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A nail is located at a certain distance vertically below the point of suspension of a simple pendulum. The pendulum bob is released from a position where the string makes an angle of \[\text{6}0{}^\circ \] with the vertical. Calculate the distance of nail from the point of suspension such that the bob will just perform revolutions with the nail as centre. Assume the length of the pendulum to be one meter.

Answer:

                In Fig.5(HT).11. S is suspension N is the nail. SN = y; SL' = L = 1 m For performing revolution about N, minimum vel. Required  at L' is \[\sqrt{5gr}\] This is provided by conversion of P.E. of pendulum bob at  A into K.E. AT L'. i.e. \[\frac{1}{2}m\upsilon _{L}^{2}=mgh=mg\left( L-SB \right)=mg\left( L-L\cos \text{ }\!\!\theta\!\!\text{ } \right)\] or \[\upsilon _{L}^{2}=2gL\left( 1-\cos \text{ }\!\!\theta\!\!\text{ } \right)\]or \[5\,g\,r=2\,g\,L\left( 1-\cos \text{ }\!\!\theta\!\!\text{ } \right)\] or \[r=\frac{2}{5}\times \left( 1-\cos {{60}^{\text{o}}} \right)=\frac{1}{5}=0.2\] \[\therefore \]  \[y=L-r=1-0.2=0.8m\]