Question

How can you find the angle at which gravity potential energy of a pendulum system is equal to the kinetic energy of the “bob” or “datum”?

Answer 1

$\begin{gathered} \emptyset=\sin ^{-1}(\frac{\frac{v^2}{2g}-l}{l}) \\ \text{where} \\ \emptyset\text{ is the angle} \\ l\text{ is the length of the pendulum} \\ \text{v is the velocity } \\ and\text{ g is the acceleation of gravity} \end{gathered}$

Explanation

Step 1

diagram

Step 2

the kinetick energy is given by

$E_{_k}=\frac{1}{2}mv^2$

and potenital energy is given by:

$E_p=\text{mgh}$

so, the points where the energy is the same,

$\begin{gathered} \frac{1}{2}mv^2=mgh \\ m\text{ is canceled} \\ so \\ \frac{1}{2}v^2=gh \\ \text{divide both sides by g} \\ \frac{\frac{1}{2}v^2}{g}=\frac{gh}{g} \\ \frac{v^2}{2g}=h \\ \end{gathered}$

but h is

$h=l-l\sin (\emptyset)$

hence,replacing

$\begin{gathered} l-l\sin (\emptyset)=\frac{v^2}{2g} \\ now,\text{ solve for angle(}\emptyset) \\ subtract\text{ l in both sides} \\ l-l\sin (\emptyset)-l=\frac{v^2}{2g}-l \\ l\sin (\emptyset)=\frac{v^2}{2g}-l \\ \text{divide both sides by l} \\ \frac{l\sin (\emptyset)}{l}=\frac{\frac{v^2}{2g}-l}{l} \\ \sin (\emptyset)=\frac{\frac{v^2}{2g}-l}{l} \\ \text{ inverse sin function} \\ \sin ^{-1}(\sin (\emptyset)=\sin ^{-1}\frac{\frac{v^2}{2g}-l}{l} \\ \emptyset=\sin ^{-1}(\frac{\frac{v^2}{2g}-l}{l}) \end{gathered}$

therefore, the answer is

$\begin{gathered} \emptyset=\sin ^{-1}(\frac{\frac{v^2}{2g}-l}{l}) \\ \text{where} \\ \emptyset\text{ is the angle} \\ l\text{ is the length of the pendulum} \\ \text{v is the velocity } \\ and\text{ g is the acceleation of gravity} \end{gathered}$

I hope this helps you