The moment of inertia of a solid cylinder about its axis is given by 0.5 mR2. If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is

Question

BlackTom AI · Accepted Answer

v = ωR for rolling without slipping, and rotational kinetic energy K_rot = (1/2) I ω^2 while translational kinetic energy K_trans = (1/2) m v^2.

1. Relevant knowledge points and concepts
- Moment of inertia of a solid cylinder about its symmetry axis: I = (1/2) m R^2.
- Kinetic energies: K_trans = (1/2) m v^2, K_rot = (1/2) I ω^2.
- Rolling without slipping constraint: v = ω R (so ω = v/R).

2. Complete reasoning process step by step
- Substitute I into the rotational KE formula, then eliminate ω using v = ωR to express both energies in terms of v.

3. Calculations with intermediate steps
- I = (1/2) m R^2.
- K_rot = (1/2) I ω^2 = (1/2) × [(1/2) m R^2] × ω^2 = (1/4) m R^2 ω^2.
- Use ω = v/R → ω^2 = v^2 / R^2, so K_rot = (1/4) m R^2 × (v^2 / R^2) = (1/4) m v^2.
- K_trans = (1/2) m v^2.
- Ratio K_rot : K_trans = (1/4 m v^2) : (1/2 m v^2). Cancel common factor m v^2:
  (1/4) : (1/2) = 1/4 ÷ 1/2 = (1/4) × (2/1) = 1/2.
- Expressed as a ratio in simplest integer form: 1 : 2.

4. Why the correct answer is correct
Because for this cylinder I/(mR^2) = 1/2, and the general relation K_rot/K_trans = I/(mR^2) (derived above) gives 1/2. That means the cylinder’s rotational kinetic energy is half its translational kinetic energy, i.e., 1:2.

5. Why the other options are incorrect
- 1:1 would require I = mR^2 (so K_rot = K_trans), but for a solid cylinder I = (1/2)mR^2, not mR^2.
- 2:1 would require I = 2 mR^2, inconsistent with the given I.
- 1:3 would require I = (1/3) mR^2, also inconsistent.
Thus none of those correspond to the given moment of inertia.

6. Conclusion
The ratio of rotational to translational kinetic energy for the given solid cylinder rolling without slipping is 1:2, which matches the provided answer.

The moment of inertia of a solid cylinder about its axis is given by 0.5 mR2. If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is单项选择题

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