Rotational vs Translational Kinetic Energy

Translational kinetic energy is one half m v squared for the motion of the center of mass. Rotational kinetic energy is one half I omega squared for the spin about it. They are not rival quantities: a rolling object has both at once, and adding them gives the total.

AP Physics: Unit 6 (topics 3.1 Translational Kinetic Energy, 6.1 Rotational Kinetic Energy, 6.5 Rolling). Translational kinetic energy is AP Physics 1 Unit 3, Work, Energy, and Power, weighted at 18 to 23 percent of the multiple-choice section, at Topic 3.1, learning objective 3.1.A, describe the translational kinetic energy of an object in terms of the object's mass and velocity. EK 3.1.A.1 gives K = (1/2)mv^2, EK 3.1.A.2 states that it is a scalar, and EK 3.1.A.3 states that different observers may measure different values depending on the observer's frame of reference. Suggested skills for Topic 3.1 are 1.C, 2.B, 3.B and 3.C. Rotational kinetic energy is Unit 6, Energy and Momentum of Rotating Systems, weighted at 5 to 8 percent, at Topic 6.1, learning objective 6.1.A, describe the rotational kinetic energy of a rigid system in terms of the rotational inertia and angular velocity of that rigid system. EK 6.1.A.1 gives K = (1/2)I omega^2; EK 6.1.A.1.i states that the rotational inertia of an object about a fixed axis can be used to show that its rotational kinetic energy is equivalent to its translational kinetic energy, which is its total kinetic energy; EK 6.1.A.1.ii states that the total kinetic energy of a rigid system is the sum of the rotational kinetic energy due to rotation about its center of mass and the translational kinetic energy due to the linear motion of its center of mass; EK 6.1.A.2 states that a rigid system can have rotational kinetic energy while its center of mass is at rest; and EK 6.1.A.3 states that rotational kinetic energy is a scalar quantity. Suggested skills for Topic 6.1 are 1.A, 2.B, 2.C and 3.C. The two terms are combined in Topic 6.5, Rolling, at EK 6.5.A.1, K_tot = K_trans + K_rot, with suggested skills 1.A, 2.A, 2.C and 3.C. On the sheets: all four print K = (1/2)mv^2, the work-energy theorem, I = sum m_i r_i^2 and the parallel axis theorem I' = I_cm + Md^2. The AP Physics 1 and AP Physics 2 sheets print the rotational energy as a bare K = (1/2)I omega^2, sharing the symbol with the translational one; the two Physics C sheets print K_rot = (1/2)I omega^2. Rotational work is printed on all four, as W = tau delta theta on the algebra-based sheets and as the integral of tau with respect to theta on the Physics C ones. None of the four sheets prints P = tau omega.

The second one is the first one, summed over the parts

Most treatments introduce these as two formulas that resemble each other. They are closer than that. Rotational kinetic energy is not a new kind of energy. It is the ordinary kinetic energy of the bits of a spinning object, added up.

The AP Physics 1 CED says so at EK 6.1.A.1.i: the rotational inertia of an object about a fixed axis can be used to show that the rotational kinetic energy of that object is equivalent to its translational kinetic energy, which is its total kinetic energy. And again from the other direction at EK 6.1.A.2: a rigid system can have rotational kinetic energy while its center of mass is at rest due to the individual points within the rigid system having linear speed and, therefore, kinetic energy.

The derivation is three lines. Chop a rigid body into particles of mass $m_i$ at distance $r_i$ from the axis. Every particle is doing ordinary translation at $v_i = r_i\omega$ , so add up ordinary kinetic energies:

K = \sum \frac{1}{2}m_i v_i^2 = \sum \frac{1}{2}m_i (r_i\omega)^2 = \frac{1}{2}\left(\sum m_i r_i^2\right)\omega^2 = \frac{1}{2}I\omega^2

The middle step is where the rotational inertia appears, and it appears because $\sum m_i r_i^2$ is exactly what the AP equation sheets define $I$ to be. Nothing was assumed and nothing new was introduced. The whole of rotational kinetic energy is a bookkeeping device for not having to sum over particles every time.

That is why the two are so easy to confuse and why the confusion matters: they are the same physical thing described at two levels. What separates them is not what they are but what each one accounts for.

