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

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6.System of Particles and Rotational Motion

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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 $\omega$ about the pivot. The maximum angular speed $\omega_M$ is achieved for $x=x_M$. Then
$(A)$ $\omega=\frac{3 v x}{ L ^2+3 x^2}$
$(B)$ $\omega=\frac{12 v x}{L^2+12 x^2}$
$(C)$ $x_M=\frac{L}{\sqrt{3}}$
$(D)$ $\omega_M=\frac{v}{2 L} \sqrt{3}$

A$A,B,C$

B$A,B,D$

C$A,C,D$

D$A,C$

(IIT-2020)

Solution

by the angular momentum conservation about the suspension point.
$mvx =\left(\frac{ m \ell^2}{3}+ mx ^2\right) \omega$
$\therefore \omega=\frac{ mvx }{\frac{ m \ell^2}{3}+ mx ^2}=\frac{2 vx }{\ell^2+3 x }$
For maximum $\omega \Rightarrow \frac{ d \omega}{ dx }=0$
$\therefore x _{ M }=\frac{\ell}{\sqrt{3}}$
So the $\omega=\frac{ V }{2 \ell} \sqrt{3}$

Standard 11

Physics

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