Physics, asked by samsunniharbegum1973, 9 months ago

An ice skater spins with arms outstretched at 1.9 rps.
Her moment of inertia at this instant is 1.33 kg m²?.
She pulls in her arms to increase her rate of spin. If
the moment of inertia is 0.48 kg m-after she pulls in
her arms, what is her new rate of rotation ?

Answers

Answered by phanihrushik890
2
5.26 rps roughly...........


Attachments:
Answered by NirmalPandya
7

Given :

  • Initial moment of inertia (I₁) = 1.33 kg m²
  • Final moment of inertia (I₂) = 0.48 kg m²
  • Initial rate of rotation = 1.9 r.p.s.

To find :

  • Final rate of rotation

Solution :

  • Let initial angular velocity be ω₁ and final angular velocity be ω₂
  • We first have to convert r.p.s. to radians /second.
  • ∴ ω₁ = 1.9 × 2π

           ω₁ = 3.8π rad/sec.

  • We use the principle of conservation of angular momentum.
  • Initial momentum = Final momentum
  • I₁ω₁ = I₂ω₂
  • ∴ ω₂ = I₁ω₁/I₂

                 = (1.33 × 3.8π)/ 0.48

           ω₂ = 10.529π rad/sec

  • This is the final angular velocity.
  • New rate of rotation = ω₂/2π

                                           = 10.529π/2π

                                           = 5.2645 r.p.s.

Answer : The new rate of rotation of the ice skater is 5.2645 r.p.s.

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