An object is tied to a string and rotated in a vertical circle of radius $r$. Constant speed is maintained along the trajectory. If $T_{\max}/T_{\min} = 2$, then $v^{2}/rg$ is

Question

Dr. Abubacker · Accepted Answer

(c) At the lowest point, $\frac{mv^{2}}{r} = T_{L} - mg$ …(i)
At the highest point, $\frac{mv^{2}}{r} = T_{H} + mg$ …(ii)
As $\frac{T_{\max}}{T_{\min}} = \frac{T_{L}}{T_{H}} = 2$
$	herefore T_{L} = 2T_{H}$
From Eqs. (i) and (ii),
$2T_{H} - mg = T_{H} + mg$
$T_{H} = 2mg$
From Eq. (ii), $\frac{mv^{2}}{r} = 3mg$ or $\frac{v^{2}}{rg} = 3$

Q. An object is tied to a string and rotated in a vertical circle of radius $r$. Constant speed is maintained along the trajectory. If $T_{\max}/T_{\min} = 2$, then $v^{2}/rg$ is

Explanation

Reviewed by Dr. Abubacker

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