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

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

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

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

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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 𝜔_𝑀 is achieved for 𝑥 = 𝑥_𝑀. Then

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$ω = \frac{3 v x}{L^{2} + 3 x^{2}}$

$ω = \frac{12 v x}{L^{2} + 12 x^{2}}$

$x_{M} = \frac{L}{\sqrt{3}}$

$ω_{M} = \frac{V}{2 L} \sqrt{3}$

answer is A, C, D.

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Detailed Solution

by the angular momentum conservation about the suspension point.

$\begin{array}{l} mvx = (\frac{m ℓ^{2}}{3} + {mx}^{2}) ω \\ ∴ ω = \frac{mvx}{\frac{m ℓ^{2}}{3} + {mx}^{2}} = \frac{3 vx}{ℓ^{2} + 3 x^{2}} \\ For maximum ω \Rightarrow \frac{d ω}{dx} = 0 \\ \Rightarrow 3 v \frac{d}{d x} (\frac{x}{ℓ^{2} + 3 x^{2}}) = 0 \\ \Rightarrow \frac{d}{d x} (\frac{1}{\frac{ℓ^{2}}{x} + \frac{3 x^{2}}{x}}) = 0 \\ \Rightarrow \frac{d}{dx} {(\frac{ℓ^{2}}{x} + 3 x)}^{- 1} = 0 \\ \Rightarrow (- 1) {(\frac{ℓ^{2}}{x} + 3 x)}^{- 2} (- \frac{ℓ^{2}}{x^{2}} + 3) = 0 \\ \Rightarrow \frac{- ℓ^{2}}{x^{2}} + 3 = 0 \\ ∴ x_{M} = \frac{ℓ}{\sqrt{3}} \\ T h u s, ω_{max} = \frac{3 v (\frac{ℓ}{\sqrt{3}})}{ℓ^{2} + 3 (ℓ^{2} / 3)} = \frac{\sqrt{3} V}{2 ℓ} \\ So the ω = \frac{V}{2 ℓ} \sqrt{3} \end{array}$