IIT JEE , AIEEE Forum - Physics , Mechanics

pinnaka

Here is the solution:
here is how it goes
Let us calculate the center of mass when the particle strikes and stikes to the rod.
We will callcuate the center of mass from the end where the particle collide
x_cm = m(l/2)+m(0)/2m
x_cm = 1/4l from the end considered
Now we have to apply conservation of angular momentum about the x_cm. There is rule hich says this
mv(l/2) = I*omega+ mvr
since the center of mass upon collision is going to move away from it mvr= 0
mv(l/2) = I*omega----1
I = I_cm+md^2+I_particle
I = 1/3ml^2+ml^2/16 +ml^2/16-----2
now plug 2 in 1 to get the required answer for omega

----Hope this helps

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