JEE Challenger
More from System of Particles and Rotational Motion
PhyMedAdv 2020 • Paper 2

Angular Speed of Pivoted Rod Hit by Embedded Bullet

A uniform rod of mass mm and length LL, suspended vertically from a top pivot, is struck horizontally by a bullet of identical mass mm moving with speed vv at a distance xx from the pivot. The bullet gets embedded into the rod. The combined assembly begins to rotate with angular speed ω\omega around the pivot. If the peak angular speed ωM\omega_M is attained when x=xMx = x_M, select the correct option(s):

Question Diagram 1

Options

A

ω=3vxL2+3x2\omega = \frac{3vx}{L^2 + 3x^2}

Correct
B

ω=12vxL2+12x2\omega = \frac{12vx}{L^2 + 12x^2}

C

xM=L3x_M = \frac{L}{\sqrt{3}}

Correct
D

ωM=v2L3\omega_M = \frac{v}{2L}\sqrt{3}

Correct

Topics & Concepts

System of Particles and Rotational MotionAngular MomentumMoment of InertiaCollision

Step-by-Step Solution

To determine the correct statements, we analyze the collision of the bullet with the pivoted rod using the principle of conservation of angular momentum about the pivot.

1. Conservation of Angular Momentum

Since no external horizontal torque acts on the system about the pivot during the collision, the total angular momentum about the pivot is conserved.

  • Initial Angular Momentum (LiL_i): Only the bullet contributes to the angular momentum about the pivot before the collision: Li=mvxL_i = m v x

  • Moment of Inertia of the Combined System (II): The system consists of the rod of mass mm and length LL, and the embedded bullet of mass mm at distance xx from the pivot: I=Irod+Ibullet=13mL2+mx2I = I_{\text{rod}} + I_{\text{bullet}} = \frac{1}{3} m L^2 + m x^2

  • Final Angular Momentum (LfL_f): Lf=Iω=(13mL2+mx2)ωL_f = I \omega = \left(\frac{1}{3} m L^2 + m x^2\right) \omega

Equating LiL_i and LfL_f: mvx=m(L23+x2)ωm v x = m \left(\frac{L^2}{3} + x^2\right) \omega

vx=(L2+3x23)ωv x = \left(\frac{L^2 + 3x^2}{3}\right) \omega

ω=3vxL2+3x2\omega = \frac{3v x}{L^2 + 3x^2}

This confirms that Option (A) is correct and Option (B) is incorrect.

2. Maximum Angular Speed (ωM\omega_M) and Corresponding Distance (xMx_M)

To find the distance x=xMx = x_M that maximizes the angular speed ω\omega, we differentiate ω\omega with respect to xx and set the derivative to zero:

dωdx=3v⋅(L2+3x2)⋅1−x⋅(6x)(L2+3x2)2=0\frac{d\omega}{dx} = 3v \cdot \frac{(L^2 + 3x^2) \cdot 1 - x \cdot (6x)}{(L^2 + 3x^2)^2} = 0

L2+3x2−6x2=0L^2 + 3x^2 - 6x^2 = 0

L2−3x2=0  ⟹  xM=L3L^2 - 3x^2 = 0 \implies x_M = \frac{L}{\sqrt{3}}

Since xM=L3≈0.577Lx_M = \frac{L}{\sqrt{3}} \approx 0.577L, it lies within the physical range of the rod (0≤x≤L0 \le x \le L).

This confirms that Option (C) is correct.

3. Value of Maximum Angular Speed (ωM\omega_M)

Substituting xM=L3x_M = \frac{L}{\sqrt{3}} into the expression for ω\omega:

ωM=3v(L3)L2+3(L3)2=3vLL2+L2=3vL2L2=v2L3\omega_M = \frac{3v \left(\frac{L}{\sqrt{3}}\right)}{L^2 + 3\left(\frac{L}{\sqrt{3}}\right)^2} = \frac{\sqrt{3} v L}{L^2 + L^2} = \frac{\sqrt{3} v L}{2L^2} = \frac{v}{2L}\sqrt{3}

This confirms that Option (D) is correct.

Conclusion

The correct options are A, C, and D.

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