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 \[r=\frac{2}{5}\times
\left( 1-\cos {{60}^{\text{o}}} \right)=\frac{1}{5}=0.2\]
\[\therefore
\] \[y=L-r=1-0.2=0.8m\]
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