Maximum and minimum tension in string whirling in a circle of radius 2.5 ,m with constant velocity a

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5.Work, Energy, Power and Collision

hard

The maximum and minimum tension in the string whirling in a circle of radius $2.5 \,m$ with constant velocity are in the ratio $5 : 3$ then its velocity is

A

$\sqrt {98} \,\,m/s$

B

$7\,\,m/s$

C

$\sqrt {490} \,\,m/s$

D

$\sqrt {4.9} $

Solution

(a) In this problem it is assumed that particle although moving in a vertical loop but its speed remain constant.

Tension at lowest point ${T_{\max }} = \frac{{m{v^2}}}{r} + mg$

Tension at highest point ${T_{\min }} = \frac{{m{v^2}}}{r} – mg$

$\frac{{{T_{\max }}}}{{{T_{\min }}}} = \frac{{\frac{{m{v^2}}}{r} + mg}}{{\frac{{m{v^2}}}{r} – mg}} = \frac{5}{3}$

by solving we get,$v = \sqrt {4gr} $$ = \sqrt {4 \times 9.8 \times 2.5} $ $ = \sqrt {98} \,m/s$

Standard 11

Physics

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