Classical-Mechanics · Unit 24 · Video 5 · Interactive Practice

Move the Center, Then Turn About It: Rolling Kinetic Energy and Chasles's Theorem

IKey Formulas

Formula	Name	What you need
$K = \sum_i \tfrac{1}{2} m_i v_{cm,i}^2 + \tfrac{1}{2} m_T V_{cm}^2$	Kinetic energy split	Any system of particles; the cross term vanishes because $\sum_i m_i \vec{v}_{cm,i} = \vec{0}$
$K_{\text{total}} = \tfrac{1}{2} m_T V_{cm}^2 + \tfrac{1}{2} I_{cm}\,\omega_{cm}^2$	Rigid body: translation + rotation	Rotation about an axis through the cm that keeps its direction
$V_{cm} = R\,\omega_{cm},\ \ I_{cm} = \tfrac{1}{2}MR^2 \;\Rightarrow\; K = \tfrac{3}{4}MV_{cm}^2$	Rolling uniform disc	Rolling without slipping
$d\vec{r}_{cm,1} = \dfrac{\mu}{m_1}\,d\vec{r}_{1,2},\ \ d\vec{r}_{cm,2} = -\dfrac{\mu}{m_2}\,d\vec{r}_{1,2}$	Chasles, central step	$\mu = \dfrac{m_1 m_2}{m_1+m_2}$ , $\ d\vec{r}_i = d\vec{R}_{cm} + d\vec{r}_{cm,i}$

Key Insight: Motion and energy split along the same line. A rigid body's small displacement is a translation of its cm plus a rotation about it, and its kinetic energy is $\tfrac{1}{2}m_T V_{cm}^2$ for the first plus $\tfrac{1}{2}I_{cm}\omega_{cm}^2$ for the second, with no cross term. The energy split alone needs no rigidity.

IIVisualization 1 — The Split for Any System

For any cloud of particles, $K$ is the energy about the cm plus $\tfrac{1}{2}m_T V_{cm}^2$ , with no cross term.

Observer frame velocity u (m/s, along x) 0 Arrows

💡 Nothing here holds the particles together: the same split covers a gas, pool balls after the break, or a firecracker's fragments — and only the $\tfrac{1}{2}m_T V_{cm}^2$ term depends on the inertial frame.

IIIVisualization 2 — The Rolling Disc

Rolling without slipping gives every point $\vec{V}_{cm}$ plus a turn about the cm, so a disc's energy splits two thirds to one third.

Arrows Body Distance rolled 0.00 m

IVVisualization 3 — Chasles: One Pair, One Angle

Any step of the pair is the cm's step $d\vec{R}_{cm}$ plus each particle's step about the cm; fixed length makes that a turn through one $d\theta$ .

Masses

💡 The steps are drawn enlarged; every relation is first order in the displacements. The full theorem still needs every particle of the body to share one angle and one axis through the body's cm.

VQuiz Questions

Problem 1 · Kinetic Energy of a Rolling Disc

Given: A uniform solid disc of mass $M = 4\ \text{kg}$ and radius $R = 0.3\ \text{m}$ rolls without slipping with its center moving at $V_{cm} = 3\ \text{m/s}$ . Find its total kinetic energy.

✅ Correct!

K = \tfrac{1}{2}MV_{cm}^2 + \tfrac{1}{4}MV_{cm}^2 = \tfrac{3}{4}(4)(3)^2 = 27\ \text{J}

; the radius never enters.

❌ Only the translation.

\tfrac{1}{2}MV_{cm}^2 = 18\ \text{J}

is the cm term; the disc also turns about its cm, adding

\tfrac{1}{2}I_{cm}\omega_{cm}^2

❌ Only the rotation.

\tfrac{1}{2}I_{cm}\omega_{cm}^2 = \tfrac{1}{4}MV_{cm}^2 = 9\ \text{J}

is the energy about the cm; add the cm term

\tfrac{1}{2}MV_{cm}^2

❌ Check the moment of inertia.

