Consider a motorcycle of mass 150 kg, one wheel of which has a mass of 10 kg and a radius of 30 cm. What is the ratio of the rotational kinetic energy of the wheels to the total translational kinetic energy of the bike? Assume the wheels are uniform discs.

Question

BlackTom AI · Accepted Answer

For a rolling uniform disc, the rotational kinetic energy is related to the translational speed by ω = v/R, and the moment of inertia is I = (1/2) m_w R^2.

1) Relevant formulas and concepts
- Rotational KE of one wheel: K_rot = (1/2) I ω^2.
- For a uniform disc: I = (1/2) m_w R^2.
- Rolling without slipping: ω = v/R → ω^2 = v^2 / R^2.
- Translational KE of the motorcycle (given mass M = 150 kg): K_trans = (1/2) M v^2.
- There are two wheels, so include rotational energy of both.

2) Step-by-step calculation
- Rotational KE of one wheel:
  I = (1/2) m_w R^2 → K_rot,one = (1/2) I ω^2 = (1/2)*(1/2 m_w R^2)*(v^2 / R^2)
                     = (1/4) m_w v^2.
- Total rotational KE of two identical wheels:
  K_rot,total = 2 * K_rot,one = 2 * (1/4 m_w v^2) = (1/2) m_w v^2.
- Substitute wheel mass m_w = 10 kg:
  K_rot,total = (1/2) * 10 * v^2 = 5 v^2 (in joules units implicit).
- Translational KE of the bike (mass M = 150 kg):
  K_trans = (1/2) M v^2 = (1/2) * 150 * v^2 = 75 v^2.
- Ratio (rotational of wheels) : (translational of bike) =
  K_rot,total / K_trans = (5 v^2) / (75 v^2) = 5 / 75 = 1 / 15 ≈ 0.066666... ≈ 0.067:1.

3) Why this is correct
- The R^2 factors cancel because I contains R^2 and ω^2 brings a 1/R^2, so the result depends only on masses, not on v or R for uniform discs rolling without slipping.
- Two wheels were included because the problem asks for the rotational kinetic energy of the wheels (plural) for the motorcycle.

4) Why the other choices are incorrect
- 0.033:1 would correspond to including only one wheel’s rotation (half the correct value).
- 0.33:1 and 0.67:1 are much larger and would require unrealistically larger wheel mass fractions (e.g., m_w comparable to the whole bike mass), inconsistent with the given numbers.

Conclusion: The calculated ratio is 1/15 ≈ 0.067:1, which matches the provided correct answer 0.067:1.

Consider a motorcycle of mass 150 kg, one wheel of which has a mass of 10 kg and a radius of 30 cm. What is the ratio of the rotational kinetic energy of the wheels to the total translational kinetic energy of the bike? Assume the wheels are uniform discs.单项选择题

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