$\frac{1}{2}mv_{\text{cm}}^2$ accounts for the motion of the object as a whole, treating it as a point at its center of mass.
$\frac{1}{2}I\omega^2$ accounts for the motion the parts have in addition to that, because they are also circling the center.

EK 6.1.A.1.ii puts the two together: the total kinetic energy of a rigid system is the sum of its rotational kinetic energy due to its rotation about its center of mass and the translational kinetic energy due to the linear motion of its center of mass. Two terms, no overlap, no double counting.

Rotational vs translational kinetic energy, side by side

Property	Translational kinetic energy	Rotational kinetic energy
Equation	$K = \frac{1}{2}mv^2$ , EK 3.1.A.1	$K = \frac{1}{2}I\omega^2$ , EK 6.1.A.1
Inertia term	Mass $m$ , in kg	Rotational inertia $I = \sum m_i r_i^2$ , in $\text{kg}\cdot\text{m}^2$
Speed term	$v$ in m/s	$\omega$ in rad/s, radians compulsory
Unit of the result	Joules	Joules, the same unit
Scalar or vector	Scalar, EK 3.1.A.2	Scalar, EK 6.1.A.3
Depends on the axis chosen	No	Yes, because $I$ does
Zero when	The center of mass is at rest	The object is not spinning
Can be nonzero with the object's center of mass at rest	No	Yes, EK 6.1.A.2
Work that changes it	$W = F_{\parallel}d = Fd\cos\theta$	$W = \tau\Delta\theta$
Frame dependence stated in the CED	Yes, EK 3.1.A.3	No matching statement under Topic 6.1
Symbol on the AP Physics 1 and 2 sheets	$K$	$K$ , the same letter
Symbol on the two Physics C sheets	$K$	$K_{\text{rot}}$
Which CED unit	Unit 3, Work, Energy, and Power	Unit 6, Energy and Momentum of Rotating Systems

Two rows are worth pausing on.

"Depends on the axis chosen." Translational kinetic energy needs a reference frame and nothing more. Rotational kinetic energy needs a frame and an axis, because $I$ is a property of the object and the axis together. Change the axis and the same spin carries a different rotational kinetic energy, which is one of the reasons mass vs rotational inertia is a real distinction rather than a relabelling.

"Symbol on the AP Physics 1 and 2 sheets." Both sheets print $K = \frac{1}{2}mv^2$ in the translational block and $K = \frac{1}{2}I\omega^2$ in the rotational block, with no subscript on either. The two Physics C sheets write the second as $K_{\text{rot}}$ . So on an algebra-based exam the two energies share a letter and you have to keep them apart yourself, which is a small typographic fact with a real cost in a rolling problem.

The case that separates them: a cylinder that is doing both

A solid cylinder of mass $3.0$ kg and radius $0.20$ m rolls without slipping at $6.0$ m/s. Its rotational inertia about its center is $I = \frac{1}{2}mr^2 = 0.060\ \text{kg}\cdot\text{m}^2$ , and the rolling condition gives $\omega = \frac{v}{r} = 30$ rad/s.

Quantity	Value
Translational kinetic energy, $\frac{1}{2}mv^2$	$54$ J
Rotational kinetic energy, $\frac{1}{2}I\omega^2$	$27$ J
Total, by EK 6.5.A.1	$81$ J
Rotational share	$33.3\%$

Both terms are live, at the same instant, for the same object. That is the case the whole page is about, and it is why "is this translational or rotational kinetic energy" is usually the wrong question.

Now change one thing at a time.

Take the same cylinder and spin it at $30$ rad/s with its center held still. Translational kinetic energy is zero and rotational kinetic energy is still $27$ J. EK 6.1.A.2 licenses this directly: the individual points within the rigid system have linear speed and therefore kinetic energy, even though the object as a whole is going nowhere.

Take the same cylinder and slide it at $6.0$ m/s without letting it turn. Translational kinetic energy is $54$ J and rotational is zero.

Take a hoop of the same mass and radius rolling at $6.0$ m/s. Now $I = mr^2 = 0.12\ \text{kg}\cdot\text{m}^2$ , the rotational term doubles to $54$ J, the translational term is unchanged at $54$ J, and the split is even. Same mass, same speed, same radius, twice the rotational energy, because the mass sits further from the axis.