36\ \text{J}

uses

I_{cm} = MR^2

, a hoop; a uniform disc has

I_{cm} = \tfrac{1}{2}MR^2

Show solution

Rolling without slipping gives $\omega_{cm} = V_{cm}/R = 10\ \text{rad/s}$ , and $I_{cm} = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(4)(0.3)^2 = 0.18\ \text{kg}\cdot\text{m}^2$ .

K = \tfrac{1}{2}MV_{cm}^2 + \tfrac{1}{2}I_{cm}\omega_{cm}^2 = \tfrac{1}{2}(4)(3)^2 + \tfrac{1}{2}(0.18)(10)^2 = 18 + 9 = 27\ \text{J}

Equivalently $K = \tfrac{3}{4}MV_{cm}^2 = \tfrac{3}{4}(4)(9) = 27\ \text{J}$ : two thirds translational, one third rotational.

Problem 2 · The Firecracker

Given: A $2\ \text{kg}$ firecracker slides along a frictionless floor at $5\ \text{m/s}$ and explodes into two $1\ \text{kg}$ pieces. Just afterwards both move along the original line, at $8\ \text{m/s}$ and $2\ \text{m/s}$ . Find the kinetic energy of the motion about the center of mass just after the explosion.

✅ Correct! In the cm frame the pieces move at

\pm 3\ \text{m/s}

, so

\sum_i \tfrac{1}{2}m_i v_{cm,i}^2 = 9\ \text{J}

, and

9 + 25 = 34\ \text{J}

is the total.

❌ Momentum is not energy. The total momentum in the cm frame is zero, which kills the cross term; the pieces still move in that frame, so their kinetic energy there is positive.

❌ That is the other term.

\tfrac{1}{2}m_T V_{cm}^2 = \tfrac{1}{2}(2)(5)^2 = 25\ \text{J}

is the energy of the cm motion, unchanged by the internal explosion.

❌ That is the total.

\tfrac{1}{2}(1)(8)^2 + \tfrac{1}{2}(1)(2)^2 = 34\ \text{J}

includes the cm motion; subtract

\tfrac{1}{2}m_T V_{cm}^2

Show solution

The explosion is internal, so the cm velocity is unchanged:

V_{cm} = \frac{(1)(8) + (1)(2)}{2} = 5\ \text{m/s}

Relative to the cm the pieces move at $v_{cm,1} = 8 - 5 = 3\ \text{m/s}$ and $v_{cm,2} = 2 - 5 = -3\ \text{m/s}$ :

\sum_i \tfrac{1}{2}m_i v_{cm,i}^2 = \tfrac{1}{2}(1)(3)^2 + \tfrac{1}{2}(1)(3)^2 = 9\ \text{J}

Check: $\tfrac{1}{2}m_T V_{cm}^2 = 25\ \text{J}$ and $25 + 9 = 34\ \text{J} = \tfrac{1}{2}(1)(8)^2 + \tfrac{1}{2}(1)(2)^2$ . Before the explosion $K = 25\ \text{J}$ with nothing about the cm; the explosion added $9\ \text{J}$ , all of it to the term about the cm.

Problem 3 · One Angle for the Pair

Given: Two particles of a rigid body, $m_1 = 3\ \text{kg}$ and $m_2 = 1\ \text{kg}$ , are $|\vec{r}_{1,2}| = 0.8\ \text{m}$ apart. In a small displacement, particle 1 moves $|d\vec{r}_{cm,1}| = 1.0\ \text{mm}$ about the pair's center of mass. Find particle 2's distance from the cm and the length of its step about the cm.

What is $|\vec{r}_{cm,2}|$ ?

What is $|d\vec{r}_{cm,2}|$ ?

✅ Correct! Both particles turn through

d\theta = 1.0\ \text{mm}/0.2\ \text{m} = 3.0\ \text{mm}/0.6\ \text{m} = 0.005\ \text{rad}

about the cm.

❌ Check the distance. Use

|\vec{r}_{cm,2}| = \dfrac{\mu}{m_2}|\vec{r}_{1,2}|

with

\mu = \dfrac{m_1 m_2}{m_1 + m_2}

❌ That is particle 1's distance. The heavier particle sits nearer the cm:

m_1|\vec{r}_{cm,1}| = m_2|\vec{r}_{cm,2}|

, so the light particle 2 is three times farther out.

