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114 學年度臺灣大學轉學生招生考試|第 57 題

普通物理學(A)

Suppose an ice skater is spinning at 0.8 revolutions per second with her arms extended. She has a moment of inertia of 2.34 kg · m2 with her arms extended and of 0.363 kg · m2 with her arms close to her body (appropriate for a 60-kg skater). What is her angular velocity in revolutions per second after she pulls in her arms?
  1. 2.2 rev/s
  2. 3.2 rev/s
  3. 4.2 rev/s
  4. 5.2 rev/s
  5. 6.2 rev/s
提示
No external torque acts during the pull-in, so angular momentum L = Iω is conserved. The smaller moment of inertia must spin faster.
正確答案

正解:D

詳解
A. (A) 2.2 rev/s — incorrect. Too small; this would require the moment of inertia to barely change, contradicting the given values.
B. (B) 3.2 rev/s — incorrect. Still too small relative to the ratio I₁/I₂ ≈ 6.4.
C. (C) 4.2 rev/s — incorrect. Below the conserved-momentum result; does not match I₁ω₁ = I₂ω₂.
D. (D) 5.2 rev/s — correct. Conservation of angular momentum: I₁ω₁ = I₂ω₂, so ω₂ = (2.34/0.363) × 0.8 = 6.45 × 0.8 ≈ 5.16 ≈ 5.2 rev/s.
E. (E) 6.2 rev/s — incorrect. Too large; this overestimates the ratio of the moments of inertia.

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