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Q. A particle tied to a string describes a vertical circular motion of radius $r$ continually. If it has a velocity $\sqrt{3 gr}$ at the highest point, then the ratio of the respective tensions in the string holding it at the highest and lowest points is

KEAMKEAM 2013Work, Energy and Power

Solution:

Tension at highest point
$T_{H}=\frac{m v^{2}}{r}-m g=3 m g-m g=2 m g$
and tension at lowest point
$T_{L}=\frac{m v^{2}}{r}+m g$
Here, $v_{L}^{2}=3 g r+2 g \cdot 2 t=7 \,gr$
So, $ T_{L}=7 m g+m g=8\, mg$
Hence,$\frac{T_{H}}{T_{L}}=\frac{2 m g}{8 mg}=\frac{1}{4}$