Read those four together and the division of labour is clear. The translational term only ever asks how fast the center of mass is moving. The rotational term asks how the mass is arranged and how fast it turns, and it does not care whether the object is going anywhere.

For a rolling object the two are locked, because $\omega = \frac{v}{r}$ ties them together, and the ratio comes out as a pure number:

\frac{K_{\text{rot}}}{K_{\text{trans}}} = \frac{\frac{1}{2}(\beta m r^2)\left(\frac{v}{r}\right)^2}{\frac{1}{2}mv^2} = \beta

where $\beta = \frac{I}{mr^2}$ is the shape factor. So a solid cylinder always splits one third rotational and two thirds translational when it rolls, whatever its size, mass or speed. A solid sphere splits $\frac{2}{7}$ and $\frac{5}{7}$ . A hoop splits it evenly. That constancy is what drives the ramp race in rolling vs sliding.

The analogy, term for term, and where it stops

The CED states the analogy as a required piece of knowledge rather than a study aid. EK 5.1.A.4: angular displacement, angular velocity, and angular acceleration around one axis are analogous to linear displacement, velocity, and acceleration in one dimension and demonstrate the same mathematical relationships.

Extended to energy, the correspondence is exact everywhere it is printed:

Translational	Rotational	Both on the sheets
$m$	$I$	Yes, $I = \sum m_i r_i^2$
$v$	$\omega$	Yes, linked by $v = r\omega$
$K = \frac{1}{2}mv^2$	$K = \frac{1}{2}I\omega^2$	Yes, both blocks
$\vec{F}_{\text{net}} = m\vec{a}$	$\tau_{\text{net}} = I\alpha$	The sheets give $\alpha_{\text{sys}} = \frac{\Sigma\tau}{I_{\text{sys}}}$
$W = Fd\cos\theta$	$W = \tau\Delta\theta$	Yes, both
$\Delta K = \sum W_i$	Same statement, applied to $\frac{1}{2}I\omega^2$	The work-energy theorem is printed once
$P = Fv\cos\theta$	$P = \tau\omega$	No. $P = \tau\omega$ is on none of the four sheets
$\vec{p} = m\vec{v}$	$L = I\omega$	Yes, both

The rotational power relationship is a genuine gap worth knowing about. All four sheets print an instantaneous power for the translational case, and none prints $P = \tau\omega$ . It follows in one line from $P = \frac{W}{\Delta t}$ with $W = \tau\Delta\theta$ , so it is a derivation rather than a memorised extra, but it is not handed to you.

Where the analogy stops is not in the algebra, it is in what the symbols mean.

$m$ cannot change during a problem; $I$ can, because a system can rearrange its own mass relative to the axis. That single asymmetry is why a skater speeds up.
$m$ needs no axis; $I$ needs one, so a rotational kinetic energy without a stated axis is not yet a number.
A rolling body needs both terms; a sliding body needs one. There is no situation in which you add two translational kinetic energies for a single rigid object, and there is a very common one in which you add a translational and a rotational.
The CED attaches a frame-dependence clause to the translational case and not to the rotational one. EK 3.1.A.3 reads: different observers may measure different values of the translational kinetic energy of an object, depending on the observer's frame of reference. Topic 6.1's three essential knowledge statements say the rotational energy is a scalar and how it combines, and print no matching sentence.

Adding them without double counting

The sum in EK 6.1.A.1.ii is precise about which rotation and which translation, and the precision is doing work.

K_{\text{tot}} = \underbrace{\frac{1}{2}Mv_{\text{cm}}^2}_{\text{motion of the center of mass}} + \underbrace{\frac{1}{2}I_{\text{cm}}\omega^2}_{\text{rotation about the center of mass}}

Two conditions are hiding in that line.

The rotational inertia must be about the center of mass, not about some other axis. Use $I$ about a different point and you will have counted part of the translation twice, because a rotation about an off-center axis already carries the center of mass around. If a problem hands you $I$ about the end of a rod and you want the two-term form, shift it with the parallel axis theorem $I' = I_{\text{cm}} + Md^2$ , which is printed on all four sheets, and use $I_{\text{cm}}$ .

