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Rotational Motion question

2020 · Shift 2 · Q45
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Rotational Motion question

2020 · Shift 2 · Q45

JEE AdvancedPhysicsRotational MotionMultiple correct+4 / −2
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\omegaω about the pivot. The maximum angular speed ω\omegaω M is achieved for x = xM. Then JEE Advanced 2020 Paper 2 Offline Physics - Rotational Motion Question 42 English
  1. A
    ω=3vxL2+3x2\omega = {{3vx} \over {{L^2} + 3{x^2}}}ω=L2+3x23vx​
  2. B
    ω=12vxL2+12x2\omega = {{12vx} \over {{L^2} + 12{x^2}}}ω=L2+12x212vx​
  3. C
    xM=L3{x_M} = {L \over {\sqrt 3 }}xM​=3​L​
  4. D
    ωM=v2L3{\omega _M} = {v \over {2L}}\sqrt 3ωM​=2Lv​3​
View written solutionFree

Correct answer: A, C, D

  1. Use conservation of angular momentum about the pivot during collision

Since the collision is instantaneous, external impulsive torque about the pivot is zero. Hence angular momentum about the pivot is conserved.

  • Bullet mass =m=m=m
  • Rod mass =m=m=m
  • Bullet speed =v=v=v
  • Bullet hits at distance xxx from pivot and embeds

Initial angular momentum about pivot: Li=mvxL_i = mvxLi​=mvx

(Here velocity is horizontal and rod is vertical, so the perpendicular distance is exactly xxx.)

  1. Moment of inertia of the combined system after collision

  • Rod about one end: Irod=13mL2I_{\text{rod}} = \frac{1}{3}mL^2Irod​=31​mL2

  • Embedded bullet treated as point mass at distance xxx: Ibullet=mx2I_{\text{bullet}} = mx^2Ibullet​=mx2

So total moment of inertia is I=13mL2+mx2I = \frac{1}{3}mL^2 + mx^2I=31​mL2+mx2

  1. Find angular speed ω\omegaω just after collision

By angular momentum conservation, mvx=Iω=(13mL2+mx2)ωmvx = I\omega = \left(\frac{1}{3}mL^2 + mx^2\right)\omegamvx=Iω=(31​mL2+mx2)ω

Cancelling mmm: vx=(L23+x2)ωvx = \left(\frac{L^2}{3}+x^2\right)\omegavx=(3L2​+x2)ω

Therefore,

= \frac{3vx}{L^2+3x^2}$$ So **Option A is correct**. Option B gives a different expression, so **B is incorrect**. --- 4. **Maximize $\omega$ with respect to $x$** We need to maximize $$\omega(x)=\frac{3vx}{L^2+3x^2}$$ Since $3v$ is constant, maximize $$f(x)=\frac{x}{L^2+3x^2}$$ Differentiate: $$f'(x)=\frac{(L^2+3x^2)-x(6x)}{(L^2+3x^2)^2} =\frac{L^2-3x^2}{(L^2+3x^2)^2}$$ Set numerator zero for extremum: $$L^2-3x^2=0$$ $$3x^2=L^2$$ $$x=\frac{L}{\sqrt3}$$ Thus, $$x_M=\frac{L}{\sqrt3}$$ So **Option C is correct**. --- 5. **Find maximum angular speed $\omega_M$** Substitute $x=\dfrac{L}{\sqrt3}$ into $$\omega = \frac{3vx}{L^2+3x^2}$$ Now, $$x^2=\frac{L^2}{3}$$ Hence denominator: $$L^2+3x^2=L^2+3\cdot\frac{L^2}{3}=2L^2$$ Numerator: $$3vx=3v\cdot\frac{L}{\sqrt3}=\sqrt3\,vL$$ Therefore, $$\omega_M=\frac{\sqrt3\,vL}{2L^2}=\frac{\sqrt3\,v}{2L}$$ So **Option D is correct**. --- 6. **Final option check** - **A:** Correct - **B:** Incorrect - **C:** Correct - **D:** Correct Thus the correct options are: $$\boxed{A,\ C,\ D}$$
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