Rotational kinetic energy = ½Iω²: Worked examples

Before any example, three tools we will reuse. Each is stated in plain words so a symbol never appears un-earned:

#	Case class	What is special	Example
A	Pure spin (axis fixed, centre still)	only $\frac{1}{2} I ω^{2}$	Ex 1
B	Spin + roll (translation added)	$\frac{1}{2} M v^{2} + \frac{1}{2} I ω^{2}$	Ex 2
C	Shape race (same $M, R$ , different $I$ )	who reaches bottom first	Ex 3
D	Unit trap ( $ω$ in rev or deg)	must convert to rad/s	Ex 4
E	Axis trap (off-centre axis)	parallel-axis theorem needed	Ex 5
F	Degenerate / limiting ( $R \to 0$ , all mass at centre, $ω \to 0$ )	formula's edge behaviour	Ex 6
G	Real-world word problem (flywheel powering a device)	energy → time / work	Ex 7
H	Exam twist (two bodies coupled, find shared $ω$ )	conservation + geometry	Ex 8

Example 1 — Cell A: pure spin

Worked example Grinding wheel, centre fixed

A solid disc grinding wheel: $M = 4 kg$ , $R = 0.10 m$ , spinning at $ω = 150 rad/s$ on a fixed spindle. Its centre does not move. Find $K_{r o t}$ .

Forecast: small radius, so $I$ is tiny — but $ω$ is squared. Guess the order of magnitude.

Step 1 — Get $I$ for a solid disc about its centre. $I = \frac{1}{2} M R^{2} = \frac{1}{2} (4) (0.10)^{2} = 0.02 kg\cdotm^{2}$ Why this step? The centre is still, so this is pure rotation — the only energy is $\frac{1}{2} I ω^{2}$ , and we need $I$ first. The disc formula $\frac{1}{2} M R^{2}$ comes from Moment of inertia.

Step 2 — Plug into the energy formula. $K_{r o t} = \frac{1}{2} I ω^{2} = \frac{1}{2} (0.02) (150)^{2} = 225 J$ Why this step? Direct substitution; nothing translates so there is no $\frac{1}{2} M v^{2}$ term.

Verify: Units: $kg\cdotm^{2} \cdot (rad/s)^{2} = kg\cdotm^{2} / s^{2} = J$ ✓ (radian is dimensionless). Sanity: $ω^{2} = 22500$ is large, so even a $0.02$ moment of inertia gives a respectable $225$ J.

Example 2 — Cell B: spin + roll

Figure — Rotational kinetic energy = ½Iω²

Worked example Cylinder rolling down a ramp

A uniform solid cylinder ( $I_{c m} = \frac{1}{2} M R^{2}$ ) rolls without slipping from rest down a height $h = 1.4 m$ . Find its centre speed $v$ at the bottom. Take $g = 9.8 m/s^{2}$ .

Forecast: a sliding block would reach $2 g h$ . Will the cylinder be faster or slower? Why?

Step 1 — Write energy conservation with BOTH kinetic terms. $M g h = translation \frac{1}{2} M v^{2} + rotation \frac{1}{2} I_{c m} ω^{2}$ Why this step? Look at the figure: the centre moves forward (translation) and the body spins (rotation). Gravity's stored energy splits between the two. From Conservation of energy, no slipping means friction does no net work.

Step 2 — Kill $ω$ using the rolling link $ω = v / R$ . $\frac{1}{2} I_{c m} ω^{2} = \frac{1}{2} (\frac{1}{2} M R^{2}) \frac{v ^{2}}{R ^{2}} = \frac{1}{4} M v^{2}$ Why this step? We want one unknown. The rolling link $v_{c m} = R ω$ trades $ω$ for $v$ ; the $R^{2}$ cancels, so the answer never depends on the radius.

