11.1 Rotational Kinetic Energy

Energy Hidden in Spin

A flywheel in a Formula 1 car spins at 60,000 RPM. That flywheel stores enough energy to accelerate the car out of a curve --- energy that the team's engineers harvest and deploy at precisely the right moment. But this energy is not in the car's forward motion. The car might be momentarily slowing for a turn. The energy is in the rotation itself, locked in the spinning mass of the flywheel.

Where exactly is this energy? You already know that a moving object carries kinetic energy $\frac{1}{2}mv^2$. But every point on the flywheel is moving --- each tiny piece of mass traces a circle at some speed that depends on its distance from the axis. The energy is distributed across all those moving pieces. The question is: can we write a single, clean expression for the total kinetic energy stored in rotation?

Before you read on: Two wheels spin at the same angular velocity $\omega$. Wheel A has twice the moment of inertia of Wheel B. How much more rotational kinetic energy does Wheel A have --- twice as much, or four times as much?

Commit to your answer before continuing.

Building the Formula from What You Know

Start with something familiar. A single particle of mass $m$ moving at speed $v$ has kinetic energy:

$$KE = \frac{1}{2}mv^2$$

Now imagine that particle is not moving in a straight line but traveling in a circle of radius $r$ around a fixed axis. Its speed is related to the angular velocity by $v = r\omega$ (from Section 9.5). Substituting:

$$KE = \frac{1}{2}m(r\omega)^2 = \frac{1}{2}mr^2\omega^2$$

That factor $mr^2$ should look familiar. For a single particle, the moment of inertia about the axis is $I = mr^2$ (Section 10.3). So the kinetic energy of a single particle moving in a circle is:

$$KE = \frac{1}{2}I\omega^2$$

A rigid body is just many particles, all sharing the same angular velocity $\omega$. The total kinetic energy is the sum of the kinetic energies of all the particles:

$$KE_{\text{rot}} = \sum \frac{1}{2}m_i r_i^2 \omega^2 = \frac{1}{2}\left(\sum m_i r_i^2\right)\omega^2 = \frac{1}{2}I\omega^2$$

The $\omega^2$ factors out because every particle in a rigid body has the same angular velocity. What differs from particle to particle is $m_i r_i^2$ --- each particle's contribution to the moment of inertia. The sum $\sum m_i r_i^2$ is just $I$, the total moment of inertia you computed in Chapter 10.

This is the rotational kinetic energy:

$$KE_{\text{rot}} = \frac{1}{2}I\omega^2$$

Now check your prediction. The formula is $KE_{\text{rot}} = \frac{1}{2}I\omega^2$. Since $KE$ is directly proportional to $I$ (not $I^2$), doubling $I$ at constant $\omega$ doubles the kinetic energy. Wheel A has twice the rotational kinetic energy of Wheel B.

If you predicted "four times," you may have been thinking of the effect of doubling $\omega$ instead --- we will see that next.

Exploring the Dependence on $I$ and $\omega$

[Interactive: Flywheel Energy Explorer. A flywheel is shown spinning about its center. Two sliders control the system:

1. Mass distribution slider: Adjusts the moment of inertia $I$ by moving mass between the hub (small $I$) and the rim (large $I$). The current value of $I$ is displayed in kg$\cdot$m$^2$. A visual animation shows the mass moving outward or inward.
2. Angular velocity slider: Adjusts $\omega$ from 0 to 20 rad/s. The flywheel visibly spins faster or slower.

A real-time readout displays $KE_{\text{rot}} = \frac{1}{2}I\omega^2$, both as a number and as a bar graph. The bar changes height continuously as the student adjusts either slider.

Guided exploration prompts:

• "Set $I = 2$ kg$\cdot$m$^2$ and $\omega = 4$ rad/s. Record the kinetic energy."
• "Now double $\omega$ to 8 rad/s, keeping $I$ the same. What happened to the KE? By what factor did it change?"
• "Reset $\omega$ to 4 rad/s. Now double $I$ to 4 kg$\cdot$m$^2$. What happened to the KE? By what factor did it change?"
• "Which has a bigger effect on KE --- doubling $\omega$ or doubling $I$?"
• "Can you find two different combinations of $I$ and $\omega$ that give the same KE?"]

If you worked through those prompts, you discovered something important:

• Doubling $I$ (at constant $\omega$) doubles $KE$. The relationship is linear.
• Doubling $\omega$ (at constant $I$) quadruples $KE$. The relationship is quadratic.

This asymmetry is built into the formula: $KE_{\text{rot}} = \frac{1}{2}I\omega^2$. The $\omega$ is squared; the $I$ is not. Angular velocity has a stronger effect on kinetic energy than moment of inertia does.