The velocity must be that of the center of mass, not of some convenient point on the rim.

Get both right and the two terms are genuinely independent contributions with no overlap. That is a theorem rather than an approximation, and it is what makes energy problems about rolling objects tractable at all.

The alternative for a rolling object is to work about the contact point, where the whole motion is a pure rotation, because the contact point is instantaneously at rest. Then

K_{\text{tot}} = \frac{1}{2}I_{\text{contact}}\omega^2 \quad\text{with}\quad I_{\text{contact}} = I_{\text{cm}} + Mr^2

and expanding gives $\frac{1}{2}(I_{\text{cm}} + Mr^2)\omega^2 = \frac{1}{2}I_{\text{cm}}\omega^2 + \frac{1}{2}M(r\omega)^2 = \frac{1}{2}I_{\text{cm}}\omega^2 + \frac{1}{2}Mv_{\text{cm}}^2$ , which is the two-term form again. Two routes, one answer, and the parallel axis theorem is what connects them. Use one or the other. Using both, by adding a translational term to a rotational term computed about the contact point, is the double count and it inflates the energy.

Worked example three does that consistency check with numbers.

Where the confusion costs a mark

Writing only $\frac{1}{2}mv^2$ for a rolling object. EK 6.5.A.1 requires both terms. For a solid cylinder that omission understates the kinetic energy by a third and makes the object arrive too fast in a ramp problem.
Writing only $\frac{1}{2}I\omega^2$ for a rolling object. The other half of the same mistake, and it understates the energy by two thirds for a cylinder.
Using $I$ about the wrong axis in the two-term sum. It must be $I_{\text{cm}}$ . Using $I$ about the contact point in that sum double counts the translation.
Using degrees per second for $\omega$ . $\frac{1}{2}I\omega^2$ requires radians, because it descends from $v = r\omega$ , and the error is a factor of $57.2958$ squared, which is about $3283$ .
Treating rotational kinetic energy as having a direction. EK 6.1.A.3 says it is a scalar. Clockwise and counterclockwise at the same rate carry identical energy, and there is no sign to track.
Assuming an object with its center of mass at rest has no kinetic energy. EK 6.1.A.2 says a spinning system does, because its parts are moving.
Adding the two kinetic energies as vectors, or worrying about their directions. They are scalars and they add as numbers.
Assuming a heavier or larger object stores more rotational energy at a given rolling speed. The ratio to the translational term is $\beta$ , the shape factor, and mass and radius cancel out of it.
Quoting a rotational inertia from memory. No AP equation sheet prints a table for common shapes, so an exam supplies the $I$ it wants you to use.
Confusing the two on an algebra-based sheet because they share the symbol $K$ . The AP Physics 1 and AP Physics 2 sheets print both as a bare $K$ . Subscript them yourself in your working.

When one term dominates, and why that lulls you

Several everyday setups make one of the two invisible, and each teaches a habit that fails on the next problem.

All of Unit 3 has no rotational term. Work, energy and power in AP Physics 1 Unit 3 treat objects as points, so $\frac{1}{2}mv^2$ is the whole of the kinetic energy for the entire unit. That is a hundred problems' worth of practice at ignoring the other term, three units before you meet it.

A flywheel or a turntable has no translational term. Anything spinning about a fixed axle has $v_{\text{cm}} = 0$ , so the rotational term is the whole story and the two never have to be added.

A sliding block has no rotational term. Same lesson as Unit 3 with friction added.

An object whose mass is concentrated near its axis has a small rotational share. A car's forward kinetic energy hugely exceeds the rotational energy of its wheels, so ignoring the wheels is a fair approximation in an engineering estimate and a wrong answer on an AP question about a rolling wheel.

The places where both terms are compulsory:

Any rolling object. This is the big one, and it is the whole of Topic 6.5.
A yo-yo, or anything unwinding from a string. The falling motion and the spin are locked together and both carry energy.
A rod pivoted at one end and released. Depending on which axis you work about, either one term about the pivot or two terms about the center of mass, and the parallel axis theorem reconciles them.
A collision between a moving object and something that can spin, where the energy accounting has to name where each joule went.

