Energy of a car’s wheels while moving — what forms are present?
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AKinetic energy of translation and rotation both
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BPotential energy only
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CKinetic energy of translation only
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DKinetic energy of rotation only
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ENo energy because wheels roll without slipping
Answer
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