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MHT CET 2019 (2 May, Shift 1) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from MHT CET 2019 (2 May, Shift 1)

Q65MHT CET 2019 · 2 MAY · SHIFT 1Difficult

Angular Momentum & its Conservation

A uniform rod of length ' 6 L ' and mass ' 8 m ' is pivoted at its centre ' CC '. Two masses ' mm ' and ' 2m2m with speed 2v,v2v, v as shown strikes the rod and stick to the rod. Initially the rod is at rest. Due to impact, if it rotates with angular velocity ' ω\omega ' then ' ω\omega ' will be

Figure

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Answer: (a)

  1. Rod: Irod=112(8m)(6L)2=24mL2I_{rod} = \frac{1}{12}(8m)(6L)^2 = 24mL^2.
  2. From the figure both masses give anticlockwise angular momentum about C: L=2m(v)(L)+m(2v)(2L)=6mvLL = 2m(v)(L) + m(2v)(2L) = 6mvL.
  3. After sticking: I=24mL2+2mL2+m(2L)2=30mL2I = 24mL^2 + 2m L^2 + m(2L)^2 = 30mL^2.
  4. ω=6mvL30mL2=v5L\omega = \frac{6mvL}{30mL^2} = \frac{v}{5L}
  5. So the answer is (a) v5L\frac{v}{5L}.
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