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A thin uniform bar of length L and mass 8m lies on a smooth horizontal table . Two point masses m and 2m are moving in the same horizontal plane from opposite sides of the bar with speeds 2v and v respectively . The masses stick to the bar after collision at a distance L3and L6  respectively from the centre of the bar. If the bar starts rotating about its center of mass as a result of collision, the angular speed of the bar will be :

 

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a
υ5L 
b
6υ5L 
c
3υ5L 
d
υ6L

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detailed solution

Correct option is C

Here, the two masses crash with the rod and become locked together; as a result, the masses enter the system and the imposed external torque is therefore required to be zero. Therefore, the system's angular momentum must be conserved. We are aware that a body's centre of mass is determined by,

R=mirimi

After the collision, the system's centre of mass will be located around point O.

R=8m×0+2m×L6-m×L38m+2m+m=0
We also know that a body's angular momentum around a point is determined by,

L=mvr=

Now, the system's angular momentum before impact is,

Li=m×2v×L3+2m×v×L6

Li=2mvL3+mvL3=mvL

Let's assume that the system's angular speed upon collision is,
Thus, the system's angular momentum around O after the impact is,

Lf=(Irod+Im+I2m)ω

Now, a rod's moment of inertia around its centre of mass is given by,

I=ML212

Irod=8mL212

The mass m's moment of inertia is Im=mL32=mL29

The mass 2m moment of inertia is  I2m=2mL62=mL218

thus we get,

Lf=(Irod+Im+I2m)ω

Lf=8mL212+mL29+mL218ω=56mL2ω

Thus, Li=Lf

mvL=56mL2ω

ω=6v5L

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