Energy of a car’s wheels while moving — what forms are present?

Difficulty: Easy

Correct Answer: Kinetic energy of translation and rotation both

Explanation:


Introduction / Context:
Rolling motion combines translation of the center of mass and rotation about that center. Recognizing both energy contributions is critical for braking, fuel economy, and vehicle dynamics analyses.



Given Data / Assumptions:

  • Wheels are rigid and roll without slipping.
  • Vehicle moves with speed v; wheel has angular speed ω with v = ω r.
  • Gravitational potential energy changes are not under consideration on level ground.


Concept / Approach:
Total kinetic energy of a rolling rigid wheel equals translational kinetic energy of its center of mass plus rotational kinetic energy about its center. Thus, KE_total = (1/2) m v^2 + (1/2) I ω^2. For pure rolling, v and ω are linked through the radius r.



Step-by-Step Solution:

Write translational KE: K_trans = (1/2) m v^2.Write rotational KE: K_rot = (1/2) I ω^2.Relate rolling: v = ω r; substitute if needed to compare magnitudes.Add contributions: K_total = (1/2) m v^2 + (1/2) I ω^2.


Verification / Alternative check:
Special case: A thin hoop (I = m r^2) gives K_total = (1/2) m v^2 + (1/2) m r^2 (v^2 / r^2) = m v^2, demonstrating substantial rotational contribution.



Why Other Options Are Wrong:

  • Potential energy only: Irrelevant on level motion; energy form is kinetic.
  • Translation only / Rotation only: Neglect one essential component of rolling motion.
  • No energy: Rolling without slipping does not eliminate kinetic energy; it simply avoids sliding friction losses.


Common Pitfalls:
Forgetting rotational inertia effects; underestimating braking distances by ignoring rotational energy of wheels and drivetrain components.



Final Answer:
Kinetic energy of translation and rotation both


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