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

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 ωM is achieved for x=xM. Then (A) ω=(3 vX / L 2+3 x 2) (B) ω=(12 vx / L 2+12 x 2) (C) xM=(L/√3) (D) ω M =( V /2 L ) √3

Tardigrade Admin · Accepted Answer

Conserving angular momentum about pivot. ⇒ mvx =(( mL 2/3)+ mx 2) ω ⇒ ω=(3 vx / L 2+3 x 2) For ω to be maximum ⇒ (d ω/d x)=0, this gives x=(L/√3) and ω max =(√3 v/2 L)

$ω = \frac{3 v X}{L ^{2} + 3 x ^{2}}$

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

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

$ω_{M} = \frac{V}{2 L} 3$

Conserving angular momentum about pivot.
$\Rightarrow m vx = (\frac{m L ^{2}}{3} + m x^{2}) ω$
$\Rightarrow ω = \frac{3 vx}{L ^{2} + 3 x ^{2}}$
For $ω$ to be maximum
$\Rightarrow \frac{d ω}{d x} = 0$ ,
this gives $x = \frac{L}{3}$
and $ω_{m a x} = \frac{3 v}{2 L}$

ω=L2+3x23vX​

ω=L2+12x212vx​

xM​=3​L​

ωM​=2LV​3​

Conserving angular momentum about pivot. ⇒mvx=(3mL2​+mx2)ω ⇒ω=L2+3x23vx​ For ω to be maximum ⇒dxdω​=0, this gives x=3​L​ and ωmax​=2L3​v​

$ω = \frac{3 v X}{L ^{2} + 3 x ^{2}}$

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

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

$ω_{M} = \frac{V}{2 L} 3$

Conserving angular momentum about pivot.
$\Rightarrow m vx = (\frac{m L ^{2}}{3} + m x^{2}) ω$
$\Rightarrow ω = \frac{3 vx}{L ^{2} + 3 x ^{2}}$
For $ω$ to be maximum
$\Rightarrow \frac{d ω}{d x} = 0$ ,
this gives $x = \frac{L}{3}$
and $ω_{m a x} = \frac{3 v}{2 L}$