1169. A chain hanging on a smooth hook slides down. At the beginning of the movement, 10 m of the chain hangs on one side of the hook, and 8 m on the other. Neglecting resistances, find: 1) in what time the entire chain will slide off the hook and 2) what will be the speed of the chain at the initial moment of its free fall.
Solution. If at time $t$ the length of the part of the chain moving downward is $s$ (m), then at this moment the force $F$ driving the chain is the difference between the weights of the parts of the chain hanging on either side of the hook, $F=\delta g s-\delta g(18-s)=2 \delta g(s-9)$; where $\delta$ is the mass of 1 m of the chain, $g$ is the acceleration due to gravity.

According to the second law of mechanics, the differential equation of motion of the chain will be

$$
18 \delta \frac{d^{2} s}{d t^{2}}=2 \delta g(s-9) \text { or } 9 \frac{d^{2} s}{d t^{2}}=g(s-9)
$$

We solve it as an incomplete second-order equation, not explicitly containing the independent variable $t(\$ 6)^{*}$.

Assuming $\frac{d s}{d t}=v$, we have $\frac{d^{2} \mathrm{~s}}{d l^{2}}=\frac{d v}{d t} \cdot \frac{d s}{d s}=v \frac{d v}{d s}$,

$$
\begin{gathered}
9 v \frac{d v}{d s}=g(s-9) ; 9 v d v=g(s-9) d s \\
\frac{9}{2} v^{2}=\frac{g}{2}(s-9)^{2}+\frac{c_{1}}{2} ; 9 v^{2}=g(s-9)^{2}+c_{1}
\end{gathered}
$$[^48]

The value of the constant $c_{1}$ is determined from the initial condition given in the problem: when $s=10, v=0$ :

$$
\begin{gathered}
0=g(10-9)^{2}+c_{1} ; c_{1}=-g \\
9 v^{2}:=g(s-9)^{2}-g .
\end{gathered}
$$

Replacing $v=\frac{d s}{d t}$ in equation (9), separating variables, and integrating, we get:

$$
\begin{gathered}
9\left(\frac{d s}{d t}\right)^{2}=g(s-9)^{2}-g ; \quad \frac{d s}{d t}=\frac{\sqrt{g}}{3} \sqrt{(s-9)^{2}-1} \\
\frac{d(s-9)}{\sqrt{(s-9)^{2}-1}}=\frac{\sqrt{g}}{3} d t ; \ln \left[s-9+\sqrt{(s-9)^{2}-1}\right]=\frac{\sqrt{g}}{3} t+c_{2}
\end{gathered}
$$

The value of the constant $c_{: 2}$ is determined from the initial condition: when $t==0, s=10$.

$$
\begin{gathered}
\ln \mid=c_{2} ; \quad c_{2}=0 \\
t=\frac{3}{\sqrt{g}} \ln \left[s-9+\sqrt{(s-9)^{2}-1}\right]
\end{gathered}
$$

The time it takes for the entire chain to slide off the hook is found from equation (10), by setting $s=18$ in it:

$$
t=\frac{3}{\sqrt{g}} \ln (9+\sqrt{80}) \approx 2.9 \text { sec }
$$

The speed of the chain at the initial moment of its free fall is found from equation (9), by setting $s=18$ in it:

$$
v=\frac{\sqrt{80}}{3} \approx 9.3 \mathrm{m} / \mathrm{sec}
$$