The habit worth building: when an object can both move and turn, write two terms and set one to zero if it deserves to be zero. Writing two and cancelling one costs a line. Writing one and forgetting the other costs the question.

What the CED asks, and how the exam frames it

Translational kinetic energy is AP Physics 1 Unit 3, Work, Energy, and Power, weighted at 18 to 23 percent of the multiple-choice section, at Topic 3.1, Translational Kinetic Energy. One learning objective, 3.1.A: describe the translational kinetic energy of an object in terms of the object's mass and velocity. Three essential knowledge statements support it: EK 3.1.A.1 gives $K = \frac{1}{2}mv^2$ , EK 3.1.A.2 says translational kinetic energy is a scalar quantity, and EK 3.1.A.3 says different observers may measure different values of the translational kinetic energy of an object depending on the observer's frame of reference. Suggested skills for Topic 3.1 are 1.C, 2.B, 3.B and 3.C.

Rotational kinetic energy is Unit 6, Energy and Momentum of Rotating Systems, weighted at 5 to 8 percent, at Topic 6.1, Rotational Kinetic Energy. One learning objective, 6.1.A: describe the rotational kinetic energy of a rigid system in terms of the rotational inertia and angular velocity of that rigid system. EK 6.1.A.1 gives $K = \frac{1}{2}I\omega^2$ ; EK 6.1.A.1.i says the rotational inertia about a fixed axis can be used to show that the object's rotational kinetic energy is equivalent to its translational kinetic energy, which is its total kinetic energy; EK 6.1.A.1.ii gives the two-term sum for a system whose center of mass is moving; EK 6.1.A.2 says a rigid system can have rotational kinetic energy while its center of mass is at rest; and EK 6.1.A.3 says rotational kinetic energy is a scalar quantity. Suggested skills for Topic 6.1 are 1.A, 2.B, 2.C and 3.C.

The two terms are put together in Topic 6.5, Rolling, at EK 6.5.A.1: the total kinetic energy of a system is the sum of the system's translational and rotational kinetic energies, $K_{\text{tot}} = K_{\text{trans}} + K_{\text{rot}}$ . Suggested skills for Topic 6.5 are 1.A, 2.A, 2.C and 3.C.

The three topics sit in two different units three units apart, which is a fair part of why the pairing is learned late.

On the sheets, checked across all four: every one prints $K = \frac{1}{2}mv^2$ , the work-energy theorem $\Delta K = \sum W_i = \sum F_{\parallel,i}d_i$ , and a rotational kinetic energy. The AP Physics 1 and AP Physics 2 sheets print the rotational one as a bare $K = \frac{1}{2}I\omega^2$ ; the AP Physics C: Mechanics and AP Physics C: Electricity and Magnetism sheets print $K_{\text{rot}} = \frac{1}{2}I\omega^2$ . All four print rotational work: $W = \tau\Delta\theta$ on the algebra-based sheets and $W = \int \tau \cdot d\theta$ on the Physics C ones. All four print $I = \sum m_i r_i^2$ and the parallel axis theorem $I' = I_{\text{cm}} + Md^2$ , with the Physics C sheets adding $I = \int r^2\,dm$ . None of the four prints $P = \tau\omega$ .

For the surrounding material: Topic 3.1 and Topic 6.1 carry the CED framing, conservation of energy and the work-energy theorem have the routines, and the work, energy and power practice set has problems. The definitions live at rotational kinetic energy and kinetic energy.

Deriving one half I omega squared by adding up ordinary kinetic energies

Two point masses of $0.50$ kg each sit at the ends of a massless rod, $0.40$ m from the midpoint on either side. The rod spins about its midpoint at $6.0$ rad/s with the midpoint held fixed. Find the total kinetic energy twice: once by adding the translational kinetic energies of the two masses, and once from $\frac{1}{2}I\omega^2$ . Then repeat with the masses moved to $0.20$ m and comment.

