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 an angular speed ω about the pivot. The…

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 an angular speed ω about the pivot. The maximum angular speed ω M is achieved for x = x M . Then

Aicharya · Accepted Answer

The net torque on the system (rod+bullet) will be zero about hinge point. Therefore, we can apply the principle of angular momentum conservation about hinge point.

From angular momentum conservation,

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

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

For maximum angular velocity,

$\frac{d ω}{d x} = 0$

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

$\Rightarrow ω_{\max} = \frac{3 v}{(\frac{\sqrt{3} L^{2}}{L} + \sqrt{3} L)} = \frac{\sqrt{3} v}{2 L}$

A rod of mass m and length L , pivoted at one of its ends, is hanging vertically. A bullet of the same mass…

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