A particle is whirled in a vertical circle of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 ......... $ms^{-1}$.

Similar Questions

$A$ heavy mass is attached at one end of a thin wire and whirled in a vertical circle. The chances of breaking the wire are maximum when:

$A$ particle is kept at rest at the top of a sphere of diameter $42\, m$. When disturbed slightly,it slides down. At what height $H$ from the bottom will the particle leave the sphere? $(...\, m)$

$ABCDE$ is a channel in the vertical plane,part $BCDE$ being circular with radius $r$. $A$ block is released from $A$ and slides without friction and without rolling. The block will complete the loop if $h$ is

Length of a simple pendulum is $1 \,m$. When its bob is at its lowest point its velocity is $7 \,ms^{-1}$. If the bob leaves its circular path at a height $h$ above the centre of the circle, then the value of $h$ is $(g=10 \,ms^{-2})$. (in $\,m$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module