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# If a rod of mass m is fixed at one end and a particle of mass m is attached to its other end then what is the minimum velocity required for the system to perform circular motion?Why is moment of inertia of the rod taken without any influence of particle and,  the Rotational energy and KE is initial energy in Energy conservation equation. why?

10 years ago

if rod is fixed about one end & it is rotated in verticle plane then , moment of inertia about that end will be sum of both of particle as well as rod ...

I = Imass + Irod

= mL2 + mL2/3  = 4mL2/3            ...............1

now from energy conservation

[PE + KE ]i = [PE + KE]f         ..................2

your question is why we take only KE at initial moment ???see if we are taking lowest point as refrence for potential energy then at bottom most point PE will be 0 & at topmost point it will be 2mgL ...

now i m taking the point of refrence at H meter below the lowest point then , initial PE will be

2mgH + mgL/2 (rod + mass)  & final PE will be  2mgH + mg7L/2  (rod + mass) ....

indivisually both values are different but when we put these values in

enrgy conservation equation then result is same always ...so take any point as refrence & there is no need to assume that

at lowest point potential energy is 0 ...

now from eq 2 ,

(2mgH + mgL/2) + (KE)i  =  (2mgH+7mgL/2)  +  (KE)f

(KE)f = 0 , at top so

(KE)i =  3mgL

now , KE = IW2/2 = 2mL2W2/3           so ,        (from eq 1)

2mL2 W2/3 = 3mgL

W = 3(g/2L)1/2

V = Wr

=WL/2 = (3/2) (gL/2)1/2  m/s

approve if u like my ans