Rotational kinetic energy = ½Iω²: Question bank

True or false — justify

A body whose center of mass is stationary has zero kinetic energy.

False. If it spins, every particle except those on the axis is moving, so

K = \frac{1}{2} I ω^{2} > 0

even with

v_{c m} = 0

. Motion of the whole is not required — motion of the parts is enough.

Two objects of equal mass and equal

ω

always have equal rotational KE.

False.

K

depends on

I = \sum m_{i} r_{i}^{2}

(with

r_{i}

= distance from the axis), not just mass. Spread that same mass to larger

r

and

I

(hence

K

) grows — a hoop stores double a disc of equal

M, R, ω

Doubling

ω

doubles the rotational kinetic energy.

False.

K \propto ω^{2}

, so doubling

ω

quadruples

K

. The square is the whole reason flywheels are spun as fast as materials allow.

Doubling the moment of inertia doubles the rotational kinetic energy (at fixed

ω

True.

K = \frac{1}{2} I ω^{2}

is linear in

I

. Only

ω

is squared —

I

enters to the first power.

For a body rolling without slipping, all of its kinetic energy is rotational.

False. It has both:

K = \frac{1}{2} M v_{c m}^{2} + \frac{1}{2} I_{c m} ω^{2}

. The center of mass genuinely translates, so translational KE is a real, separate term.

The moment of inertia

I

is a fixed property of an object, like its mass.

False.

I

depends on the chosen axis. The same rod has

\frac{1}{12} M L^{2}

about its center but

\frac{1}{3} M L^{2}

about its end. Change the axis, change

I

Rotational KE can be negative if the body spins "backwards."

False.

K = \frac{1}{2} I ω^{2}

contains

ω^{2} \geq 0

and

I > 0

, so

K \geq 0

regardless of spin direction. Reversing the sign of

ω

leaves

ω^{2}

unchanged.

A sliding block and a uniform solid sphere (

I_{c m} = \frac{2}{5} M R^{2}

) released from the same height reach the bottom at the same speed.

False. The sphere's gravitational energy splits into translation and spin. For that specific inertia the algebra gives

v = 10 g h /7

, which is less than the block's

2 g h

. Storing energy in rotation slows the descent. (A different shape, e.g. a hoop, gives a different, even slower speed.)

If two flywheels store the same energy at the same

ω

, they must have the same mass.

False. Equal

K

at equal

ω

forces equal

I

, not equal

M

. A light ring far from the axis can match a heavy disc close to it.

Spot the error

ω = 60

deg/s, so

K = \frac{1}{2} I (60)^{2}

." — where is the mistake?

ω

must be in radians/s, because the derivation used

v = r ω

which only holds in radians. Convert first:

6 0^{\circ} / s = 60 \times π /180 \approx 1.05 rad/s

, then square that.

K = \frac{1}{2} (I ω)^{2}

." — what went wrong?

Only

ω

is squared, not

I

: it is

\frac{1}{2} I ω^{2}

, not

\frac{1}{2} (I ω)^{2}

. (The quantity

I ω

is angular momentum — a different thing entirely.)

"The ball is rolling, so

K = \frac{1}{2} M v_{c m}^{2}

." — what's missing?

The rotational term

\frac{1}{2} I_{c m} ω^{2}

. A rolling body spins as well as translates; dropping the spin term underestimates its energy and overestimates its speed.

I

about the center is

\frac{1}{2} M R^{2}

, so I'll use that even though the object spins about a point on its rim." — fix it.

You must shift the axis first with the Parallel axis theorem:

I = I_{c m} + M d^{2}

, where

d

is the distance between axes. Using the center value for a rim axis leaves out the

M d^{2}

term and undercounts

I

"To get total KE of many particles I average their speeds and use

\frac{1}{2} M v_{a v g}^{2}

." — why is this wrong?

KE is not built from an average speed; it's the sum

\sum \frac{1}{2} m_{i} v_{i}^{2}

. Because speed is squared, fast outer (large-

r

) particles contribute disproportionately — averaging first destroys the

r^{2}

weighting that becomes

I

"Spin rate is the same, mass is the same, so the two shapes clearly have equal KE." — what was ignored?

