moment of inertia of rod about point  O is I_R = mL²/3 

moment of inertia of particle about O is I_p = mL²

total moment of inertia of ststem = 2I_R + 2I_m

                                        I = 8mL²/3          ..............1

now , since no external torque is acting on the system so angular momentam will remain conserved ,

L_i = L_{f     .........2}

L_i = muL + muL = 2muL           (sum of angular momentam of indivisual particles)

finally let this system ( rod + 2 masses) is rotating with W angular velocity then

L_f = IW 

putting these values in eq 2 we get

W = 2muL/I =  3u/4L

this is the required ans

Dear student,

Its a simple problem, just apply principle of conservation of angular momentum about point O and you will get the result..

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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Sagar Singh

B.Tech, IIT Delhi