Question

Derive an expression for difference in tensions at highest and lowest point for a particle performing vertical circular motion.

Answer 1

vertical circular motion (ref. image)

use energy conservation at pt A & C

Assume A as ground level

$\frac{1}{2}m{n}^2=2mgR+\frac{1}{2}m{v}^{\prime 2}$

$v^2=4gR+v^{\prime 2}$

$T-mg\cos \theta =\frac{m{v}^2}{R}$

at $ptA\theta =0^o$

At $pt.C\theta =\pi \left({180}^o\right)$

$T_A-mg=\frac{m{V}^2}{R}$ ...(1)

$T_C+mg=\frac{m{V}^{\prime 2}}{R}$ ....(2)

subtract (2) from (1)

$T_A-T_C-2mg=\frac{m}{R}(V^2-V^{\prime 2})$

$T_A-T_C=2mg+\frac{m}{R}(V^2-V^{\prime 2})$
$V^2-V^{\prime 2}=4gR$

$\therefore T_A-T_C=2mg+\frac{m}{R}\cdot 4gR$

$=6mg$

For complete vertical circular motion.