Kinetic energy of rolling objects
A rolling object has both translational and rotational kinetic energy.
The common mistake
Setting mgh = ½mv² for a ball rolling down a ramp and forgetting the rotational energy, so the speed comes out too high.
Instead: For rolling without slipping, total KE = ½mv² + ½Iω², with ω = v/r. Some of the energy goes into spinning, so a rolling object is slower at the bottom than a sliding one.
Worked example with Guide me
A solid cylinder (I = ½mr²) rolls from rest down a 1.2 m high ramp. Find v at the bottom (g = 10 m/s²).
- Sugar · What's the rotational KE in terms of v?Student · ½ · ½mr² · (v/r)² = ¼mv².
- Sugar · So total KE is?Student · ½mv² + ¼mv² = ¾mv².
- Sugar · Set mgh = ¾mv². What's v?Student · v² = 4gh/3 = 16, so v = 4 m/s.
Answer: 4 m/s (sliding without friction would give about 4.9 m/s).
Try 3 practice questions
1. A hoop (I = mr²) rolls. What fraction of its KE is rotational?
2. Rotational KE of I = 2 kg·m² spinning at 3 rad/s?
3. Which reaches the bottom first: a solid sphere or a hoop?
Quick answers
- What is the most common mistake with kinetic energy of rolling objects?
- Setting mgh = ½mv² for a ball rolling down a ramp and forgetting the rotational energy, so the speed comes out too high.
- How do I get kinetic energy of rolling objects right?
- For rolling without slipping, total KE = ½mv² + ½Iω², with ω = v/r. Some of the energy goes into spinning, so a rolling object is slower at the bottom than a sliding one.
- Can you show a worked example?
- A solid cylinder (I = ½mr²) rolls from rest down a 1.2 m high ramp. Find v at the bottom (g = 10 m/s²). 4 m/s (sliding without friction would give about 4.9 m/s).
Related concepts
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