Rolling KE = ½mv² + ½Iω²: Worked examples — Physics deep dive

This page is the exhaustive workbook for Rolling KE. The parent gave you the formula $K E = \frac{1}{2} M v_{c m}^{2} + \frac{1}{2} I_{c m} ω^{2} = \frac{1}{2} M v_{c m}^{2} (1 + β), β = \frac{I _{c m}}{M R ^{2}} .$ Here we hit every kind of problem this formula can appear in — every shape, the slipping case where $v \neq = ω R$ , the degenerate "point mass" limit, an energy-loss twist, and a real-world word problem. Nothing is left as "you'll figure it out."

The scenario matrix

Every rolling-KE problem is one (or a blend) of these case classes. Each cell is covered by at least one worked example below.

Cell	Case class	What makes it special	Covered by
A	Pure energy split, given $v$	Both KEs from a known speed	Ex 1 (ring)
B	All four standard shapes	Compare $β$ across shapes	Ex 2
C	Incline from rest	PE → rolling KE, solve for $v$	Ex 3
D	Race / ordering	Mass & radius cancel	Ex 4
E	Slipping (constraint broken)	$v \neq = ω R$ , need both separately	Ex 5
F	Degenerate limits	$β \to 0$ (point mass) and $β \to \infty$	Ex 6
G	Energy-loss twist	Rolling then hits a rough patch / friction removes energy	Ex 7
H	Real-world word problem	Units, given power/mass, back out speed	Ex 8
I	Reverse problem	Given the KE split, find the shape ( $β$ )	Ex 9

The examples

Worked example Ex 5 — (Cell E) SLIPPING: constraint broken

A disc ( $M = 2 kg$ , $R = 0.1 m$ ) is spun up on ice to $ω = 20 rad/s$ while its center moves at only $v_{c m} = 1 m/s$ . On ice it slips: $v \neq = ω R$ . Find total KE.

Forecast: Because $ω R = 2 m/s \neq = v = 1 m/s$ , we cannot use the compact $\frac{1}{2} M v^{2} (1 + β)$ formula. We must use the two energies separately.

Step 1 — Check the constraint. Rolling Without Slipping requires $v = ω R$ . Here $ω R = (20) (0.1) = 2 \neq = 1$ . Why this step? The compact formula secretly assumes $v = ω R$ ; violating it makes that formula wrong. Step 2 — Translational KE. $\frac{1}{2} M v^{2} = \frac{1}{2} (2) (1^{2}) = 1 J$ . Why? This term only ever uses $v_{c m}$ , slipping or not. Step 3 — Rotational KE. $I_{c m} = \frac{1}{2} M R^{2} = \frac{1}{2} (2) (0. 1^{2}) = 0.01 kg\cdotm^{2}$ ; $\frac{1}{2} I ω^{2} = \frac{1}{2} (0.01) (2 0^{2}) = 2 J$ . Why? Rotational KE always uses the true $ω$ , which here is independent of $v$ . Step 4 — Add (König's theorem still holds). $K E = 1 + 2 = 3 J$ .

Verify: If we'd wrongly used the compact formula, $\frac{1}{2} (2) (1) (1.5) = 1.5 J$ — wrong by half, because it forced $ω R = v$ ✓ (shows why you must split when slipping).

Worked example Ex 6 — (Cell F) Degenerate limits of

$β$ What does the rolling speed $v = 2 g h / (1 + β)$ become in the two extremes: (a) $β \to 0$ (a "point mass" bead), and (b) $β \to \infty$ (all mass on a massless rim spun far out)?

Forecast: $β \to 0$ should behave like a frictionless block ( $2 g h$ ); $β \to \infty$ should crawl toward zero speed.

Step 1 — Case $β \to 0$ . $v \to \frac{2 g h}{1 + 0} = 2 g h$ . Why? No moment of inertia means no energy is diverted to spin — it's a pure sliding block. Step 2 — Case $β \to \infty$ . $v = \frac{2 g h}{1 + β} \to 0$ . Why? All the PE is swallowed by rotation; almost nothing is left to move the center forward. Step 3 — Sanity ordering. For finite shapes, $0 < β < \infty$ , so $0 < v < 2 g h$ always — rolling is always between these bounds.