Pause and think: A figure skater pulls her arms in, decreasing her moment of inertia but increasing her angular velocity (you will see why in Section 11.5). Based on what you just discovered about the $I$ vs. $\omega^2$ dependence, would you expect her kinetic energy to increase, decrease, or stay the same?

The Structural Analogy

Look at these two equations side by side:

Translational	Rotational
$KE_{\text{trans}} = \frac{1}{2}mv^2$	$KE_{\text{rot}} = \frac{1}{2}I\omega^2$

Every piece maps across:

• Mass $m$ (resistance to translational acceleration) $\longleftrightarrow$ Moment of inertia $I$ (resistance to angular acceleration)
• Speed $v$ (translational rate) $\longleftrightarrow$ Angular speed $\omega$ (rotational rate)
• Both are multiplied by $\frac{1}{2}$. Both involve the square of the rate quantity.

This is not a coincidence. The rotational formula was derived from the translational one by substituting $v = r\omega$ and recognizing $\sum mr^2 = I$. The structural analogy reflects the physics: both formulas count the kinetic energy of moving mass, but one tracks motion along a line and the other tracks motion around an axis.

The analogy also tells you something about units. Since $I$ has units of kg$\cdot$m$^2$ and $\omega$ has units of rad/s:

$$[KE_{\text{rot}}] = \text{kg} \cdot \text{m}^2 \cdot \frac{1}{\text{s}^2} = \text{kg} \cdot \frac{\text{m}^2}{\text{s}^2} = \text{J}$$

Rotational kinetic energy is measured in joules, the same unit as translational kinetic energy. Energy is energy --- the unit does not care whether the motion is translational or rotational.

When Objects Both Translate and Rotate

So far, we have considered pure rotation about a fixed axis. But many objects do both: a bowling ball rolling down a lane, a wheel rolling along a road, a cylinder rolling down a ramp.

Consider a ball rolling across a table. Its center of mass moves forward with speed $v_{\text{cm}}$, and simultaneously the ball spins about its center with angular velocity $\omega$. These are two separate contributions to kinetic energy:

• The translational kinetic energy of the center of mass: $\frac{1}{2}mv_{\text{cm}}^2$
• The rotational kinetic energy about the center of mass: $\frac{1}{2}I_{\text{cm}}\omega^2$

The total kinetic energy is the sum:

$$KE_{\text{total}} = \frac{1}{2}mv_{\text{cm}}^2 + \frac{1}{2}I_{\text{cm}}\omega^2$$

This decomposition is exact, not an approximation. It comes from a theorem you can prove by carefully summing the kinetic energies of all the particles in the body: the total KE of a rigid body equals the KE of a particle with the body's total mass located at the center of mass, plus the KE of rotation about the center of mass.

For rolling without slipping, there is a kinematic constraint linking translation and rotation:

$$v_{\text{cm}} = R\omega$$

where $R$ is the radius. This constraint means that once you know $v_{\text{cm}}$ (or $\omega$), you know both. The two terms in the energy are not independent --- they are locked together by the rolling condition.

Pause and think: For a solid sphere ($I_{\text{cm}} = \frac{2}{5}mR^2$) rolling without slipping, what fraction of the total kinetic energy is rotational? What fraction is translational? You have all the information you need --- try it before reading on.

Check your answer

Substitute $I_{\text{cm}} = \frac{2}{5}mR^2$ and $v_{\text{cm}} = R\omega$: $$KE_{\text{rot}} = \frac{1}{2}\left(\frac{2}{5}mR^2\right)\omega^2 = \frac{1}{5}mR^2\omega^2$$ $$KE_{\text{trans}} = \frac{1}{2}m(R\omega)^2 = \frac{1}{2}mR^2\omega^2$$ $$KE_{\text{total}} = \frac{1}{2}mR^2\omega^2 + \frac{1}{5}mR^2\omega^2 = \frac{7}{10}mR^2\omega^2$$ The rotational fraction is: $$\frac{KE_{\text{rot}}}{KE_{\text{total}}} = \frac{\frac{1}{5}mR^2\omega^2}{\frac{7}{10}mR^2\omega^2} = \frac{1/5}{7/10} = \frac{2}{7} \approx 29\%$$ The translational fraction is $\frac{5}{7} \approx 71\%$. For a rolling solid sphere, about 29% of the kinetic energy is rotational. This fraction depends only on the shape (through $I_{\text{cm}}$), not on the mass or size.

Why Mass Distribution Matters

Two objects can have the same total mass and the same angular velocity, yet store very different amounts of rotational kinetic energy. The difference is in how the mass is distributed.

