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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)

```
6 years ago

510 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
```
6 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..

All the best.
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Sagar Singh
B.Tech, IIT Delhi

```
6 years ago
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