Step 3 — Combine and solve. $M g h = \frac{1}{2} M v^{2} + \frac{1}{4} M v^{2} = \frac{3}{4} M v^{2} \Rightarrow v = \frac{4 g h}{3}$ Numerically $v = \frac{4 ( 9.8 ) ( 1.4 )}{3} = 18.29 \dots = 4.28 m/s$ . Why this step? $M$ cancels — mass doesn't matter, only the mass distribution (through the $\frac{1}{2}$ in $I$ ).

Verify: A slider would get $2 g h = 27.44 = 5.24 m/s$ . The cylinder is slower (4.28 < 5.24) because a quarter of the energy went into spin. ✓

Example 3 — Cell C: the shape race

Worked example Hoop vs disc vs sphere, same ramp

Three bodies of the same mass and radius are released together from the same height and roll without slipping. $I_{h oo p} = M R^{2}$ , $I_{d i sc} = \frac{1}{2} M R^{2}$ , $I_{s p h er e} = \frac{2}{5} M R^{2}$ . Rank their bottom speeds.

Forecast: more of the mass sitting far out means more $I$ . Does bigger $I$ mean faster or slower?

Step 1 — Write the general result with $I = c M R^{2}$ . Repeating Ex 2 with $I = c M R^{2}$ (so $c$ is the shape factor): $M g h = \frac{1}{2} M v^{2} + \frac{1}{2} (c M R^{2}) \frac{v ^{2}}{R ^{2}} = \frac{1}{2} (1 + c) M v^{2} \Rightarrow v = \frac{2 g h}{1 + c}$ Why this step? One formula covers all shapes; only $c$ changes. Bigger $c$ (mass farther out) means a bigger denominator, hence smaller $v$ .

Step 2 — Insert each $c$ .

Sphere $c = \frac{2}{5}$ : $v = \frac{2 g h}{1.4} = \frac{10 g h}{7}$ — fastest.
Disc $c = \frac{1}{2}$ : $v = \frac{2 g h}{1.5} = \frac{4 g h}{3}$ — middle.
Hoop $c = 1$ : $v = \frac{2 g h}{2} = g h$ — slowest. Why this step? The hoop's mass is all at radius $R$ , giving the largest $I$ , so it stores the most in spin and has the least left for forward motion.

Verify (numbers as ratios of $g h$ ): sphere $10/7 = 1.195$ , disc $4/3 = 1.155$ , hoop $1.000$ . Order sphere > disc > hoop ✓. Notice radius and mass never appear — a marble beats a truck tyre of the same shape. ✓

Example 4 — Cell D: the unit trap

Worked example Motor spec given in rev/min

A flywheel disc, $M = 12 kg$ , $R = 0.30 m$ , spins at $600 rev/min$ . Find $K_{r o t}$ .

Forecast: if you plug $600$ straight in, will your answer be too big or too small?

Step 1 — Convert rev/min to rad/s. One revolution is $2 π$ radians; one minute is $60$ s: $ω = 600 \cdot \frac{2 π}{60} = 20 π = 62.83 rad/s$ Why this step? The formula was derived from $v = r ω$ , which is only true in radians (see the Angular velocity note and the [!mistake] in the parent). Skipping this makes the number meaningless.

Step 2 — Moment of inertia and energy. $I = \frac{1}{2} (12) (0.30)^{2} = 0.54 kg\cdotm^{2}, K = \frac{1}{2} (0.54) (62.83)^{2} = 1066 J$ Why this step? Standard disc $I$ , then $\frac{1}{2} I ω^{2}$ with the converted $ω$ .

Verify: Had we wrongly used $ω = 600$ , we'd get $\frac{1}{2} (0.54) (600)^{2} = 97, 200 J$ — about $91 \times$ too large. The conversion is not optional. $K \approx 1.07 kJ$ ✓.

Example 5 — Cell E: the axis trap (parallel axis)

Worked example Rod swinging about one end

A uniform rod, $M = 2 kg$ , length $L = 1.2 m$ , is pivoted at one end and swings so that at the bottom of its swing $ω = 3 rad/s$ . Find $K_{r o t}$ .

