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(A) For a bob to complete a vertical circle,the minimum velocity at the lowest point is $v_0 = \sqrt{5gL}$.Applying the law of conservation of energy

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$\sqrt{10} mg$

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(A) For a bob to complete a vertical circle,the minimum velocity at the lowest point is $v_0 = \sqrt{5gL}$.Applying the law of conservation of energy between the lowest point and the point where the string is horizontal (at height $L$):$\frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + mgL$$\frac{1}{2}m(5gL) = \frac{1}{2}mv^2 + mgL$$\frac{5}{2}gL = \frac{1}{2}v^2 + gL \implies \frac{3}{2}gL = \frac{1}{2}v^2 \implies v^2 = 3gL$.At the horizontal position,the forces acting on the bob are:$1$. Centripetal force (horizontal): $F_x = \frac{mv^2}{L} = \frac{m(3gL)}{L} = 3mg$.$2$. Gravitational force (vertical): $F_y = mg$.The net force is $F_{	ext{net}} = \sqrt{F_x^2 + F_y^2} = \sqrt{(3mg)^2 + (mg)^2} = \sqrt{9m^2g^2 + m^2g^2} = \sqrt{10}mg$.

$A$ simple pendulum of length $L$ carries a bob of mass $m$. When the bob is at its lowest position,it is given the minimum horizontal speed necessary for it to move in a vertical circle about the point of suspension. When the string is horizontal,the net force on the bob is

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