A particle has velocity √3 r g at the highest point in vertical circle. Find the ratio of tensions at the highest and lowest point

Question

A particle has velocity √3 r g at the highest point in vertical circle. Find the ratio of tensions at the highest and lowest point (A) 1: 6 (B) 1: 4 (C) 1: 3 (D) 1: 2

Tardigrade Admin · Accepted Answer

In vertical circular motin tension at top is mg - mv2/r and tension at bottom is mg + mv2/r Tt o p=(m v2/r)-mg=2mg (Tt o p/Tb o t t o m)=(2 m g/2 m g + 6 m g)=(1/4)

Q. A particle has velocity $3 r g$ at the highest point in vertical circle. Find the ratio of tensions at the highest and lowest point

1 : 6

1 : 4

1 : 3

1 : 2

In vertical circular motin tension at top is mg - mv²/r and tension at bottom is mg + mv²/r

$T_{t o p} = \frac{m v ^{2}}{r} - m g = 2 m g$
$\frac{T _{t o p}}{T _{b o tt o m}} = \frac{2 m g}{2 m g + 6 m g} = \frac{1}{4}$

Q. A particle has velocity 3rg​ at the highest point in vertical circle. Find the ratio of tensions at the highest and lowest point