❌ Check the step. Both particles turn through the same

d\theta

, so each step about the cm is proportional to that particle's distance from the cm.

❌ Equal steps would be a translation. Steps about the cm satisfy

m_1\,d\vec{r}_{cm,1} = -m_2\,d\vec{r}_{cm,2}

, so they are not equal when the masses differ.

❌ The ratio is inverted. Particle 2 is farther from the cm, so for the same angle it moves farther:

|d\vec{r}_{cm,2}| = \dfrac{m_1}{m_2}|d\vec{r}_{cm,1}|

Show solution

The reduced mass is $\mu = \dfrac{(3)(1)}{3+1} = \tfrac{3}{4}\ \text{kg}$ , so

|\vec{r}_{cm,1}| = \frac{\mu}{m_1}|\vec{r}_{1,2}| = \frac{1}{4}(0.8) = 0.2\ \text{m}, \qquad |\vec{r}_{cm,2}| = \frac{\mu}{m_2}|\vec{r}_{1,2}| = \frac{3}{4}(0.8) = 0.6\ \text{m}

Check: $m_1|\vec{r}_{cm,1}| = 0.6 = m_2|\vec{r}_{cm,2}|$ . From $d\vec{r}_{cm,1} = \dfrac{\mu}{m_1}d\vec{r}_{1,2}$ and $d\vec{r}_{cm,2} = -\dfrac{\mu}{m_2}d\vec{r}_{1,2}$ ,

|d\vec{r}_{cm,2}| = \frac{m_1}{m_2}\,|d\vec{r}_{cm,1}| = 3(1.0\ \text{mm}) = 3.0\ \text{mm}

Both turn through one angle: $d\theta = \dfrac{1.0\ \text{mm}}{200\ \text{mm}} = \dfrac{3.0\ \text{mm}}{600\ \text{mm}} = 0.005\ \text{rad}$ .

Problem 4 · Rolling Down a Ramp

Given: A uniform solid disc starts from rest and rolls without slipping down a ramp, its center dropping $h = 0.6\ \text{m}$ . Static friction at the contact point does no work, because that point is momentarily at rest. With $g = 9.8\ \text{m/s}^2$ , find the speed $V_{cm}$ at the bottom.

✅ Correct!

\tfrac{3}{4}MV_{cm}^2 = Mgh

gives

V_{cm} = \sqrt{4gh/3} = 2.80\ \text{m/s}

; one third of the released energy goes into turning.

❌ That is a sliding block.

\sqrt{2gh}

puts all of

Mgh

into translation; the rolling disc must also carry

\tfrac{1}{4}MV_{cm}^2

of rotation.

❌ That is a hoop.

\sqrt{gh}

follows from

K = MV_{cm}^2

, i.e.

I_{cm} = MR^2

; a disc has

I_{cm} = \tfrac{1}{2}MR^2

❌ That is a solid sphere.

\sqrt{10gh/7}

follows from

I_{cm} = \tfrac{2}{5}MR^2

; a disc has

I_{cm} = \tfrac{1}{2}MR^2

Show solution

Gravity is the only force doing work, so mechanical energy is conserved. Rolling gives $\omega_{cm} = V_{cm}/R$ , and the disc's kinetic energy at the bottom is

K = \tfrac{1}{2}MV_{cm}^2 + \tfrac{1}{2}\big(\tfrac{1}{2}MR^2\big)\Big(\frac{V_{cm}}{R}\Big)^2 = \tfrac{3}{4}MV_{cm}^2

Setting $\tfrac{3}{4}MV_{cm}^2 = Mgh$ :

V_{cm} = \sqrt{\frac{4gh}{3}} = \sqrt{\frac{4(9.8)(0.6)}{3}} = \sqrt{7.84} = 2.80\ \text{m/s}

Mass and radius cancel, so every uniform disc arrives at the same speed.

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