Rolling KE = ½mv² + ½Iω²

WHY split motion into translation + rotation?

Derivation from scratch (König's theorem)

Step 1 — Write total KE as a sum. $KE = \sum_i \tfrac12 m_i v_i^2 = \sum_i \tfrac12 m_i\,(\vec v_{cm}+\vec u_i)\cdot(\vec v_{cm}+\vec u_i)$ Why this step? Speed-squared is just the dot product of velocity with itself; this lets us expand cleanly.

Step 2 — Expand the dot product. $KE = \sum_i \tfrac12 m_i\Big(v_{cm}^2 + 2\,\vec v_{cm}\cdot\vec u_i + u_i^2\Big)$ Why this step? Standard $(a+b)^2 = a^2+2ab+b^2$ , but with vectors so the middle term is a dot product.

Step 3 — Handle each of the three terms.

Term 1: $\sum_i \tfrac12 m_i v_{cm}^2 = \tfrac12\Big(\sum_i m_i\Big)v_{cm}^2 = \tfrac12 M v_{cm}^2$ . → Translational KE.
Term 2: $\sum_i m_i\,\vec v_{cm}\cdot\vec u_i = \vec v_{cm}\cdot\sum_i m_i \vec u_i = \vec v_{cm}\cdot M\vec u_{cm,\,rel}$ . But $\vec u_{cm,\,rel}$ is the velocity of the cm as seen from the cm — which is zero! So this term vanishes.
Term 3: $\sum_i \tfrac12 m_i u_i^2$ . For pure rotation about the cm, $u_i = \omega r_i$ , so this becomes $\tfrac12\big(\sum_i m_i r_i^2\big)\omega^2 = \tfrac12 I_{cm}\,\omega^2$ . → Rotational KE.

Why the cross term dies (the key trick): the center of mass cannot move relative to itself. That zero is why the two energies cleanly separate — it is the whole reason the formula is a clean sum.

The rolling-without-slipping condition

Substitute $\omega = v_{cm}/R$ and write $I_{cm}=\beta MR^2$ (where $\beta$ is a shape number): $KE = \tfrac12 M v_{cm}^2 + \tfrac12(\beta MR^2)\frac{v_{cm}^2}{R^2} = \tfrac12 M v_{cm}^2(1+\beta)$

Shape	$I_{cm}$	$\beta=\frac{I_{cm}}{MR^2}$	Rotational fraction $\frac{\beta}{1+\beta}$
Ring/hollow cylinder	$MR^2$	$1$	$1/2$ (50%)
Disc/solid cylinder	$\tfrac12 MR^2$	$1/2$	$1/3$ (33%)
Solid sphere	$\tfrac25 MR^2$	$2/5$	$2/7$ (29%)
Hollow sphere	$\tfrac23 MR^2$	$2/3$	$2/5$ (40%)

Worked examples

Worked example Example 1 — Energy split of a rolling disc

A solid disc of mass $2\,\text{kg}$ rolls with $v_{cm}=3\,\text{m/s}$ . Find total KE and the rotational part.

Step 1: $\beta=\tfrac12$ for a disc. Why? $I_{cm}=\tfrac12MR^2$ . Step 2: $KE=\tfrac12 M v^2(1+\beta)=\tfrac12(2)(9)(1.5)=13.5\,\text{J}$ . Why? Direct plug-in of compact formula. Step 3: Rotational fraction $=\frac{\beta}{1+\beta}=\frac{1/2}{3/2}=\tfrac13$ , so $KE_{rot}=13.5\times\tfrac13=4.5\,\text{J}$ . Check: Translational $=\tfrac12(2)(9)=9\,\text{J}$ ; $9+4.5=13.5$ ✓.

Worked example Example 2 — Speed at the bottom of an incline (Forecast-then-Verify)

A sphere rolls without slipping from rest down a height $h$ . Find $v_{cm}$ at the bottom.

Forecast: It should be slower than free-slide $\sqrt{2gh}$ because some PE goes into spin.

Step 1 — Energy conservation: $Mgh = \tfrac12 M v^2(1+\beta)$ . Why? No slipping ⇒ no friction work lost; all PE → rolling KE. Step 2 — Solve: $v=\sqrt{\dfrac{2gh}{1+\beta}}$ . Step 3 — Sphere ( $\beta=2/5$ ): $v=\sqrt{\dfrac{2gh}{1.4}}=\sqrt{\tfrac{10gh}{7}}\approx 0.845\sqrt{2gh}$ . Verify: Indeed slower than $\sqrt{2gh}$ ✓, and a ring ( $\beta=1$ ) gives $\sqrt{gh}$ — slowest, as predicted.

