HCVERMA -WORK&ENER.-PG136-PRB58,64

A simple pendulum of length L having a bob of mass 'm' is deflected frm its rest position by angle  and released. The string hits the peg which is fixed at a distance 'x' below pt. of suspension and the bob starts going in a circle  centred at the peg. (1)Assuming that initially the bob has a height less than peg,show that the max. ht. reached by bob equals its initial ht.(2)if pendulum is released with  = 90' and "'x'=L/2 ,find max. ht. reached by bob above its lowest position before the string becomes slack. (3) Find min. value of x/L for which bob goes in a complete circle about the peg when the pendulum is released from  =90'

plz write the Q.

58)

conservation of energy

U1+K1=U2+K2

K1=K2=0

thsu U1=U2

hence h needs 2 b d same

b) conservation of energy

U1+K1=U2+K2

1/2mv^2=mgl

thus v=

l=2r

thus v2=

then again conservation of energy

1/2m(v2)^2=mg(r+rcos@)+1/2m(v3)^2

4mgr/2=mg(r+rcos@)+1/2*mgrcos@                      mgcos@=m(v3)^2/r    thus (v3)^2=grcos@

thus solvin u get cos@=2/3

height aboove d lowest point=r+2/3r