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(C) For a particle in a vertical circle,the tension at the bottom $(T_{max})$ and top $(T_{min})$ are given by:$T_{max} = \frac{mv_{max}^2}{R} + mg$$T

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) For a particle in a vertical circle,the tension at the bottom $(T_{max})$ and top $(T_{min})$ are given by:$T_{max} = \frac{mv_{max}^2}{R} + mg$$T_{min} = \frac{mv_{min}^2}{R} - mg$Using conservation of energy between the top and bottom points:$\frac{1}{2}mv_{max}^2 = \frac{1}{2}mv_{min}^2 + mg(2R)$$v_{max}^2 = v_{min}^2 + 4gR$Substituting $v_{max}^2$ into the $T_{max}$ equation:$T_{max} = \frac{m(v_{min}^2 + 4gR)}{R} + mg = \frac{mv_{min}^2}{R} + 5mg$The ratio is given as $\frac{T_{max}}{T_{min}} = \frac{\frac{mv_{min}^2}{R} + 5mg}{\frac{mv_{min}^2}{R} - mg} = \frac{5}{3}$Let $x = \frac{v_{min}^2}{Rg}$. Then $\frac{x+5}{x-1} = \frac{5}{3} \Rightarrow 3x + 15 = 5x - 5 \Rightarrow 2x = 20 \Rightarrow x = 10$.Since $x = \frac{v_{min}^2}{Rg} = 10$,we have $v_{min}^2 = 10gR = 10 	imes 10 	imes 1.0 = 100$.Thus,$v_{min} = 10 \ ms^{-1}$.

$A$ particle is whirled in a vertical circle of radius $1.0 \ m$ using a string with one end fixed. If the ratio of maximum and minimum tension in the string is $\frac{5}{3}$,the minimum velocity of the particle during circular motion is ......... $ms^{-1}$.

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