4. A small weightless ring is hinged to a light rod of length $2 l$. At the midpoint and end of the rod, point masses of $m$ and $2 m$ are fixed. The ring is placed on a smooth horizontal rod. Initially, the rod was held horizontally and then released. Find the speed of the ring at the moment when the rod passes the vertical position. Assume the mass of the rod is zero.

![](https://cdn.mathpix.com/cropped/2024_05_06_0209b43e87d803cc3743g-1.jpg?height=109&width=349&top_left_y=1182&top_left_x=1593)
4. Since the external forces on the system of bodies and the spoke act only in the vertical direction (the gravitational forces of the bodies and the reaction force of the rod), the projection of the momentum of the system of bodies on the horizontal direction is conserved. Therefore, if the lower body has a speed $v$ (which is obviously directed to the left) at the moment the spoke passes through the vertical position, the middle body has a speed $2 v$, directed to the right. The speeds of the bodies can be found using the law of conservation of mechanical energy

$$
\frac{2 m v^{2}}{2}+\frac{m(2 v)^{2}}{2}=5 m g l
$$

From this, we find

$$
v=\sqrt{\frac{5 g l}{3}}
$$

The speed of the ring is easiest to find by transitioning to a reference frame associated with the central body. In this reference frame, the middle body is at rest, while the lower body and the ring move in opposite directions with the same speeds. On the other hand, the speed of the lower body in this reference frame is directed to the left and equals $3 v$. Therefore, the speed of the ring is also $3 v$, but directed to the right. Therefore, in the reference frame associated with the Earth, the speed of the ring is

$$
v_{\kappa}=5 v=5 \sqrt{\frac{5 g l}{3}}=\sqrt{\frac{125 g l}{3}}
$$

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