114 學年度臺灣大學轉學生招生考試｜第 64 題
普通物理學(A)
- 5.2 m/s
- 3.5 m/s
- 2.4 m/s
- 1.7 m/s
- 0.2 m/s
提示
A rolling disk stores energy in both translation and rotation, so its total kinetic energy is ¾mv². Also remember the disk already has speed at the top, and convert 18 cm and 80 cm/s to SI units.
正確答案
正解：D
詳解
A. (A) 5.2 m/s — incorrect. Far too fast; this ignores that a rolling body's KE is ¾mv² (extra rotational share) and that the drop is only 18 cm.
B. (B) 3.5 m/s — incorrect. Still too fast for an 18 cm drop with the rolling constraint.
C. (C) 2.4 m/s — incorrect. This is what you would get from a sliding (non-rolling) block using v = √(v₀² + 2gh); rolling reduces the speed gain.
D. (D) 1.7 m/s — correct. For a rolling solid disk, KE = ½mv² + ½Iω² = ¾mv². Energy conservation: ¾v² = ¾v₀² + gh, so v = √(v₀² + (4/3)gh) = √(0.8² + (4/3)(9.8)(0.18)) = √(0.64 + 2.35) ≈ 1.73 m/s.
E. (E) 0.2 m/s — incorrect. Far too slow; the disk speeds up as it descends, it cannot drop below its initial 0.8 m/s.
