Rotational Kinetic Energy: Study Guide (Physics C: Mechanics 6.1)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 6.1 called Rotational Kinetic Energy. This guide is the calculus-based one. It builds K = ½Iω² by integrating over a continuous body and proves the translation-plus-rotation split. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

Where the energy of a spinning body lives

Picture a grinding wheel spinning on a fixed axle. Its centre of mass does not move. Yet press a tool against it and sparks fly: energy is leaving the wheel. So the wheel had kinetic energy.

The energy is in the pieces. Every small piece of the wheel moves in a circle, so every piece has a speed and an ordinary kinetic energy. Pieces on opposite sides move in opposite directions, so their momenta cancel and the centre of mass stays still. Their kinetic energies do not cancel, because kinetic energy is a scalar that is never negative.

From ½v² dm to ½Iω²

Model a rigid body as many small pieces of mass dm. The body turns about a fixed axis with angular velocity ω. Every piece shares the same ω. A piece a perpendicular distance r from the axis moves in a circle with speed

v = rω (ω in rad/s)

Its kinetic energy is dK = ½v² dm = ½r²ω² dm. Add (integrate) over the whole body:

K = ∫½v² dm = ∫½r²ω² dm

ω is the same for every piece, so it comes outside the integral:

K = ½ω² ∫r² dm

The integral is the rotational inertia about the axis from Topic 5.4, I = ∫r² dm. So:

K = ½Iω²

Three things follow from this derivation.

1. It is the whole kinetic energy. For rotation about a fixed axis, ½Iω² is just the sum of ½v² dm for every piece. It is not extra energy on top of a ½mv² term.
2. The unit is the joule. kg·m² × (rad/s)² = kg·m²/s² = J. The radian is dimensionless and drops out. But ω must be in rad/s, because v = rω only holds in radians.
3. Distant mass dominates. Each piece contributes in proportion to r² dm. Double the distance of a piece from the axis and its share of the energy goes up four times.

The same argument works for a collection of point objects, where the integral becomes a sum: K = ½ω² Σmᵢrᵢ².

Moving and spinning: translation plus rotation

Many bodies move and spin at once: a thrown baton, a tumbling satellite, a wheel skidding along a road. For any rigid body, the total kinetic energy splits into two scalar parts:

K_total = ½Mv_cm² + ½I_cm ω²

• ½Mv_cm² is the translational part: the whole mass M moving with the velocity of the centre of mass.
• ½I_cm ω² is the rotational part: spinning about an axis through the centre of mass, using I_cm.

Why the split works

Write the velocity of each piece as the centre-of-mass velocity plus its velocity relative to the centre of mass: v = v_cm + v′. Then

v² = v_cm² + 2v_cm·v′ + v′²

Integrate ½v² dm over the body:

K = ½Mv_cm² + v_cm·∫v′ dm + ½∫v′² dm

• The middle integral is zero. ∫v′ dm is the total momentum measured in the centre-of-mass frame, and by the definition of the centre of mass that is zero.
• In the last integral, each piece circles the centre of mass, so v′ = r′ω, where r′ is its distance from an axis through the centre of mass. That integral is ½ω²∫r′² dm = ½I_cm ω².

So the cross term vanishes, and K = ½Mv_cm² + ½I_cm ω² exactly. There is no “mixed” energy.

A fixed axis that is not through the centre of mass

A rod swinging about one end turns about a fixed axis, so K = ½I_pivot ω² is already the total. You can also split it: the centre of mass moves at v_cm = dω, where d is its distance from the pivot, and the rod turns about its centre of mass at the same ω. The parallel axis theorem, I_pivot = I_cm + Md², shows that the two methods agree:

½I_pivot ω² = ½(I_cm + Md²)ω² = ½I_cm ω² + ½M(dω)² = ½I_cm ω² + ½Mv_cm²

Use one method or the other. Adding ½I_pivot ω² and ½Mv_cm² counts the centre-of-mass motion twice.

