$2.3.16.$ A weight suspended on a thread of length $l$ was deflected by a distance $r$ from the equilibrium point and released. What is its maximum velocity?
Solution
Let's consider the geometry of the system. Let the maximum deflection angle of the thread be $2\alpha$. The distance $r$ is the chord between the initial deflection point $D$ and the lowest equilibrium point $C$. The vertical height change $h$ (which corresponds to $BC$ in the diagram) can be expressed in two ways: $$h = l(1 - \cos 2\alpha) = r \sin\alpha$$
Using the trigonometric identity $1 - \cos 2\alpha = 2 \sin^2\alpha$, we get: $$2l \sin^2\alpha = r \sin\alpha$$ From this, we express the sine of half the deflection angle $\alpha$: $$\sin\alpha = \frac{r}{2l} \quad \text{(a)}$$
From the law of conservation of energy, the kinetic energy $E_{\text{kin}}$ will be maximal at the lowest point $C$, where the potential energy is minimal. The maximum kinetic energy equals the initial potential energy: $$\frac{mv_{\text{max}}^2}{2} = mgh = mgr \sin\alpha$$ Canceling the mass $m$: $$\frac{v_{\text{max}}^2}{2} = gr \sin\alpha$$
Expressing the maximum velocity of the weight: $$v_{\text{max}} = \sqrt{2gr \sin\alpha} \quad \text{(b)}$$
Substitute the angle expression from (a) into equation (b): $$v_{\text{max}} = \sqrt{2g \cdot r \frac{r}{2l}} = r\sqrt{\frac{g}{l}}$$
