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
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a
ω=3vxL2+3x2
b
ω=12vxL2+12x2
c
xM=L3
d
ωM=V2L3
answer is A.
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Detailed Solution
by the angular momentum conservation about the suspension point.mvx=mℓ23+mx2ω∴ω=mvxmℓ23+mx2=3vxℓ2+3x2 For maximum ω⇒dωdx=0⇒3vddxxℓ2+3x2=0⇒ddx1ℓ2x+3x2x=0⇒ddxℓ2x+3x−1=0⇒(−1)ℓ2x+3x−2−ℓ2x2+3=0⇒-ℓ2x2+3=0∴xM=ℓ3Thus, ωmax=3vℓ3ℓ2+3ℓ2/3=3V2ℓ So the ω=V2ℓ3