Route one, particle by particle. Each mass moves in a circle of radius $0.40$ m at $v = r\omega = (0.40)(6.0) = 2.4$ m/s.
Kinetic energy of one mass, from EK 3.1.A.1: $K = \frac{1}{2}mv^2 = \frac{1}{2}(0.50)(2.4)^2 = \frac{1}{2}(0.50)(5.76) = 1.44$ J.
Two of them: $K_{\text{total}} = 2(1.44) = 2.88$ J. Note that nothing rotational was used. This is two ordinary kinetic energies added together.
Route two, through the rotational inertia. $I = \sum m_i r_i^2 = (0.50)(0.40)^2 + (0.50)(0.40)^2 = 0.080 + 0.080 = 0.16\ \text{kg}\cdot\text{m}^2$ .
$K = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.16)(6.0)^2 = \frac{1}{2}(0.16)(36) = 2.88$ J. The same number, and it had to be: the second route is the first route with the algebra done in advance, which is what EK 6.1.A.1.i asserts.
Confirm that the centre of mass is at rest. The two masses are equal and opposite about the midpoint, so $\vec{v}_{\text{cm}} = 0$ and the translational term $\frac{1}{2}Mv_{\text{cm}}^2$ is zero. All $2.88$ J of it is rotational, with the object going nowhere, which is exactly EK 6.1.A.2.
Now move the masses to $0.20$ m, same rod, same $6.0$ rad/s. Each speed becomes $v = (0.20)(6.0) = 1.2$ m/s, each kinetic energy $\frac{1}{2}(0.50)(1.44) = 0.36$ J, and the total is $0.72$ J.
Check through the rotational route: $I = 2(0.50)(0.20)^2 = 0.040\ \text{kg}\cdot\text{m}^2$ and $K = \frac{1}{2}(0.040)(36) = 0.72$ J. Agreement again.
Halving the radius quartered the energy at fixed $\omega$ , because $I$ carries $r^2$ . Nothing similar happens on the translational side: moving mass around inside an object does not change $\frac{1}{2}mv_{\text{cm}}^2$ at all. That is the asymmetry the whole comparison rests on.

At $0.40$ m the total is $2.88$ J by both routes, all of it rotational since the center of mass is at rest. At $0.20$ m it is $0.72$ J by both routes, a quarter as much for the same angular velocity. The two routes agree because $\frac{1}{2}I\omega^2$ is the sum of the parts' translational kinetic energies with the algebra done once.

A rolling cylinder, and the same cylinder doing one thing at a time

A solid cylinder of mass $3.0$ kg and radius $0.20$ m has $I = \frac{1}{2}mr^2$ about its center. Find its translational, rotational and total kinetic energy when it rolls without slipping at $6.0$ m/s; when it slides at $6.0$ m/s without turning; and when it spins at $30$ rad/s with its center held still. Then repeat the rolling case for a hoop of the same mass and radius.

Rotational inertia: $I = \frac{1}{2}(3.0)(0.20)^2 = \frac{1}{2}(3.0)(0.040) = 0.060\ \text{kg}\cdot\text{m}^2$ .
Rolling at $6.0$ m/s. The rolling condition from EK 6.5.B.1 gives $\omega = \frac{v}{r} = \frac{6.0}{0.20} = 30\ \text{rad/s}$ .
Translational term: $\frac{1}{2}Mv_{\text{cm}}^2 = \frac{1}{2}(3.0)(6.0)^2 = \frac{1}{2}(3.0)(36) = 54$ J.
Rotational term: $\frac{1}{2}I\omega^2 = \frac{1}{2}(0.060)(30)^2 = \frac{1}{2}(0.060)(900) = 27$ J.
Total, by EK 6.5.A.1: $54 + 27 = 81$ J, of which $\frac{27}{81} = 33.3\%$ is rotational. Check against the shape factor: $\beta = \frac{I}{mr^2} = \frac{0.060}{(3.0)(0.040)} = 0.500$ , and the predicted ratio $\frac{K_{\text{rot}}}{K_{\text{trans}}} = \beta = 0.500$ , which matches $\frac{27}{54}$ .
Sliding at $6.0$ m/s without turning. $\omega = 0$ , so the rotational term is zero and the total is $54$ J. The object carries two thirds of the energy it had while rolling at the same speed.
Spinning at $30$ rad/s with the center at rest. $v_{\text{cm}} = 0$ , so the translational term is zero and the total is $27$ J, all rotational, as EK 6.1.A.2 allows.
Now the hoop, same mass, same radius, rolling at $6.0$ m/s. $I = mr^2 = (3.0)(0.040) = 0.12\ \text{kg}\cdot\text{m}^2$ , twice the cylinder's, and $\omega$ is still $30$ rad/s because the rolling condition depends only on $v$ and $r$ .
Hoop terms: translational $54$ J, unchanged; rotational $\frac{1}{2}(0.12)(900) = 54$ J; total $108$ J, split exactly evenly. Same mass, same radius, same speed, and the hoop is carrying a third more energy than the cylinder, all of it in the extra rotation.
The comparison in one line. The translational term was $54$ J in every rolling and sliding case here, because it only ever asks how fast the center of mass moves. The rotational term went $27$ , $0$ , $27$ and $54$ J, because it asks about the arrangement of the mass and the rate of spin.

