Rotational kinetic energy = ½Iω²

WHAT is it?

WHY does it matter? Energy methods sidestep forces and torques. If you only care about speeds at the start and end (a rolling ball, a flywheel, a yo-yo), conservation of energy is far faster than solving $\tau = I\alpha$ step by step.

HOW — Derivation from scratch

We build it from $K=\tfrac12 mv^2$ for a single point mass. No formula is assumed.

Step 1 — Chop the body into tiny pieces. Imagine the rigid body as many particles $m_1, m_2, \dots$ at distances $r_1, r_2, \dots$ from the axis. Why this step? Total KE is just the sum of each piece's KE — KE is additive.

$K = \sum_i \tfrac{1}{2} m_i v_i^2$

Step 2 — Use the rigid-body link between $v$ and $\omega$ . For circular motion about the axis, a particle at radius $r_i$ moves with speed $v_i = r_i\,\omega$ Why this step? In a rigid body every particle sweeps the same angle in the same time, so they all share one $\omega$ . The linear speed differs only through $r_i$ .

Step 3 — Substitute and pull out the common factor. $K = \sum_i \tfrac{1}{2} m_i (r_i\omega)^2 = \tfrac{1}{2}\,\omega^2 \sum_i m_i r_i^2$ Why this step? Since $\omega$ is the same for all particles, $\tfrac12\omega^2$ is a constant and factors out of the sum.

Step 4 — Name the leftover sum. The quantity $\sum_i m_i r_i^2$ depends only on how mass is distributed about the axis. We call it the moment of inertia: $I \equiv \sum_i m_i r_i^2 \quad\Longrightarrow\quad \boxed{K_{rot} = \tfrac{1}{2}I\omega^2}$ Why this step? This isolates everything geometric into one symbol $I$ , leaving a formula that looks exactly like $\tfrac12 mv^2$ with $m\!\to\!I$ and $v\!\to\!\omega$ . That analogy is the whole point.

The linear ↔ rotational analogy

Linear	Rotational
mass $m$	moment of inertia $I$
velocity $v$	angular velocity $\omega$
$K=\tfrac12 mv^2$	$K=\tfrac12 I\omega^2$

Worked examples

Worked example 2 — Ball rolling down a ramp (energy method)

A solid sphere ( $I_{cm}=\tfrac25 MR^2$ ) rolls without slipping from height $h$ . Find $v_{cm}$ at the bottom.

Step 1: Energy conservation: $Mgh = \tfrac12 Mv^2 + \tfrac12 I\omega^2$ . Why? All gravitational PE converts to translational + rotational KE (no slipping ⇒ friction does no net work). Step 2: Rolling condition $\omega=v/R$ , so $\tfrac12 I\omega^2 = \tfrac12(\tfrac25 MR^2)(v/R)^2 = \tfrac15 Mv^2$ . Why? Eliminate $\omega$ to leave one unknown $v$ . Step 3: $Mgh = \tfrac12 Mv^2 + \tfrac15 Mv^2 = \tfrac{7}{10}Mv^2 \Rightarrow v=\sqrt{\tfrac{10gh}{7}}$ . Why? Note $v$ is less than $\sqrt{2gh}$ (a sliding block) — some energy went into spin.

Common mistakes (steel-manned)

Recall Feynman: explain to a 12-year-old

Imagine a merry-go-round full of kids. Each kid is whizzing around, so each has "moving energy." Kids near the edge zoom fast; kids near the middle barely move. To find the total energy you'd add up every kid's energy. Instead of doing that every time, we note all kids turn together at the same spin rate $\omega$ . We bundle "how heavy and how spread out" into one number $I$ . Then the spinning energy is just half of $I$ times the spin rate squared — exactly like a moving truck's energy is half its mass times its speed squared. Spread the kids farther out → harder to spin → more energy stored.

Active-recall flashcards

#flashcards/physics

Define rotational kinetic energy.

The KE of a rigid body spinning about an axis,

K=\tfrac12 I\omega^2

Why does

\omega

factor out of the particle sum?

In a rigid body every particle shares the same

\omega

, so

\tfrac12\omega^2

is constant across the sum.

What is

I

in terms of particles?

I=\sum_i m_i r_i^2

— sum of (mass × distance-from-axis²).

What replaces

m

and

v

in the rotational analogue of

\tfrac12 mv^2

m\to I

and

v\to\omega

Total KE of an object rolling without slipping?

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

with

v_{cm}=R\omega

Must

\omega

be in degrees or radians in

\tfrac12 I\omega^2

Radians per second (because the derivation used

v=r\omega

Why does a solid sphere reach the bottom of a ramp slower than a sliding block?

Part of the PE becomes rotational KE, leaving less for translation:

v=\sqrt{10gh/7}<\sqrt{2gh}

Hoop vs disc, same M,R,ω: which has more rotational KE and why?

The hoop (double), because all its mass sits at radius

R

, giving a larger

I=\sum mr^2

Does

I

depend on the chosen axis?

Yes — e.g. rod about center is

\tfrac{1}{12}ML^2

, about end

\tfrac13 ML^2

KE of disc M=20kg, R=0.5m, ω=300 rad/s?

I=2.5

K=\tfrac12(2.5)(300)^2=1.125\times10^5

Connections

Moment of inertia — the $I$ in the formula; computed from mass distribution.
Kinetic energy of a particle — the $\tfrac12 mv^2$ we summed up.
Rolling motion — combines translational + rotational KE.
Work-energy theorem (rotational) — $W=\Delta K_{rot}$ via $\tau\,d\theta$ .
Parallel axis theorem — shifts $I$ to off-center axes before energy calc.
Angular velocity — defines $\omega$ in rad/s.
Conservation of energy — the principle powering examples 2.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab koi cheez ghoom rahi hoti hai — jaise fan, wheel, ya spinning disc — to uske andar ka har particle move kar raha hota hai, isliye usme energy hoti hai. Is energy ko bolte hain rotational kinetic energy. Idea simple hai: har particle ki KE $\tfrac12 m v^2$ hoti hai, aur rigid body me sabka angular velocity $\omega$ same hota hai, sirf $v=r\omega$ alag hota hai. Sab particles ki energy add karo to formula nikal aata hai: $K=\tfrac12 I\omega^2$ , jahan $I=\sum m r^2$ ko moment of inertia kehte hain.

Yahan magic ye hai ki ye bilkul $\tfrac12 mv^2$ jaisa dikhta hai — bas $m$ ki jagah $I$ aur $v$ ki jagah $\omega$ . Iska matlab linear aur rotational world me ek hi pattern repeat hota hai. $I$ batata hai ki mass axis se kitni door spread hai. Agar mass door hai (jaise hoop), to $I$ bada, aur same $\omega$ pe zyada energy store hoti hai — isiliye flywheels me mass edge pe rakhte hain.

Do cheezein hamesha yaad rakho: pehla, $\omega$ ko radians per second me hi daalna hai, degrees me nahi, kyunki derivation $v=r\omega$ se aayi hai. Doosra, agar object roll kar raha hai (slip nahi), to total energy do parts me hoti hai — $\tfrac12 Mv^2$ (aage badhne ki) aur $\tfrac12 I\omega^2$ (ghoomne ki). Isiliye rolling ball, sliding block se dheere bottom pe pahunchti hai, kyunki uski thodi energy spin me chali jaati hai. Energy method exam me bahut fast hota hai jab sirf start aur end ki speed chahiye.

WHAT is it?

HOW — Derivation from scratch

The linear ↔ rotational analogy

Worked examples

Common mistakes (steel-manned)

Active-recall flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Rotational Mechanics

Connections