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