Kinetic energy with the centre of mass at rest

If v_cm = 0, the translational part is zero, but the rotational part need not be. A flywheel on a fixed axle through its centre of mass has total momentum zero and kinetic energy ½I_cm ω². This is how flywheels store energy: the energy sits in the moving rim while the body as a whole goes nowhere.

The reason is the difference between a vector and a scalar. Momentum is a vector, so the momenta of opposite sides cancel. Kinetic energy is a scalar that is never negative, so the contributions of all the pieces add up.

A scalar, never negative

Because ω is squared, a wheel turning clockwise at 20 rad/s has exactly the same rotational kinetic energy as the same wheel turning counterclockwise at 20 rad/s. Two identical wheels spinning in opposite directions have a total kinetic energy twice that of one wheel, not zero. When you add rotational kinetic energies, you never need a sign convention.

Graphs and comparisons

K = ½Iω² gives you functional dependences to reason with:

• K against ω for a fixed body is a parabola through the origin, opening upward. Doubling ω multiplies K by 4.
• K against I at fixed ω is a straight line through the origin. A thin hoop has twice the rotational inertia of a uniform disc of the same mass and radius about their central axes (MR² against ½MR²), so at the same ω it has twice the kinetic energy.
• Constant angular acceleration from rest: ω = αt, so K = ½Iα²t². K against t is a parabola starting flat at the origin. In terms of angle, ω² = 2αθ, so K = Iαθ: K against θ is a straight line. That straight line is a first hint of Topic 6.2, where a torque does work through an angle.

When a question asks you to compare two scenarios, write K = ½Iω² for each, cancel what is the same, and state the ratio with a reason, not just the equation.

Worked example 1: a rod with non-uniform density

Question. A thin rod of length L = 0.60 m lies along the x-axis from a pivot at x = 0. Its linear mass density is λ(x) = Cx with C = 5.0 kg/m², so it gets heavier towards the far end (Figure 1). It turns about the pivot, in the plane of the page, at ω = 4.0 rad/s. (a) Find the rod’s kinetic energy by integrating ½v² dm. (b) Check the answer by splitting it into translation and rotation. (c) What fraction of the kinetic energy is in the outer half of the rod?

(a) Integrate directly.

1. The element at x has mass dm = λ dx = Cx dx and speed v = xω.
2. dK = ½(xω)² Cx dx = ½Cω²x³ dx.
3. K = ½Cω² ∫₀ᴸ x³ dx = ½Cω² × L⁴/4 = ½ × 5.0 × 16 × (0.60)⁴/4 = 1.296 J ≈ 1.3 J.
4. Same thing via I: I = ∫₀ᴸ x² Cx dx = CL⁴/4 = 5.0 × 0.1296/4 = 0.162 kg·m², and ½Iω² = ½ × 0.162 × 16 = 1.296 J. ✓

(b) Translation plus rotation.

1. Mass: M = ∫₀ᴸ Cx dx = CL²/2 = 0.90 kg.
2. Centre of mass: x_cm = (∫₀ᴸ x · Cx dx)/M = (CL³/3)/(CL²/2) = 2L/3 = 0.40 m.
3. I_cm = I_pivot − Mx_cm² = 0.162 − 0.90 × 0.16 = 0.018 kg·m².
4. v_cm = x_cm ω = 0.40 × 4.0 = 1.6 m/s, so ½Mv_cm² = ½ × 0.90 × 2.56 = 1.152 J.
5. ½I_cm ω² = ½ × 0.018 × 16 = 0.144 J.
6. Total: 1.152 + 0.144 = 1.296 J. ✓ The two methods agree, as the parallel axis theorem says they must.

(c) Outer half. Integrate from L/2 to L: K_outer = ½Cω²(L⁴ − (L/2)⁴)/4 = (15/16) of the total = 1.215 J, about 94 %. The outer half holds 3/4 of the mass but nearly all the energy, because those pieces are both heavier and faster.

