6.1 - Rotational Kinetic Energy

Energy in Rotation

Continuing the parallel between linear and rotational motion, rotational kinetic energy is the energy an object has because it’s rotating, just as translating objects have kinetic energy from linear motion, rotating objects have kinetic energy from spinning.

Rotational Kinetic Energy Formula

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

where

K_{\text{rot}}

is rotational kinetic energy in Joules,

I

is rotational inertia in kg·m², and

\omega

is angular velocity in rad/s. The units are Joules, the same as every other form of energy.

The Perfect Analogy

This formula mirrors translational kinetic energy exactly, with direct substitutions:

Translational	→	Rotational
$K_{\text{trans}} = \frac{1}{2}mv^2$		$K_{\text{rot}} = \frac{1}{2}I\omega^2$
Mass ( $m$ )	→	Rotational inertia ( $I$ )
Linear velocity ( $v$ )	→	Angular velocity ( $\omega$ )

Same structure, different variables. Mass measures resistance to linear acceleration, rotational inertia measures resistance to angular acceleration, linear velocity describes how fast something moves, and angular velocity describes how fast something spins.

Where Rotational Kinetic Energy Comes From

A spinning object can have kinetic energy even when its center of mass isn’t moving at all. Picture a bicycle wheel spinning on a fixed axle: the center stays put, but every other piece of the wheel is moving in a circle with speed

v = r\omega

, and anything with mass and speed has kinetic energy.

Add up the ordinary kinetic energy

\frac{1}{2}mv^2

of every small piece of mass. Every piece shares the same

\omega

, so

\omega^2

factors out, and what’s left in the parentheses is exactly the rotational inertia,

I = \sum m_i r_i^2

(on the formula sheet):

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

\frac{1}{2}I\omega^2

isn’t a new kind of energy. It’s the translational kinetic energy of all the moving pieces, added up. That’s why, for an object turning about a fixed axis,

\frac{1}{2}I\omega^2

(with

I

measured about that axis) is the object’s total kinetic energy.

Example: two 0.50 kg balls are attached to the ends of a light 0.80 m rod, which spins about its center at 6.0 rad/s. Each ball is

r = 0.40

m from the axis.

Ball by ball: each ball moves at

v = r\omega = (0.40)(6.0) = 2.4

m/s, so each has

\frac{1}{2}mv^2 = \frac{1}{2}(0.50)(2.4)^2 = 1.44

J, and the pair has

2(1.44) = 2.88

As a rotating system:

I = \sum mr^2 = 2(0.50)(0.40)^2 = 0.16

kg·m², so

K_{\text{rot}} = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.16)(6.0)^2 = 2.88

Both ways give

K \approx 2.9

J (two significant figures). The center of mass sits on the axis and never moves, so the system has no translational kinetic energy, yet it still carries 2.9 J of kinetic energy because each ball is moving.

Rolling Objects: Combined Motion

Rolling objects, wheels, balls, cylinders, both translate and rotate at once, so they carry both types of kinetic energy.

K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

For rolling without slipping,

v = r\omega

, coupling the two energies together.

Expressing total energy in linear terms: starting from

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

, and substituting

I = cMR^2

(where

c

is the shape coefficient):

K_{\text{rot}} = \frac{1}{2}(cMR^2)\omega^2

Substituting

\omega = v/r

K_{\text{rot}} = \frac{1}{2}cMR^2\left(\frac{v}{R}\right)^2 = \frac{1}{2}cMR^2 \cdot \frac{v^2}{R^2}

Canceling

R^2

K_{\text{rot}} = \frac{1}{2}cMv^2

So total kinetic energy becomes:

K_{\text{total}} = \frac{1}{2}Mv^2 + \frac{1}{2}cMv^2 = \frac{1}{2}Mv^2(1 + c)

where

c

is the shape coefficient from the rotational inertia formula.

Object	$c$	$(1 + c)$ Factor
Solid sphere	2/5	1.4
Solid disk	1/2	1.5
Thin spherical shell	2/3	1.67
Thin hoop	1	2.0

The

(1+c)

factor tells you how much more energy a rolling object has compared to a sliding object moving at the same linear speed. For example, a rolling hoop has twice the kinetic energy of a sliding block at the same speed.

Why this matters for a race down a ramp: starting from the same height, every object has the same initial potential energy,

U = Mgh

, which converts into kinetic energy:

Mgh = \frac{1}{2}Mv^2(1 + c) \implies v = \sqrt{\frac{2gh}{1 + c}}

A smaller

(1+c)

gives a larger

v

, reaching the bottom first: the solid sphere (

1+c = 1.4

) is fastest, followed by the solid disk (1.5), the thin shell (1.67), and the hoop (2.0), slowest. Objects with more mass concentrated near the center (smaller

c

) put more of their potential energy into translational motion, rather than spinning, so they move faster down the ramp.

Rotational Kinetic Energy is a Scalar

Like every form of energy, rotational kinetic energy is a scalar, no direction, always positive or zero, and energies add arithmetically without needing vector components. It’s always positive because

I

is always positive (mass and distance squared can’t be negative) and

\omega^2

is always positive (squaring removes any sign). So whether something spins clockwise or counterclockwise, its rotational kinetic energy comes out positive either way.

Fixed Axis vs. Rolling

Rotating about a fixed external axis: like a door rotating about its hinges. Only rotational kinetic energy exists:

K = \frac{1}{2}I\omega^2

Don’t add a separate translational term here, even when the center of mass moves (a door’s center of mass swings around with it). Using

I

about the fixed axis already counts the motion of every piece of the object, so

\frac{1}{2}I\omega^2

is its total kinetic energy.

