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
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
$\sqrt {98} \,\,m/s$
$7\,\,m/s$
$\sqrt {490} \,\,m/s$
$\sqrt {4.9} $
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
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$
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