Forecast: would $I$ about the end be bigger or smaller than about the centre? Which stores more energy at the same $ω$ ?

Step 1 — Recall $I$ about the centre, then shift the axis. About its centre a rod has $I_{c m} = \frac{1}{12} M L^{2}$ . The pivot is a distance $d = L /2$ away, so by the Parallel axis theorem: $I_{e n d} = I_{c m} + M d^{2} = \frac{1}{12} M L^{2} + M (\frac{L}{2})^{2} = \frac{1}{3} M L^{2}$ Why this step? $I$ is not a fixed property — it depends on the axis (parent [!mistake]). The theorem lets us slide from the centre axis to the end axis by adding $M d^{2}$ .

Step 2 — Numbers. $I_{e n d} = \frac{1}{3} (2) (1.2)^{2} = 0.96 kg\cdotm^{2}, K = \frac{1}{2} (0.96) (3)^{2} = 4.32 J$ Why this step? With the correct $I$ for the actual axis, the energy formula applies as usual.

Verify: About the centre we'd get $I = \frac{1}{12} (2) (1.2)^{2} = 0.24$ , giving only $\frac{1}{2} (0.24) (9) = 1.08 J$ . The end axis holds 4× more ( $0.96/0.24 = 4$ ) — mass is farther from the pivot on average, so it moves faster and carries more KE. ✓

Example 6 — Cell F: degenerate & limiting cases

Worked example What the formula does at the edges

Test three edge inputs on a spinning body: (a) all mass collapsed onto the axis ( $r \to 0$ ), (b) $ω \to 0$ , (c) a point mass $m$ on a string of length $R$ swung at $ω$ — does $\frac{1}{2} I ω^{2}$ agree with $\frac{1}{2} m v^{2}$ ?

Forecast: guess each answer before checking. Should (a) and (b) both give zero?

Step 1 — Case (a): all mass on the axis. Then every $r_{i} = 0$ , so $I = \sum m_{i} r_{i}^{2} = 0$ and $K = \frac{1}{2} (0) ω^{2} = 0$ . Why? A pencil-thin line of mass on the spin axis isn't actually moving — points on the axis have $v = r ω = 0$ . Zero speed → zero KE. The formula agrees.

Step 2 — Case (b): $ω \to 0$ . $K = \frac{1}{2} I (0)^{2} = 0$ regardless of $I$ . Why? No spin, no motion, no kinetic energy. ✓

Step 3 — Case (c): a single point mass, the ultimate sanity check. A point mass at radius $R$ has $I = m R^{2}$ . Its speed is $v = R ω$ , so $K = \frac{1}{2} I ω^{2} = \frac{1}{2} (m R^{2}) ω^{2} = \frac{1}{2} m (R ω)^{2} = \frac{1}{2} m v^{2} .$ Why this step? The rotational formula must reduce to the plain Kinetic energy of a particle formula for one particle — because that's exactly what it was summed from. It does.

Verify (numbers): with $m = 0.5 kg$ , $R = 2 m$ , $ω = 4 rad/s$ : $\frac{1}{2} I ω^{2} = \frac{1}{2} (0.5 \cdot 4) (16) = 16 J$ and $\frac{1}{2} m v^{2} = \frac{1}{2} (0.5) (8)^{2} = 16 J$ . Identical ✓. The formula passes all three edge tests.

Example 7 — Cell G: real-world word problem

Worked example Flywheel powering a machine

A flywheel stores $K = 1.125 \times 1 0^{5} J$ (the parent's Example 1 flywheel). A machine draws a steady power of $P = 2500 W$ and the flywheel is the only source. For how long can it run before its energy is spent?

Forecast: power is energy per second — guess whether the answer is seconds or minutes.