Common mistakes

Flashcards

Total KE of a rigid body in plane motion (König's theorem)

KE=\tfrac12 Mv_{cm}^2+\tfrac12 I_{cm}\omega^2

Why does the cross term in König's derivation vanish?

Because

\sum m_i\vec u_i = M\vec u_{cm,rel}=0

— the cm has zero velocity relative to itself.

Rolling-without-slipping constraint

v_{cm}=\omega R

(contact point instantaneously at rest)

Define

\beta

and its role

\beta=I_{cm}/MR^2

; it sets the energy split and how fast a body rolls down a ramp.

Rotational fraction of KE in rolling

\dfrac{\beta}{1+\beta}

Speed of a body rolling from rest down height h

v=\sqrt{\dfrac{2gh}{1+\beta}}

For a solid sphere, what is

v

down height h?

\sqrt{10gh/7}

Order of race down incline: ring, disc, sphere

Sphere fastest, then disc, then ring (smallest β wins)

Does the winner depend on mass or radius?

No —

v\propto 1/\sqrt{1+\beta}

, both M and R cancel.

Can you use

\tfrac12 I_{contact}\omega^2

alone for rolling KE?

Yes, it equals

\tfrac12 Mv^2+\tfrac12 I_{cm}\omega^2

by the parallel axis theorem. Don't double-add.

Recall Feynman: explain to a 12-year-old

Imagine rolling a ball across the floor. It's doing two jobs at the same time: moving across the room and spinning like a top. Each job takes energy. The "moving" energy is the regular kind ( $\tfrac12 mv^2$ ). The "spinning" energy is the extra kind ( $\tfrac12 I\omega^2$ ). To know how much total go-power the ball has, you add both jobs together. A ring has lots of its weight at the edge, so spinning costs it a lot of energy — that's why a ring loses a downhill race to a solid ball that keeps most of its energy for going forward.

Connections

König's Theorem — the general parent result this is a special case of.
Moment of Inertia — supplies the $I_{cm}$ and $\beta$ values.
Parallel Axis Theorem — why the contact-point method gives the same KE.
Rolling Without Slipping — provides the $v=\omega R$ constraint.
Conservation of Energy on Inclines — where this KE formula is applied.
Angular Velocity and Angular Acceleration — defines $\omega$ .

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab koi cheez (ball, ring, cylinder) zameen pe roll karti hai, toh wo ek saath do kaam kar rahi hoti hai: ek toh aage ki taraf slide kar rahi hai (uska center aage badh raha hai), aur doosra wo apne center ke around ghoom bhi rahi hai (spin). Dono kaamon mein energy lagti hai. Isliye total kinetic energy = translation ki energy ( $\tfrac12 Mv^2$ ) + rotation ki energy ( $\tfrac12 I\omega^2$ ). Yeh "slide + spin" wala funda yaad rakho.

Yeh formula aise hi nahi aata — iske peeche König's theorem hai. Har particle ki velocity ko humne center-of-mass ki velocity plus uske around ki velocity mein toda. Jab square karke add kiya toh ek cross-term aaya jo zero ho gaya, kyunki center of mass apne aap ke relative hil hi nahi sakta. Bas isi zero ki wajah se dono energies cleanly alag ho jaati hain aur seedha jud jaati hain.

Rolling without slipping ka matlab hai contact point ek instant ke liye rukka hua hai, isse $v=\omega R$ ban jaata hai. Iske baad sab kuch ek hi variable mein aa jaata hai: $KE=\tfrac12 Mv^2(1+\beta)$ , jahan $\beta=I/MR^2$ shape ka number hai. Ring ka β = 1 (50% energy spin mein jati), sphere ka β = 0.4 (sirf 29% spin mein). Isliye incline pe race mein sphere jeetta hai, ring haarta hai — aur yeh mass ya radius pe depend nahi karta, sirf β pe. Exam mein yahi 80/20 trick kaam aati hai.

WHY split motion into translation + rotation?

Derivation from scratch (König's theorem)

The rolling-without-slipping condition

Worked examples

Common mistakes

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Rotational Mechanics

Connections