Consider two wheels, each with mass $m$ and radius $R$, spinning at the same $\omega$:

• A solid disk: $I = \frac{1}{2}mR^2$, so $KE_{\text{rot}} = \frac{1}{4}mR^2\omega^2$
• A thin-walled hoop: $I = mR^2$, so $KE_{\text{rot}} = \frac{1}{2}mR^2\omega^2$

The hoop stores twice the rotational kinetic energy of the disk, even though they have the same mass and the same angular velocity. The hoop concentrates all its mass at the maximum distance from the axis, giving it a larger moment of inertia.

This is why flywheels are designed with mass concentrated at the rim. For energy storage, you want the largest possible $I$ for a given mass. Pushing the mass outward is far more effective than simply adding more mass near the center.

Practice

Layer 1: Concrete

Problem 1. A grinding wheel has a moment of inertia of $I = 0.40$ kg$\cdot$m$^2$ and spins at $\omega = 25$ rad/s. Compute its rotational kinetic energy.

Check your answer

$$KE_{\text{rot}} = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.40)(25)^2 = \frac{1}{2}(0.40)(625) = 125 \text{ J}$$ The wheel stores 125 J of energy in its rotation.

Problem 2. A solid cylinder ($I = \frac{1}{2}mR^2$) has mass 6.0 kg, radius 0.20 m, and spins at 150 rad/s. Compute its rotational kinetic energy.

Check your answer

First, find the moment of inertia: $$I = \frac{1}{2}mR^2 = \frac{1}{2}(6.0)(0.20)^2 = \frac{1}{2}(6.0)(0.04) = 0.12 \text{ kg}\!\cdot\!\text{m}^2$$ Then the kinetic energy: $$KE_{\text{rot}} = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.12)(150)^2 = \frac{1}{2}(0.12)(22{,}500) = 1350 \text{ J}$$

Layer 2: Pattern

Problem 3. Three objects, each with mass $m = 2.0$ kg and radius $R = 0.10$ m, all spin at $\omega = 30$ rad/s. Rank them by rotational kinetic energy, from greatest to least.

Object	Moment of inertia
Thin-walled hoop	$I = mR^2$
Solid disk	$I = \frac{1}{2}mR^2$
Solid sphere	$I = \frac{2}{5}mR^2$

Check your answer

Since all three have the same $\omega$, the ranking is determined entirely by $I$. Compute each: | Object | $I$ (kg$\cdot$m$^2$) | $KE_{\text{rot}} = \frac{1}{2}I\omega^2$ (J) | |:---|:---|:---| | Thin-walled hoop | $(2.0)(0.10)^2 = 0.020$ | $\frac{1}{2}(0.020)(900) = 9.0$ | | Solid disk | $\frac{1}{2}(2.0)(0.10)^2 = 0.010$ | $\frac{1}{2}(0.010)(900) = 4.5$ | | Solid sphere | $\frac{2}{5}(2.0)(0.10)^2 = 0.008$ | $\frac{1}{2}(0.008)(900) = 3.6$ | Ranking: **Hoop > Disk > Sphere**. The pattern: at the same $\omega$, objects with more mass concentrated far from the axis store more rotational kinetic energy. The hoop, with all its mass at the rim, wins. The sphere, with mass distributed throughout (much of it near the center), stores the least.

Layer 3: Structure

Problem 4. For a rolling object (rolling without slipping), the total kinetic energy is $KE_{\text{total}} = \frac{1}{2}mv_{\text{cm}}^2 + \frac{1}{2}I_{\text{cm}}\omega^2$. Using $v_{\text{cm}} = R\omega$ and writing $I_{\text{cm}} = cmR^2$ (where $c$ is a dimensionless constant that depends on the shape), show that the fraction of total KE that is rotational depends only on $c$.

What is this fraction for:

(a) A thin-walled hoop ($c = 1$)?

(b) A solid disk ($c = \frac{1}{2}$)?

(c) A solid sphere ($c = \frac{2}{5}$)?

