Q132 / 160 · TS EAMCET 2021 (05 Aug Shift 2) · Physics

A circular hoop of radius 50 cm and mass 1 kg rotating with an angular velocity ω0 is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. Let 'v' be the velocity of the centre of the hoop when it ceases to slip. The ratio v/ω0 will be

  • A
    10 cm
  • B
    50 cm
  • Correct answer
    C
    25 cm
  • D
    12.5 cm

Solution

Initial angular momentum of the loop about contact point is, Li=mr2ω0
Final angular momentum about contact point is, Lf=mrv+mr2ω
⇒Lf=mrv+mr2vr
⇒Lf=mrv+mrv=2mrv

Since torque bout contact point is zero, so Li=Lf
⇒mr2ω0=2mrv
⇒rω0=2v
⇒vω0=r2=502=25 cm

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

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