proportionality is the same for both balls. Find the tension force of the thread at the steady-state velocity of the balls. The densities of aluminum and lead are $\rho_{1}$ and $\rho_{2}$.
Decision:
Let us first study the motion of a single free ball having density $\rho$. Let us denote the volume of the ball by $V$. Three forces act on the ball: the gravitational force $\rho Vg$ directed downward, the expulsive force $F_{a}$ directed upward, and the drag force $kv$ directed against the velocity of the ball $v$. Applying Newton's law II to the ball's motion, we get
$\rho Va=\rho Vg + F_{a}-kv$
where $a$ is the acceleration of the ball at the moment when its velocity is equal to $v$. If at the initial moment the velocity of the ball is zero, it will start moving downward with some acceleration. As the velocity increases, the drag force increases, the acceleration will decrease and when the velocity is reached
$v_{max}=\frac{\rho Vg-F_{a}}{k}$ (1).
will become zero. In this case, the equidistance of all forces applied to the ball will turn to zero, and the motion will continue with constant velocity $v=v_{max}$.
It is not difficult to show that regardless of the magnitude and direction of the initial velocity, the steady motion of the ball is a vertical fall with constant velocity. It follows from formula (1) that the greater the density of the ball, the greater the velocity of its steady motion.
Let us now consider the motion of bound balls. Since the thread is long, the magnitude of the interaction of any of the balls with glycerol will not depend on the presence of the second ball and will be the same as in its absence. The balls interact with each other only through the tension of the thread.
In steady-state motion, the balls will move vertically downward with some velocity v, the thread will be vertical and taut, and the heavier lead ball will be at the bottom.
The equations of the steady vertical motion of the balls are:
$\rho_{1} Vg-kv-F_{a}+T=0$
$\rho_{2} Vg-kv-F_{a}-T=0$.
Given the expression for the volume of the ball $V=\frac{4}{3} \pi r^{3}$ we find
$T=\frac{4}{3} \pi r^{3}g \frac{\rho_{2}-\rho_{1}}{2}$
Decision:
Let us first study the motion of a single free ball having density $\rho$. Let us denote the volume of the ball by $V$. Three forces act on the ball: the gravitational force $\rho Vg$ directed downward, the expulsive force $F_{a}$ directed upward, and the drag force $kv$ directed against the velocity of the ball $v$. Applying Newton's law II to the ball's motion, we get
$\rho Va=\rho Vg + F_{a}-kv$
where $a$ is the acceleration of the ball at the moment when its velocity is equal to $v$. If at the initial moment the velocity of the ball is zero, it will start moving downward with some acceleration. As the velocity increases, the drag force increases, the acceleration will decrease and when the velocity is reached
$v_{max}=\frac{\rho Vg-F_{a}}{k}$ (1).
will become zero. In this case, the equidistance of all forces applied to the ball will turn to zero, and the motion will continue with constant velocity $v=v_{max}$.
It is not difficult to show that regardless of the magnitude and direction of the initial velocity, the steady motion of the ball is a vertical fall with constant velocity. It follows from formula (1) that the greater the density of the ball, the greater the velocity of its steady motion.
Let us now consider the motion of bound balls. Since the thread is long, the magnitude of the interaction of any of the balls with glycerol will not depend on the presence of the second ball and will be the same as in its absence. The balls interact with each other only through the tension of the thread.
In steady-state motion, the balls will move vertically downward with some velocity v, the thread will be vertical and taut, and the heavier lead ball will be at the bottom.
The equations of the steady vertical motion of the balls are:
$\rho_{1} Vg-kv-F_{a}+T=0$
$\rho_{2} Vg-kv-F_{a}-T=0$.
Given the expression for the volume of the ball $V=\frac{4}{3} \pi r^{3}$ we find
$T=\frac{4}{3} \pi r^{3}g \frac{\rho_{2}-\rho_{1}}{2}$