4. A body of mass $m$ is attached to one end of a weightless, inextensible string, the other end of which is attached to the ceiling. The string is passed over a pulley of mass $M$, which is concentrated on its axis. At the initial moment, the body is held and then released. Find the accelerations of the body and the pulley.

![](https://cdn.mathpix.com/cropped/2024_05_06_98019744d8c01484e6a0g-1.jpg?height=171&width=1077&top_left_y=1202&top_left_x=158)
4. It is clear that the bodies will not fall freely with the acceleration due to gravity, and the string will be taut. Indeed, if the pulley falls with an acceleration of $g$, then on the side where the body is located, such a length of the string is "stretched" that the acceleration of the body should equal $2g$, which means that the string is taut, forcing the body to have an acceleration greater than $g$, and the pulley to have an acceleration less than $g$.

![](https://cdn.mathpix.com/cropped/2024_05_06_0e208de39d119289f7e1g-1.jpg?height=340&width=257&top_left_y=2189&top_left_x=1685)

The body is acted upon by the force of gravity $m \vec{g}$ and the tension force of the string $\vec{T}$. The pulley is acted upon by the force of gravity $M \vec{g}$ and two tension forces of the string $\vec{T}^{\prime}$, which are equal to each other since the mass of the pulley is concentrated on the axis, and equal in magnitude to the tension force acting on the body. Therefore, Newton's second law for the body and the pulley, projected onto the vertical axis, gives

$$
\begin{aligned}
& m a_{1}=m g+T \\
& M a_{2}=M g-2 T
\end{aligned}
$$

Considering that the acceleration of the body is twice the acceleration of the pulley, we get

$$
a_{1}=\frac{2(2 m+M) g}{4 m+M}, \quad a_{2}=\frac{(2 m+M) g}{4 m+M}
$$