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Q. $A$ ball weighing $1.0\, kg$ is tied to a string $15\, cm$ long. Initially the ball is held in position such that the string is horizontal. The ball is now released. $A$ nail $N$ its situated vertically below the support at a distance $L$. The minimum value of $L$ such that the string will be wound round the nail is
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Work, Energy and Power

Solution:

Velocity of bob at C should be $\sqrt{5gR}$
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$\sqrt{5 g(R-L)}=\sqrt{2 g R}$
$5(R-L)=2 R$
$3 R=5 L$
$L=\frac{3}{5} R=\frac{3}{5} \times 15=9\, cm$