A small bob tied at one end of a thin string | vedclass

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(C) Let the speed of the bob at the lowest position be $v_1$ and at the highest position be $v_2$.Maximum tension occurs at the lowest position and mi

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Let the speed of the bob at the lowest position be $v_1$ and at the highest position be $v_2$.Maximum tension occurs at the lowest position and minimum tension occurs at the highest position. Using the conservation of mechanical energy:$\frac{1}{2} mv_1^2 = \frac{1}{2} mv_2^2 + mg(2l)$$\Rightarrow v_1^2 = v_2^2 + 4gl$ $......(1)$At the lowest position: $T_{\max} - mg = \frac{mv_1^2}{l} \Rightarrow T_{\max} = mg + \frac{mv_1^2}{l}$At the highest position: $T_{\min} + mg = \frac{mv_2^2}{l} \Rightarrow T_{\min} = \frac{mv_2^2}{l} - mg$Given the ratio $\frac{T_{\max}}{T_{\min}} = \frac{5}{1}$:$\frac{mg + \frac{mv_1^2}{l}}{\frac{mv_2^2}{l} - mg} = 5$$mg + \frac{m}{l}(v_2^2 + 4gl) = 5(\frac{mv_2^2}{l} - mg)$$mg + \frac{mv_2^2}{l} + 4mg = \frac{5mv_2^2}{l} - 5mg$$10mg = \frac{4mv_2^2}{l}$$v_2^2 = \frac{10gl}{4} = \frac{10 	imes 10 	imes 1}{4} = 25$$v_2 = 5\, m/s$.

$A$ 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 are in the ratio $5: 1$. The velocity of the bob at the highest position is ............. $m/s$ (Take $g=10\, m/s^2$).

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