Question

In vertical circular motion, the ratio of kinetic energy of a particle at highest point to lowest point is (Particle is just completing vertical circular motion)

• A. $5:1$
• B. $3:1$
• C. $1:5$
• D. $1:3$

Answer 1

Answer: (C)

Step - 1
At top most point

$T+mg=\frac{{mv}_{top}^2}{R}$

Since, It is just completing vertical circular motion. So, T=0

$mg=\frac{m{v}_{top}^2}{R}⟹g=\frac{v_{top}^2}{R}$

$v_{top}^2=gR$

$K.E_{top}=\frac{1}{2}m{v}_{top}^2⟹K.E_{top}=\frac{1}{2}mgR.......\left(1\right)$

For the bottom most point.

Since, we knew velocity required at lowest point to just complete vertical circular motion

$v_{bottom}=\sqrt{5gR}$

$K.E_{bottom}=\frac{1}{2}m{v}_{bottom}^2=\frac{1}{2}\left(5gR\right)m.......\left(2\right)$

$\therefore \frac{K.E_{top}}{K.E_{bottom}}=\frac{\frac{1}{2}mgR..}{\frac{1}{2}\left(5gR\right)m.}=1:5$

Hence option C is the correct answer