Rolling (rotation plus translation): an object rotates about its own center of mass while that center of mass also moves through space, so both types of kinetic energy are present:

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

For a rolling wheel, the center translates forward (contributing translational KE) while the wheel spins about its center (contributing rotational KE).

Worked Example: Two Ways to Get a Door’s Kinetic Energy

A uniform 15 kg door is 0.80 m wide and swings about its hinges at 2.0 rad/s. Find its kinetic energy. A door turning about its hinges has the same rotational inertia as a rod turning about one end,

I = \frac{1}{3}Mw^2

, where

w

is the width.

Method 1, rotation about the fixed axis (the hinges):

I_{\text{hinge}} = \tfrac{1}{3}(15)(0.80)^2 = 3.2\ \mathrm{kg\cdot m^2} \qquad K = \tfrac{1}{2}I_{\text{hinge}}\omega^2 = \tfrac{1}{2}(3.2)(2.0)^2 = 6.4\ \mathrm{J}

Method 2, rotation about the center of mass plus translation of the center of mass: about its center, the door has

I_{cm} = \frac{1}{12}Mw^2

. Its center of mass is 0.40 m from the hinges, so it moves at

v_{cm} = r\omega = (0.40)(2.0) = 0.80

m/s.

K = \tfrac{1}{2}I_{cm}\omega^2 + \tfrac{1}{2}Mv_{cm}^2 = \tfrac{1}{2}\left(\tfrac{1}{12}(15)(0.80)^2\right)(2.0)^2 + \tfrac{1}{2}(15)(0.80)^2 = 1.6 + 4.8 = 6.4\ \mathrm{J}

Same answer. Method 1 treats the door as purely rotating about the hinges, so

\frac{1}{2}I\omega^2

is everything. Method 2 splits the same energy into spinning about the center of mass (1.6 J) and moving the center of mass (4.8 J). Use one method or the other, never both, or you’ll count the energy twice.

Worked Example: Rotating Sphere

A uniform solid metal sphere rotates about a frictionless axle through its center at constant angular velocity, with rotational kinetic energy of 150 J, diameter 20 cm (

= 0.20

m), and mass 1.0 kg. Find the tangential velocity at the sphere’s rim.

Finding the rotational inertia: for a solid sphere,

I = \frac{2}{5}MR^2

, with

M = 1.0

kg and

R = \frac{0.20}{2} = 0.10

I = \frac{2}{5}(1.0)(0.10)^2 = \frac{2}{5}(0.01) = 0.004 \mathrm{ kg \cdot m}^2

Finding angular velocity: from

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

\omega = \sqrt{\frac{2K_{\text{rot}}}{I}} = \sqrt{\frac{2(150)}{0.004}} = \sqrt{\frac{300}{0.004}} = \sqrt{75{,}000} \approx 274 \text{ rad/s}

That’s fast, but reasonable for a small, light sphere.

Finding tangential velocity: using

v = r\omega

v = (0.10)(274) = 27.4 \text{ m/s}

A point on the rim moves at 27.4 m/s. Notice the angular velocity is the same everywhere on the rigid sphere (274 rad/s), but linear velocity grows with distance from the axis: zero at the center, 13.7 m/s halfway out (

r = 0.05

m), and 27.4 m/s at the rim (

r = 0.10

m).

Putting 274 rad/s in context: converting to RPM,

\omega = 274 \times \frac{60}{2\pi} \approx 2616

RPM. For comparison, an idle car engine runs about 800 RPM, a highway-speed car engine runs 2500 to 3000 RPM (comparable to this sphere), a high-speed drill runs about 10,000 RPM, and an older hard drive runs about 7200 RPM. So this sphere is spinning fast, but nothing extreme, similar to a car engine at highway speed.

Energy Perspective on Rolling

A sliding object with no rotation has

K = \frac{1}{2}Mv^2

, but a rolling object with both rotation and translation has

K = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2(1+c)

. At the same linear speed, the rolling object carries more total energy since it’s also spinning.

This has real implications for conservation problems: when a rolling ball goes up a ramp, it loses both translational and rotational kinetic energy, so it climbs higher than a frictionless sliding block with the same initial linear speed would, there’s simply more kinetic energy that needs converting into potential energy.

Common Scenarios

Object rotating about a fixed axis (a door swinging on hinges, a spinning top, a ceiling fan): only rotational kinetic energy exists,

K = \frac{1}{2}I\omega^2

, whether or not the center of mass translates.

Object rolling without slipping (a ball rolling down a ramp, a wheel on a moving car, a yo-yo): both translational and rotational kinetic energy exist,

K_{\text{total}} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2

Object sliding with no rotation (a hockey puck sliding, a box pushed across the floor): only translational kinetic energy exists,

K = \frac{1}{2}Mv^2

Common Mistakes

Using the wrong $I$ formula: don’t use

I = MR^2

for a solid sphere, that’s for a hoop, a solid sphere is

I = \frac{2}{5}MR^2

Using diameter instead of radius: don’t use

D = 0.20

m as the radius, it’s

R = D/2 = 0.10

Forgetting to square $\omega$ : don’t write

K = \frac{1}{2}I\omega

, it’s

K = \frac{1}{2}I\omega^2

, omega squared.

Double-counting energy: don’t add translational KE twice for a rolling object,

K_{\text{total}} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2

, each term appears exactly once.

Wrong conversion between $v$ and $\omega$ : don’t set

v = \omega

directly, it’s

v = r\omega

, always including the radius.

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