Question

A pendulum Bob of mass m is held in horizontal position and then released show that the velocity of Bob at lowest position is under root of 2gl

Answer 1

-------------------0. U = MGL KE=0


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0 KE = 1/2 mv² U= 0
since only conservative force is acting on bob
sum of initial energy is conserved

=> mgl = 1/2 mv²
=> v =√(2gl)