Q74 / 90 · JEE Main 2019 (09 Apr Shift 2) · Physics

A thin smooth rod of length L and mass M is rotating freely with angular speed ω0 about an axis perpendicular to the rod and passing through center. Two beads of mass m and negligible size are at the center of the rod initially. The beads of mass m and negligible size are at the center of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod, will be:

  • A
    M ω0M+3m
  • B
    M ω0M+2m
  • Correct answer
    C
    M ω0M+6m
  • D
    M ω0M+m

Solution

From conservation of angular momentum.
ML212ω0=ML212+mL22+mL22ω'
ML212ω0=ML212+mL22ω'
ω'=Mω0M+6m

Hence, the correct answer is option: 'C'.

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