A rod of mass and length , 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 angula

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 $ω_{M}$ is achieved for $x = x_{M}$ . Then [2020]

Smart Achievers · Accepted Answer

(1, 3, 4)

From angular momentum conservation about the pivoted point,

$m v x = (\frac{m L^{2}}{3} + m x^{2}) ω$

[As the combined system rotates with angular speed $ω$ about the pivot]

$∴$ $ω = \frac{m v x}{\frac{m L^{2}}{3} + m x^{2}} = \frac{3 v x}{L^{2} + 3 x^{2}}$

$∴$ $ω = \frac{3 v x}{L^{2} + 3 x^{2}}$

Hence option (1) is correct.

For maximum angular velocity, $\frac{d ω}{d x} = 0$

$\frac{d}{d x} (\frac{L^{2}}{x} + 3 x) = 0 \Rightarrow \frac{L^{2}}{x^{2}} + 3 = 0 \Rightarrow x = \frac{L}{\sqrt{3}}$

$∴$ $x_{m} = \frac{L}{\sqrt{3}}$

So option (3) is correct.

$ω_{m} = \frac{3 v x}{L^{2} + 3 x^{2}} = \frac{3 v \cdot \frac{L}{\sqrt{3}}}{L^{2} + 3 {(\frac{L}{\sqrt{3}})}^{2}} = \frac{\sqrt{3}}{2 L} V$

Hence option (4) is correct.

Solution

1 $ω = \frac{3 v x}{L^{2} + 3 x^{2}}$

2 $ω = \frac{12 v x}{L^{2} + 12 x^{2}}$

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

4 $ω_{M} = \frac{v}{2 L} \sqrt{3}$

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Solution

1 ω=3vxL2+3x2

2 ω=12vxL2+12x2

3 xM=L3

4 ωM=v2L3

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1 $ω = \frac{3 v x}{L^{2} + 3 x^{2}}$

2 $ω = \frac{12 v x}{L^{2} + 12 x^{2}}$

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

4 $ω_{M} = \frac{v}{2 L} \sqrt{3}$