Rolling cylinder: $54$ J translational, $27$ J rotational, $81$ J total, $33.3\%$ rotational. Sliding without turning: $54$ J total. Spinning with the center at rest: $27$ J total. Rolling hoop of the same mass and radius: $54$ J and $54$ J, $108$ J total, split evenly.

Two routes to the same energy, and the double count that ruins it

Take the same solid cylinder, $3.0$ kg, radius $0.20$ m, $I_{\text{cm}} = 0.060\ \text{kg}\cdot\text{m}^2$ , rolling without slipping at $6.0$ m/s. Compute its total kinetic energy as a pure rotation about the contact point, using the parallel axis theorem. Show that the answer matches the two-term form, and show what the common double count would give. Then check the analogy on the work side: a torque of $12\ \text{N}\cdot\text{m}$ through $5.0$ rad on a flywheel with $I = 2.4\ \text{kg}\cdot\text{m}^2$ from rest, against a force of $12$ N through $5.0$ m on a $2.4$ kg block from rest.

Route one, about the contact point. A rolling object's contact point is instantaneously at rest, so the whole motion can be treated as a pure rotation about it. The rotational inertia about that axis comes from the parallel axis theorem, printed on all four sheets: $I' = I_{\text{cm}} + Md^2 = 0.060 + (3.0)(0.20)^2 = 0.060 + 0.12 = 0.18\ \text{kg}\cdot\text{m}^2$ .
$K = \frac{1}{2}I'\omega^2 = \frac{1}{2}(0.18)(30)^2 = \frac{1}{2}(0.18)(900) = 81$ J.
Route two, the two-term form. $\frac{1}{2}Mv_{\text{cm}}^2 + \frac{1}{2}I_{\text{cm}}\omega^2 = 54 + 27 = 81$ J. The two routes agree exactly, which they must, because expanding $\frac{1}{2}(I_{\text{cm}} + Mr^2)\omega^2$ gives $\frac{1}{2}I_{\text{cm}}\omega^2 + \frac{1}{2}M(r\omega)^2$ and $r\omega$ is $v_{\text{cm}}$ .
The double count. Adding a translational term to a rotational term computed about the contact point gives $54 + \frac{1}{2}(0.18)(900) = 54 + 81 = 135$ J, which is $67\%$ too high. The $54$ J of translation was already inside the $81$ J, carried by the $Md^2$ part of the shifted inertia. Pick one route and finish it.
Now the work side of the analogy. Rotational case. $W = \tau\Delta\theta = (12)(5.0) = 60$ J. From rest, the work-energy theorem gives $\frac{1}{2}I\omega^2 = 60$ , so $\omega^2 = \frac{120}{2.4} = 50$ and $\omega = 7.07\ \text{rad/s}$ .
Translational case. $W = Fd = (12)(5.0) = 60$ J. From rest, $\frac{1}{2}mv^2 = 60$ , so $v^2 = \frac{120}{2.4} = 50$ and $v = 7.07$ m/s.
Identical arithmetic, different units. $7.07\ \text{rad/s}$ against $7.07$ m/s, from equations that differ only by which letters stand in the slots. That is what EK 5.1.A.4 means by the same mathematical relationships, and it is the reason the analogy is worth trusting when the symbols are used carefully.
One more line to close the loop on the power gap. At that instant the rotational power is $\tau\omega = (12)(7.071) = 84.9$ W and the translational power is $Fv = (12)(7.071) = 84.9$ W. The second of those is printed on the equation sheets as $P_{\text{inst}} = F_{\parallel}v$ ; the first is not printed anywhere, and follows from $W = \tau\Delta\theta$ divided by $\Delta t$ .

