A heavy particle hanging from a fixed point by alight inexetendable string of length l is projected horizontly with speed gl
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initially total energy (TE) = mu2/2 + PE = mu2/2 (PE of ground is taken as 0)
=mu2/2
when it sustends @ with verticle then let its velocity is v ...
finally total energy = mv2/2 + PE
PE = mgR(1-cos@) , so total energy finally
TE = mv2/2 + mgR(1-cos@) .............2
since total energy is conserved so eq1 = eq2
mv2/2 + mgR(1-cos@) = mu2/2 ................3
now let tention in string at this pos is T then
T - mgcos@ = mv2/R
T = mg (given) , so
mv2 = (mg-mgcos@)R ............4
solving eq 4 & 3
3(1-cos@) = 1
cos@ = 2/3
@ = cos-1(2/3)
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