Question

IN VERTICAL CIRCULAR MOTION SHOW THAT- K.E_L= 5KE_H.

Answer 1

Lets assume that the velocity at the highest point is v. Then
$$\begin{gathered} \frac{m{v}^2}{R}=mg \\ \Rightarrow \frac{1}{2}m{v}^2=\frac{1}{2}mgR \\ \Rightarrow K{E}_H=\frac{1}{2}mgR \end{gathered}$$
At the lowest point, the KE is same as the total energy at the highest point
$$\begin{gathered} K{E}_L=K{E}_H+2mgR \\ \Rightarrow K{E}_L=\frac{1}{2}mgR+2mgR \\ \Rightarrow K{E}_L=\frac{5}{2}mgR \end{gathered}$$
Therefore,
$$\begin{gathered} \frac{K{E}_L}{K{E}_H}=\frac{5}{1} \\ \Rightarrow K{E}_L=5K{E}_H \end{gathered}$$