Unit 6 · Topic 04

Rotational energy

Rotational kinetic energy sums the motion of all particles in a rigid body.

Unit 6Rotational energyOpen navigation

Concept 01

Energy stored in rotation

Essential The minimum you should retain

K_rot=(1/2)Iω2 K_{rot}=\frac12I\omega^2 .

UnderstandInterpret and connect

At fixed ω, larger I means larger energy; doubling ω quadruples K_rot.

DeepenFormulation and conditions

I and ω must refer to the same axis.

ExploreConnections for further study

Mass distribution allows different stored energy for equal mass and outer radius.

Mathematical relation

Rotational kinetic energy

Krot=12Iω2;K=12Mvcm2+12Icmω2 K_{rot}=\tfrac12I\omega^2;\ K=\tfrac12Mv_{cm}^2+\tfrac12I_{cm}\omega^2
Represents

Rotation energy and translation–rotation composition.

Physical interpretation

The I_cm decomposition avoids double counting.

DeepenVariables, conditions, and checks

Variables

K
kinetic energy; usual unit: J

Conditions of application

  • Rigid body; I matches the axis used.

Dimensional check

J.

Two bars compare rotational K for small and large inertia.I_A, K_AI_B, K_Bsame ωI_B>I_A

At the same ω, the distribution with larger I stores more rotational kinetic energy.

Concept 02

Two contributions

Essential The minimum you should retain

A translating and rotating body may have K=(1/2)Mv_cm2+(1/2)I_cmω2 K=\frac12Mv_{cm}^2+\frac12I_{cm}\omega^2 .

UnderstandInterpret and connect

Using the center of mass avoids counting motion twice.

DeepenFormulation and conditions

Both terms are nonnegative and measured in joules.

ExploreConnections for further study

Choosing another axis requires a carefully revised balance.

Concept 03

Changing energy

Essential The minimum you should retain

Net torque work changes K_rot in the appropriate axial model.

UnderstandInterpret and connect

Energy can determine angular-speed changes without time.

DeepenFormulation and conditions

W_net=ΔK_rot connects initial and final states.

ExploreConnections for further study

Conservative forces may be included through potential energy.

Worked example

Work speeds up a flywheel

A flywheel with I=0.80 kg·m² starts at rest and receives 48 J.

Given
  • I=0.80 kg·m²
  • W_net=48 J
  • ω_i=0
Target

Find ω_f.

  1. Energy

    W_net=ΔK_rot.

  2. Substitution

    48=(1/2)(0.80)ω².

  3. Solve

    ω²=120.

  4. Result

    ω=10.95 rad/s.

Conclusion

ω_f≈10.95 rad/s.

Concept 04

Same mass, different response

Essential The minimum you should retain

Wheels with equal M and R may have different I and K_rot.

UnderstandInterpret and connect

Mass location controls how hard ω is to change.

DeepenFormulation and conditions

At fixed L, K=L²/(2I) shows another dependence on I.

ExploreConnections for further study

Energy and angular momentum are conserved under different conditions.

Concept review

Common errors

Each warning includes a concrete way to review the reasoning, not only an incorrect-answer marker.

Using I without specifying its axis.

I and K_rot must refer to the same axis.

Using contact-point I and adding CM translation without justification.

The standard split uses I_cm plus CM translation.