The mass distribution — how far each piece sits from the axis, captured by

I = \sum m_{i} r_{i}^{2}

. Same

M

and

ω

but different geometry means different

I

, hence different

K

Why questions

Why can

\frac{1}{2} ω^{2}

be pulled out of the particle-by-particle sum?

Because the body is rigid: every particle sweeps the same angle in the same time, so they share one common

ω

. A constant factor comes out of a sum.

Why does

I

grow like distance-squared rather than distance?

A particle at distance

r

from the axis has speed

v = r ω

, and KE carries

v^{2} = r^{2} ω^{2}

. The

r^{2}

in the energy is what survives into

I = \sum m_{i} r_{i}^{2}

Why does moving mass outward make an object both harder to spin up and store more energy at a given

ω

Both effects trace to the larger

I

: bigger

I

means more torque needed for the same angular acceleration (

τ = I α

) and more energy

\frac{1}{2} I ω^{2}

held at the same spin rate. One quantity governs both.

Why do we prefer energy methods over

τ = I α

for a ball rolling down a ramp?

Energy conservation only needs start and end states, sidestepping the varying friction and geometry along the path. Solving

τ = I α

would require tracking forces at every instant.

Why must the axis be stated before quoting

I

Because

I = \sum m_{i} r_{i}^{2}

measures each

r_{i}

= distance from that axis; change the axis and every

r_{i}

changes, so

I

(and any energy computed from it) changes.

Why does friction do no net work on a wheel that rolls without slipping?

The contact point is instantaneously at rest relative to the ground, so the static friction force acts through zero displacement there and transfers no energy — it only redirects motion between translation and rotation.

Why is the formula

\frac{1}{2} I ω^{2}

deliberately built to look like

\frac{1}{2} m v^{2}

To make the linear↔rotational analogy explicit:

m \to I

(resistance to motion) and

v \to ω

(rate of motion). Recognizing the parallel lets you reuse linear intuition directly.

Edge cases

What is the rotational KE of a body spinning at

ω = 0

Zero.

K = \frac{1}{2} I (0)^{2} = 0

— no spin, no rotational energy, regardless of how large

I

is.

A point mass sits exactly on the rotation axis. How much rotational KE does it carry?

Zero. Its distance from the axis is

r = 0

, so

v = r ω = 0

; it contributes nothing to

\sum m_{i} r_{i}^{2}

. Only off-axis mass stores rotational energy.

An idealized axis (a geometric line) has "zero radius." Does the on-axis material still have KE from the spin?

No. Points on the axis have

r = 0

and stay put as the body turns, so they carry no rotational KE even though the body around them is moving.

Can a very light object store more rotational energy than a heavier one at the same

ω

Yes, if its mass sits farther from the axis. Since

K = \frac{1}{2} (\sum m_{i} r_{i}^{2}) ω^{2}

, a small mass at large

r

can outweigh a large mass at small

r

ω \to \infty

, what happens to

K

, and why do real flywheels have a limit?

K \to \infty

mathematically (

K \propto ω^{2}

), but real material fails first — the outward pull on the rim (

\propto ω^{2} r

) eventually exceeds the material's strength and the wheel bursts. Energy storage is capped by strength, not by the formula.

In the limit where all of a body's mass collapses onto the axis, what is its rotational KE at any

ω

Zero. With every

r_{i} \to 0

I \to 0

and

K = \frac{1}{2} I ω^{2} \to 0

. A mass with no spread stores no spin energy.

A body translates in a straight line without any rotation. Which KE terms apply?

Only translational,

\frac{1}{2} M v_{c m}^{2}

. With

ω = 0

the rotational term vanishes — the general rolling formula reduces to the pure-translation case.

A wheel spins in place about a fixed axle (no translation of its center). Which KE terms apply?

Only rotational,

\frac{1}{2} I ω^{2}

. Since

v_{c m} = 0

, the translational term drops out — the mirror image of the pure-translation case.

Connections

Rotational kinetic energy = ½Iω² — the formula these traps probe.
Moment of inertia — why $I$ depends on axis and mass spread.
Kinetic energy of a particle — the $\frac{1}{2} m v^{2}$ pieces summed here.
Rolling motion — source of the "only translational" trap.
Parallel axis theorem — the fix for wrong-axis $I$ .
Angular velocity — the radians-not-degrees pitfall.
Conservation of energy — behind the ramp reasoning.
Work-energy theorem (rotational) — why friction does no net work while rolling.