A $1.4 \mathrm{~m}$ long rod has $3 \mathrm{~kg}$ masses at its ends. Where should the rod be pivoted so that, when released from a horizontal position, the mass on the left side passes under the pivot at a speed of $1.6 \mathrm{~m} / \mathrm{s}$?
On a rod of length $l$, we place the pivot at a distance $x$ from the left end! If we take the zero level of potential energy at the height of the pivot, then the potential energy of the horizontally stationary rod is zero. In the vertical position, the potential energy is

$$
-x m g+(l-x) m g=(l-2 x) m g
$$

where $m=3 \text{ kg}$ is the mass of the weights at the ends of the rod (we consider the rod to have negligible mass). The kinetic energy is

$$
\frac{1}{2} m v^{2}+\frac{1}{2} m\left(v \frac{l-x}{x}\right)^{2}=\frac{1}{2} m v^{2}\left[1+\left(\frac{l-x}{x}\right)^{2}\right]
$$

where $v=1.6 \text{ m/s}$ is the speed of the lower weight, and $v \frac{l-x}{x}$ is the speed of the upper weight when the vertical position is reached. Since the total energy does not change during the motion,

$$
(l-2 x) m g+\frac{1}{2} m v^{2}\left[1+\left(\frac{l-x}{x}\right)^{2}\right]=0
$$

From this, we get the following cubic equation for $x$:

$$
x^{3}-\left(\frac{v^{2}}{2 g}+\frac{l}{2}\right) x^{2}+\frac{v^{2}}{2 g} l x-\frac{v^{2}}{4 g} l^{2}=0
$$

Substituting the given data:

$$
x^{3}-0.828 \text{ m} \cdot x^{2}+0.1792 \text{ m}^{2} \cdot x-0.12544 \text{ m}^{3}=0
$$

This equation can be solved using the Cardano formula, graphically, or by some numerical approximation method.

$x$ must be chosen from the interval $[0.7 \text{ m}, 1.4 \text{ m}]$. The solution to the equation is: $x=0.8 \text{ m}$, so the rod should be pivoted at a distance of $0.8 \text{ m}$ from the left end.