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# A heavy particle hanging from a fixed point by a light inextensible sting of length 'l' is projected horizontally with speed (gl)1/2. Find the speed of the particle and the inclination of the string to the vertical at the instant of the motion when the tension in the string equal to the weight of the particle. .

509 Points
10 years ago

initially total energy (TE) = mu2/2 + PE = mu2/2           (PE of ground is taken 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)

Aiswarya Ram Gupta
35 Points
10 years ago

thnx 4 helping