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Question

A uniform rod having mass m and length L lies on a horizontal surface. A particle of mass m2 moving with speed u strikes the rod perpendicularly at one end and sticks to the rod as shown in the figure. Find the total angular momentum of the system about the origin just after the collision, if the angular velocity of the rod just after the collision is ω about its centre.


A
[muL3+mL2ω2](^k)
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B
[muL3mL2ω2](^k)
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C
Zero
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D
[muL3mL2ω](^k)
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Solution

The correct option is A [muL3+mL2ω2](^k)
COM of the (rod + particle) system w.r.t point A, just after collision is
ycom=m1y1+m2y2m1+m2=m(L2)+m2(0)m+m2
ycom=L3
Angular momentum (just after collision) about point B,
Lf=LTrans+LRot
Lf=M(r0×v0)+Lcom
Lf=(m+m2)[2L3 ^j×v ^i]+Icomω(^k) ...(1)
[ from right hand rule we obtain, direction of ω along ^k]


On the system of (rod+particle), Fext=0. Applying linear momentum conservation
i.e Pi=Pf
m2u=(m+m2)v
v=u3 ...(2)
& MOI of system about its COM,
Icom=Irod+Iparticle
where from parallel axis theorem, Irod=[mL23+m(L3)2]
Icom=[mL23+m(L3)2]+m2(L3)2
Icom=mL22 ...(3)
From Eq. (1), (2), (3),

Lf=3m2[2Lu9 ^k]+mL22×ω(^k)
Lf=[muL3+mL2ω2](^k)

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