Question

A mass is performing vertical circular motion (see figure). If the average velocity of the particle is increased, then at which point is the maximum breaking possibility of the string:

• A. A
• B. B
• C. C
• D. D

Answer 1

Answer: (B)

Tension at any point in vertical motion is given by :

$T=\frac{m{v}^2}{l}+mg\cos \theta$

where $\theta =$ angular displacement from lowest point ,

$l=$ length of string

$m=$ mass of string

It is clear that tension at the lowest point ($B$) is greatest than at other points ($A,C,D$). If we increase average velocity, tension will increase at lowest point, therefore at point B, string has maximum possibility of break.