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Question

# A rod of mass m and length L, pivoted at one of its ends, is hanging vertically. A bullet of the same mass moving at speed v strikes the rod horizontally at a distance x from its pivoted end and gets embedded in it. The combined system now rotates with angular speed ω about the pivot. The maximum angular speed ωM is achieved for x=xM.Then

A
ω=3vxL2+3x2
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B
ω=12vxL2+12x2
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C
xM=L3
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D
ωM=v2L3
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Solution

## The correct option is D ωM=v2L√3When bullet strikes the rod there is no external torque acting the system of rod & bullet (τext=0) By the angular momentum conservation about the suspension point, Li=Lf ⇒mvr⊥=Iω ∵r⊥=x, I=mL23+mx2 ⇒mvx=(mL23+mx2)ω ∴ω=mvxmL23+mx2=3vxL2+3x2 For ′ω′ to be maximum ⇒dωdx=0 ⇒(L2+3x2)(3v)−(3vx)(0+6x)(L2+3x2)2=0⇒3vL2+9vx2−18vx2=0⇒3vL2=9vx2 ⇒x=√L23=L√3 ⇒xM=L√3 ωm=3v×L√3L2+3×L23=√3vL2L2 ⇒ωm=v2L√3 Thus, options (A), (C) and (D) are correct.

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