Both routes give $81$ J: $\frac{1}{2}(0.18)(30)^2$ about the contact point, and $54 + 27$ in the two-term form. Adding a translational term on top of the contact-point rotation gives $135$ J, which double counts the translation by $54$ J. On the work side, $60$ J of work produces $\omega = 7.07\ \text{rad/s}$ on the flywheel and $v = 7.07$ m/s on the block, and the instantaneous powers are both $84.9$ W.

Frequently asked questions

What is the difference between rotational and translational kinetic energy?

Translational kinetic energy is one half the mass times the square of the speed of the center of mass, and it accounts for the object moving as a whole. Rotational kinetic energy is one half the rotational inertia times the square of the angular velocity, and it accounts for the extra motion the parts have because they are circling the center. They are not rival descriptions: the AP Physics 1 CED states at essential knowledge 6.1.A.1.ii that the total kinetic energy of a rigid system is the sum of the two. Both are scalars measured in joules, so they add as ordinary numbers.

Does a rolling object have both kinds of kinetic energy?

Yes, and that is the case most problems are built around. Essential knowledge 6.5.A.1 in the AP Physics 1 CED gives the total kinetic energy of a system with translational and rotational motion as the sum of the two. A 3.0 kilogram solid cylinder of radius 0.20 meters rolling at 6.0 meters per second has 54 joules of translational kinetic energy and 27 joules of rotational, for 81 joules in total. Writing only the first term is the standard error in a rolling energy problem, and for a solid cylinder it understates the answer by a third.

Is rotational kinetic energy the same as translational kinetic energy?

In origin, yes. Rotational kinetic energy is the ordinary kinetic energy of the individual parts of a spinning object, added up: summing one half m v squared over particles moving at r omega gives one half times the sum of m r squared times omega squared, which is one half I omega squared. Essential knowledge 6.1.A.1.i in the AP Physics 1 CED says the rotational inertia can be used to show exactly this. In use they are different, because one accounts for the motion of the center of mass and the other for the motion about it, and a rolling object needs both terms.

Can an object have rotational kinetic energy without moving?

Yes. Essential knowledge 6.1.A.2 says a rigid system can have rotational kinetic energy while its center of mass is at rest, because the individual points within it have linear speed and therefore kinetic energy. A flywheel on a fixed axle is the standard example: it goes nowhere and stores a great deal of energy. A 3.0 kilogram solid cylinder of radius 0.20 meters spinning at 30 radians per second with its center held still carries 27 joules, all of it rotational.

How much of a rolling object's energy is rotational?

The fraction is beta divided by one plus beta, where beta is the rotational inertia divided by mass times radius squared. That is 2 sevenths, about 28.6 percent, for a solid sphere; one third for a solid cylinder; 2 fifths for a thin spherical shell; and one half for a hoop. Mass, radius and speed all cancel out of the ratio, so a solid cylinder rolling at any speed always splits its kinetic energy one third rotational and two thirds translational. That constancy is why the ordering in a ramp race depends only on shape.

Is rotational kinetic energy a vector?

No. Essential knowledge 6.1.A.3 says rotational kinetic energy is a scalar quantity, exactly as essential knowledge 3.1.A.2 says translational kinetic energy is. There is no direction and no sign attached to either, so a wheel spinning clockwise and the same wheel spinning counterclockwise at the same rate carry identical rotational kinetic energy. When you add the rotational and translational terms for a rolling object you add them as plain numbers, with no components and no vector arithmetic.

Which axis do you use for the rotational kinetic energy of a rolling object?

Either the center of mass or the contact point, and you must not mix them. Using the center of mass, the total is one half M v squared plus one half I about the center times omega squared, which is the form essential knowledge 6.1.A.1.ii gives. Using the contact point, which is instantaneously at rest for a rolling object, the whole motion is a pure rotation and the total is one half I about the contact point times omega squared, where that inertia comes from the parallel axis theorem. The two give the same answer. Adding a translational term to the contact-point version double counts the translation.