Question

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 ?

Answer 1

5.26 rps roughly...........

Answer 2

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.