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)

509 Points
11 years ago

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

SAGAR SINGH - IIT DELHI
879 Points
11 years ago

Dear student,

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

All the best.

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