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)(((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}`

Step-by-step explanation

Step 1

diagram

Step 2

the kinetick energy is given by

`E_(_k)=(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} (1)/(2)mv^2=mgh \\ m\text{ is canceled} \\ so \\ (1)/(2)v^2=gh \\ \text{divide both sides by g} \\ ((1)/(2)v^2)/(g)=(gh)/(g) \\ (v^2)/(2g)=h \\ \end{gathered}`

but h is

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

hence,replacing

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

therefore, the answer is

`\begin{gathered} \emptyset=\sin ^(-1)(((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