A small bob tied at one end of a thin string of length 1 m is describing a vertical circle so that the maximum and minimum tension in the string is in the ratio 5 : 1 . The velocity of the bob at the

Question

A small bob tied at one end of a thin string of length 1 m is describing a vertical circle so that the maximum and minimum tension in the string is in the ratio 5 : 1 . The velocity of the bob at the highest position is ______ m s - 1 . (Take g = 10 m s - 2 )

TestHub · Accepted Answer

Let the speed of bob at lowest position be $v_{1}$ and at the highest position be $v_{2}$ .
Maximum tension is at lowest position and minimum tension is at the highest position.
Now, using, conservation of mechanical energy,
$\frac{1}{2} m v_{1}^{2} = \frac{1}{2} m v_{2}^{2} + m g 2 l$
$\Rightarrow v_{1}^{2} = v_{2}^{2} + 4 g l . . . . (1)$
Now, $T_{\max} - m g = \frac{m v_{1}^{2}}{l}$
$\Rightarrow T_{\max} = m g + \frac{m v_{1}^{2}}{l}$
$& T_{\min} + m g = \frac{m v_{2}^{2}}{l}$
$\Rightarrow T_{\min} = \frac{m v_{2}^{2}}{l} - m g$
$\frac{T_{\max}}{T_{\min}} = \frac{5}{1}$
$\Rightarrow \frac{m g + \frac{m v_{1}^{2}}{l}}{\frac{m v_{2}^{2}}{l} - m g} = \frac{5}{1}$
$\Rightarrow m g + \frac{m v_{1}^{2}}{l} = [\frac{m v_{2}^{2}}{l} - m g] 5$
$\Rightarrow m g + \frac{m}{l} [v_{2}^{2} + 4 g l] = \frac{5 m v_{2}^{2}}{l} - 5 m g$
$\Rightarrow m g + \frac{m v_{2}^{2}}{l} + 4 m g = \frac{5 m v_{2}^{2}}{l} - 5 m g$
$\Rightarrow 10 m g = \frac{4 m v_{2}^{2}}{l}$
$v_{2} = \frac{10 \times 10 \times 1}{4}$
Thus, the velocity of the bob at the highest position is $5 m s^{- 1} .$

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