Interpretation. Two common wrong answers show why the method matters. Using ½Mv² with the tip speed of 2.4 m/s gives 2.6 J, twice too big: most of the rod moves slower than the tip. Adding ½Mv_cm² to ½I_pivot ω² gives 2.4 J, which double-counts the centre-of-mass motion.

Worked example 2: a spinning baton thrown upward

Question. A juggler throws a uniform baton of mass M = 0.40 kg and length L = 0.70 m straight up. Its centre of mass leaves the hand at 5.0 m/s and it spins at 15 rad/s about its centre, in a vertical plane. Treat it as a thin uniform rod, so I_cm = ML²/12 (Topic 5.4). Ignore air resistance; take +y upward. (a) Find its translational, rotational and total kinetic energy at release. (b) Find its kinetic energy at the top of its flight. (c) Sketch how each kinetic energy changes with time.

(a) At release.

1. Translational: ½Mv_cm² = ½ × 0.40 × 5.0² = 5.0 J.
2. I_cm = 0.40 × 0.70²/12 = 0.01633 kg·m². Rotational: ½I_cm ω² = ½ × 0.01633 × 15² = 1.84 J.
3. Total: 5.0 + 1.84 = 6.84 J ≈ 6.8 J. About 27 % of it is rotational.

(b) At the top.

1. Gravity acts at the centre of mass, so it exerts no torque about the centre of mass. The spin stays at 15 rad/s throughout the flight, and ½I_cm ω² stays at 1.84 J.
2. The centre of mass slows under gravity: v_cm = 5.0 − 9.8t, which is zero at t = 5.0/9.8 = 0.51 s.
3. At the top, the centre of mass is momentarily at rest, but the baton still has K = 1.84 J ≈ 1.8 J, all rotational. This is the “kinetic energy with the centre of mass at rest” idea in action.

(c) The graph. K_trans = ½M(5.0 − 9.8t)² is a parabola that falls to zero at 0.51 s and rises back to 5.0 J when the baton returns to the release height at 1.02 s. K_rot is a horizontal line at 1.84 J. The total is the parabola shifted up by 1.84 J (Figure 2). The loss of total kinetic energy on the way up goes into gravitational potential energy of the baton–Earth system.

Why not use the tip speed? If the baton is horizontal at release, one tip moves at 5.0 + 0.35 × 15 = 10.25 m/s and the other at 5.0 − 5.25 = −0.25 m/s (slightly downward). No single point’s speed gives the kinetic energy; the split formula does it correctly.

Common misconceptions

• “The centre of mass is at rest, so K = 0.” A wheel on a fixed axle has rotational kinetic energy. Only its momentum is zero.
• Double counting. For a body turning about a fixed pivot, use ½I_pivot ω² or ½Mv_cm² + ½I_cm ω², never ½I_pivot ω² + ½Mv_cm².
• Using ½Mv² with the rim or tip speed. Different pieces move at different speeds. Integrate, or use I.
• Using rev/min or degrees per second. v = rω needs rad/s. Convert first: rad/s = rev/min × 2π/60.
• Giving K a sign. K is never negative; the direction of spin does not matter. Opposite spins add, they do not cancel.
• Treating ½Iω² as extra energy on top of ½mv² for a fixed axis. It is the sum of ½v² dm. It is the total.
• Using the wrong I in the split formula. The rotational term must use I about the centre of mass.
• “More mass means more kinetic energy.” At the same ω, where the mass sits matters as much as how much there is.

Where this leads

Topic 6.2, Torque and Work, explains how a torque acting through an angle changes ½Iω²: the straight-line K–θ graph above is the clue. Topic 6.5, Rolling, adds a link between v_cm and ω, so the split formula gives the energy of a rolling wheel. Next, try the practice questions, then use the revision notes and the checklist. When you are ready, move on to Topic 6.2, Torque and Work, or return to the course roadmap.