Step 1 — Relate stored energy, power, and time. Power is energy delivered per unit time, so $P = \frac{K}{t}$ , giving $t = \frac{K}{P} = \frac{1.125 \times 1 0 ^{5}}{2500} = 45 s .$ Why this step? The Work-energy theorem (rotational) says the work done on the machine equals the flywheel's loss of $K_{r o t}$ ; delivering it at constant power sets the duration.

Step 2 — Interpret. The flywheel keeps the machine running for 45 s (until $ω \to 0$ , i.e. $K \to 0$ ). Why? Once the spin energy is exhausted the flywheel stops — energy in equals energy out.

Verify: Units $J / W = J / (J/s) = s$ ✓. Cross-check: $P \cdot t = 2500 \times 45 = 1.125 \times 1 0^{5} J = K$ ✓.

Example 8 — Cell H: exam twist (coupled bodies)

Worked example Two discs clutched together

Disc 1 ( $I_{1} = 0.6 kg\cdotm^{2}$ ) spins at $ω_{1} = 40 rad/s$ . It is pressed against a stationary coaxial disc 2 ( $I_{2} = 0.2 kg\cdotm^{2}$ , $ω_{2} = 0$ ) until they lock and turn as one. Find (a) the common final $ω_{f}$ , and (b) how much kinetic energy is lost.

Forecast: in a "sticking" collision, is kinetic energy conserved? Guess before Step 3.

Step 1 — Conserve angular momentum, not energy. No external torque acts during the clutching, so total angular momentum $L = I ω$ is conserved: $I_{1} ω_{1} + I_{2} ω_{2} = (I_{1} + I_{2}) ω_{f} .$ Why this step? The clutch force is internal; friction between the faces is internal too, so it can waste energy but cannot change total $L$ . This is the rotational analogue of a perfectly inelastic collision.

Step 2 — Solve for $ω_{f}$ . $ω_{f} = \frac{I _{1} ω _{1}}{I _{1} + I _{2}} = \frac{0.6 \times 40}{0.8} = \frac{24}{0.8} = 30 rad/s .$ Why? Combined inertia $0.8$ shares the fixed momentum $24$ .

Step 3 — Compare kinetic energies. $K_{i} = \frac{1}{2} (0.6) (40)^{2} = 480 J, K_{f} = \frac{1}{2} (0.8) (30)^{2} = 360 J .$ Energy lost $= 480 - 360 = 120 J$ , turned into heat by the slipping clutch faces. Why this step? Angular momentum is conserved but kinetic energy is not — the sticking wastes $120 J$ (25% of the original).

Verify: $L_{i} = 0.6 \times 40 = 24$ , $L_{f} = 0.8 \times 30 = 24$ — momentum matches ✓. $K_{f} < K_{i}$ as expected for an inelastic lock ✓.

Recall Quick self-test

Which cell does each phrase belong to? "rev/min given" ::: Cell D (unit trap) — convert to rad/s. "pivoted at the end" ::: Cell E (axis trap) — parallel axis theorem. "two gears lock together" ::: Cell H — conserve angular momentum, energy drops. "same mass, which rolls fastest" ::: Cell C — smallest shape factor $c$ wins. "all mass on the axis" ::: Cell F — $I = 0$ , so $K = 0$ .

Connections

Parent: ½Iω² — the formula every example uses.
Moment of inertia — supplies $I$ for disc, hoop, sphere, rod.
Kinetic energy of a particle — the $\frac{1}{2} m v^{2}$ that Example 6c reduces to.
Rolling motion — the $v_{c m} = R ω$ link in Examples 2, 3.
Parallel axis theorem — the axis shift in Example 5.
Conservation of energy — powers Examples 2, 3, 7.
Work-energy theorem (rotational) — turns stored $K$ into run-time in Example 7.
Angular velocity — the radian requirement behind Example 4.

Worked examples — Rotational kinetic energy = ½Iω²

The scenario matrix