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in the figure two identical rods OA and OB each of length l and mass m are connected to each other by a massless pin connection(both the rods can rotate about the pin o which is free to move), that allowsfree rotation. The assembly is kept on a frictionless horizontal plane. Now two point masses each of mass m moving with speed u perpendicular to the AB and hit the assembly inellastically at points A and B as shown. q : find the angular speed of the rods just after the collision A)3u/4l B)6u/5l C) 3u/2l D)none of these key (B)

9 years ago

Answers : (2)

vikas askiitian expert
509 Points
							

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

moment of inertia of particle about O is Ip = mL2

total moment of inertia of ststem = 2IR + 2Im

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

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

Li = Lf     .........2

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

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

Lf = IW 

putting these values in eq 2 we get

W = 2muL/I =  3u/4L

this is the required ans

9 years ago
SAGAR SINGH - IIT DELHI
879 Points
							

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.

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Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

9 years ago
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