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Aiswarya Ram Gupta Grade: 12
        

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. . 

6 years ago

Answers : (2)

vikas askiitian expert
510 Points
										

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)


 


 

6 years ago
Aiswarya Ram Gupta
35 Points
										

thnx 4 helping

6 years ago
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