Check your answer

Write $I_{\text{cm}} = cmR^2$ and $v_{\text{cm}} = R\omega$: $$KE_{\text{trans}} = \frac{1}{2}m(R\omega)^2 = \frac{1}{2}mR^2\omega^2$$ $$KE_{\text{rot}} = \frac{1}{2}(cmR^2)\omega^2 = \frac{c}{2}mR^2\omega^2$$ $$KE_{\text{total}} = \frac{1}{2}mR^2\omega^2 + \frac{c}{2}mR^2\omega^2 = \frac{1 + c}{2}mR^2\omega^2$$ The rotational fraction is: $$\frac{KE_{\text{rot}}}{KE_{\text{total}}} = \frac{c/2}{(1+c)/2} = \frac{c}{1 + c}$$ This depends only on $c$, not on $m$, $R$, or $\omega$. (a) Hoop ($c = 1$): $\frac{1}{1+1} = \frac{1}{2} = 50\%$. Half the energy is rotational. (b) Disk ($c = \frac{1}{2}$): $\frac{1/2}{1 + 1/2} = \frac{1/2}{3/2} = \frac{1}{3} \approx 33\%$. (c) Sphere ($c = \frac{2}{5}$): $\frac{2/5}{1 + 2/5} = \frac{2/5}{7/5} = \frac{2}{7} \approx 29\%$. The more mass is concentrated at the rim (larger $c$), the larger the share of energy that goes into rotation. The hoop is the extreme case: fully half its kinetic energy is rotational. A solid sphere, with most mass near the center, devotes only about 29% to rotation. This has real consequences. If two objects roll down the same ramp starting from rest, the one that diverts more energy into rotation has less translational energy --- and moves slower. The hoop loses to the sphere in a race down a ramp.

Layer 4: Debug

Problem 5. A solid ball ($I_{\text{cm}} = \frac{2}{5}mR^2$) of mass 0.50 kg and radius 0.10 m rolls without slipping at $v_{\text{cm}} = 4.0$ m/s. A student computes the kinetic energy as:

$$KE = \frac{1}{2}mv_{\text{cm}}^2 = \frac{1}{2}(0.50)(4.0)^2 = 4.0 \text{ J}$$

The student used only the translational kinetic energy. How much energy did they miss, and what is the correct total?

Check your answer

The student's translational KE is correct: $$KE_{\text{trans}} = \frac{1}{2}(0.50)(4.0)^2 = 4.0 \text{ J}$$ Now find the rotational KE. First, compute $\omega$ from the rolling constraint: $$\omega = \frac{v_{\text{cm}}}{R} = \frac{4.0}{0.10} = 40 \text{ rad/s}$$ Then compute $I_{\text{cm}}$: $$I_{\text{cm}} = \frac{2}{5}mR^2 = \frac{2}{5}(0.50)(0.10)^2 = \frac{2}{5}(0.50)(0.01) = 0.002 \text{ kg}\!\cdot\!\text{m}^2$$ The rotational KE: $$KE_{\text{rot}} = \frac{1}{2}I_{\text{cm}}\omega^2 = \frac{1}{2}(0.002)(40)^2 = \frac{1}{2}(0.002)(1600) = 1.6 \text{ J}$$ The correct total is: $$KE_{\text{total}} = 4.0 + 1.6 = 5.6 \text{ J}$$ The student missed 1.6 J, which is $\frac{2}{7}$ of the total energy (about 29%). This is exactly the rotational fraction we derived for a solid sphere in Problem 4. For a hollow sphere or a hoop, the missed fraction would be even larger. The takeaway: ignoring rotational kinetic energy for a rolling object systematically underestimates the total energy. The error is not small --- for a hoop, you would miss half the energy.

Reflection

Think back over this section.

Why do you need to track both translational and rotational kinetic energy for rolling objects?

A rolling ball is doing two things at once: its center of mass moves through space, and it spins about its center. Each type of motion carries kinetic energy. If you use only $\frac{1}{2}mv^2$, you account for the center-of-mass motion but ignore the energy stored in the spin. That missing energy is real --- it came from somewhere (like gravity pulling the ball down a ramp) and it goes somewhere (like doing work when the ball hits something).

The fraction of energy that goes into rotation depends on the object's shape, through the moment of inertia. Objects with more mass at the rim devote a larger share of their energy to spinning. This is why a hoop rolls more slowly than a solid sphere down the same ramp --- the hoop "spends" more of its gravitational potential energy on rotation, leaving less for translation.

If you can explain this to someone without using any formulas, you understand the physics of this section.

Looking Ahead

You now know how to compute the kinetic energy stored in rotation: $KE_{\text{rot}} = \frac{1}{2}I\omega^2$. But energy does not appear from nowhere. Something has to do work to spin up a flywheel, and something does work to slow it down. In translational motion, that connection is $W = \int F\,dx$ --- force acting through a distance transfers energy. What is the rotational version?

In the next section, you will see that torque acting through an angle transfers energy into and out of rotation: $W = \int \tau\,d\theta$. This is the rotational analog of the work formula, and it leads directly to a rotational work-energy theorem. Just as $\frac{1}{2}mv^2$ was useful only after you connected it to the forces that change it, $\frac{1}{2}I\omega^2$ becomes a powerful tool once you connect it to the torques that change it.