Verify: With $h = 1.4, g = 10$ : $β = 0 \Rightarrow 28 \approx 5.29$ ; $β = 1000 \Rightarrow 28/1001 \approx 0.167 \to 0$ ✓. Every real shape (ring $4.32$ ? no— $3.74$ ; disc $4.32$ ; sphere $4.47$ ) sits inside $(0, 5.29)$ ✓.

Worked example Ex 7 — (Cell G) Energy-loss twist

A solid sphere rolling without slipping at $v = 3 m/s$ ( $M = 2 kg$ ) hits a patch where it briefly skids and loses exactly its rotational KE to friction, then re-grips. Assuming its forward speed $v$ is unchanged during the skid, what fraction of its original total KE survives?

Forecast: A sphere keeps most energy in translation ( $β = \frac{2}{5}$ is small), so it should lose only a modest slice.

Step 1 — Original total KE. $\frac{1}{2} M v^{2} (1 + β) = \frac{1}{2} (2) (9) (1.4) = 12.6 J$ . Why? Standard rolling KE before the skid. Step 2 — Rotational part lost. Fraction $\frac{β}{1 + β} = \frac{0.4}{1.4} = \frac{2}{7}$ , so $K E_{r o t} = 12.6 \times \frac{2}{7} = 3.6 J$ is removed. Why? The problem states rotational KE is dissipated by the skid. Step 3 — Surviving KE. $12.6 - 3.6 = 9.0 J$ , i.e. fraction $9.0/12.6 = 5/7 \approx 0.714$ . Why? What's left is exactly the translational share $1/ (1 + β)$ .

Verify: Translational KE $= \frac{1}{2} (2) (9) = 9 J$ matches the survivor ✓; $\frac{5}{7} + \frac{2}{7} = 1$ ✓; sphere keeps $71.4%$ , more than a ring's $50%$ would — forecast holds ✓.

Worked example Ex 8 — (Cell H) Real-world word problem

A go-kart wheel (treat as a solid disc, $M = 3 kg$ , $R = 0.25 m$ ) rolls without slipping. A motor delivers a steady $63 W$ of power to this one wheel starting from rest. How fast is the wheel's center moving after $2 s$ , if all delivered energy becomes rolling KE?

Forecast: Power $\times$ time is energy; converting to speed via the rolling-KE formula should give a few m/s.

Step 1 — Energy delivered. $E = P t = 63 \times 2 = 126 J$ . Why? Power is energy per second; steady power over time $t$ gives $E = P t$ . Step 2 — Set equal to rolling KE. $126 = \frac{1}{2} M v^{2} (1 + β)$ with disc $β = \frac{1}{2}$ : $126 = \frac{1}{2} (3) v^{2} (1.5) = 2.25 v^{2}$ . Why? All input energy becomes KE (given), so equate them. Step 3 — Solve. $v^{2} = 126/2.25 = 56$ , $v = 56 \approx 7.48 m/s$ . Why? Isolate $v$ .

Verify: Units: $J / kg = m^{2} / s^{2} = m/s$ ✓. Check back: $2.25 \times 56 = 126 J$ ✓. Radius $R$ never mattered — consistent with Cell D.

Recall Quick self-test

A body rolls with slipping. Can you use $\frac{1}{2} M v^{2} (1 + β)$ ? ::: No — that formula assumes $v = ω R$ . When slipping, use $\frac{1}{2} M v^{2} + \frac{1}{2} I_{c m} ω^{2}$ with the true, separate $ω$ . As $β \to 0$ , rolling speed down height $h$ approaches what? ::: $2 g h$ — the frictionless-slide value (no energy goes to spin). Given total KE and $M, v$ , how do you find the shape? ::: Compute $\frac{1}{2} M v^{2}$ ; then $β = \frac{K E}{\frac{1}{2} M v ^{2}} - 1$ ; match $β$ to the table. Does a heavier or larger wheel win a downhill race against an identical-shaped one? ::: Neither — it's a tie; $v$ depends only on $β$ , not $M$ or $R$ .