Classical-Mechanics · Unit 25 · Video 1 · Interactive Practice

Torque About the Center of Mass: The Orbital and Spin Equations of Motion

IKey Formulas

Formula	Name	What it says
$\vec{F}^{\,\text{ext}} = \dfrac{d}{dt}\left(m_T\vec{V}_{\text{cm}}\right)$	Translational equation	The center of mass moves like one particle of mass $m_T$ carrying every external force; it holds no information about rotation
$\vec{\tau}_S^{\,\text{ext}} = \vec{\tau}_{S,\text{cm}}^{\,\text{ext}} + \vec{\tau}_{\text{cm}}^{\,\text{ext}}$	Torque split	Orbital piece $\vec{\tau}_{S,\text{cm}}^{\,\text{ext}} = \vec{r}_{S,\text{cm}}\times\vec{F}^{\,\text{ext}}$ (all external forces placed at the cm); spin piece $\vec{\tau}_{\text{cm}}^{\,\text{ext}} = \sum_i\vec{r}_{\text{cm},i}\times\vec{F}_i^{\,\text{ext}}$ (each force where it acts, no reference to $S$ )
$\vec{\tau}_{S,\text{cm}}^{\,\text{ext}} = \dfrac{d\vec{L}_S^{\,\text{orbital}}}{dt}$	Orbital relation ( $S$ fixed)	The torque law for a single particle $m_T$ at the cm; it follows from the translational equation and adds nothing new
$\vec{\tau}_{\text{cm}}^{\,\text{ext}} = \dfrac{d\vec{L}_{\text{cm}}^{\,\text{spin}}}{dt}$	Spin relation	Holds even when $\vec{A}_{\text{cm}} \neq \vec{0}$ ; about another accelerating point $P$ an extra term $-M\,\vec{r}_{P,\text{cm}}\times\vec{a}_P$ appears

Key Insight: In uniform gravity $\vec{\tau}_{\text{cm}}^{\,\text{ext}} = \left(\sum_i m_i\vec{r}_{\text{cm},i}\right)\times\vec{g} = \vec{0}$ , so a thrown baton keeps $\vec{L}_{\text{cm}}^{\,\text{spin}} = I_{\text{cm}}\vec{\omega}_{\text{cm}}$ constant: translation fixes the parabola, spin fixes the steady turning.

IIOrbital Piece Plus Spin Piece

Moving the point $S$ changes only the orbital piece $\vec{r}_{S,\text{cm}}\times\vec{F}^{\,\text{ext}}$ ; the spin piece $\vec{\tau}_{\text{cm}}^{\,\text{ext}}$ never mentions $S$ .

External forces

S_x

(m) 0.0

S_y

(m) 0.0

IIIWhy the Center of Mass Is Special

A rod dropped level never starts turning, yet gravity has a torque about its falling end.

Point

P

P_x

(m) 2.0

P_y

(m) 4.0

IVThe Thrown Baton

With no torque about the cm, the baton's spin is constant while its orbital angular momentum about the ground keeps changing.

\omega_{\text{cm}}

(rad/s) 10

t

(s) 0.45

VQuiz Questions

Problem 1 · Split the Torque

Given: a body's center of mass is at $\vec{r}_{S,\text{cm}} = (2.0,\ 1.0)\ \text{m}$ from a fixed point $S$ at the origin. The only external force, $\vec{F} = (0,\ 5.0)\ \text{N}$ , acts at the point $(3.0,\ 1.0)\ \text{m}$ . Find the spin piece $\vec{\tau}_{\text{cm}}^{\,\text{ext}}$ ( $\hat{k}$ out of the page).

✅ Correct! The force acts

1.0\ \text{m}

to the right of the cm, so

\vec{\tau}_{\text{cm}}^{\,\text{ext}} = (1.0)(5.0)\,\hat{k} = 5.0\,\hat{k}\ \text{N}\cdot\text{m}

❌ That is the orbital piece.

\vec{r}_{S,\text{cm}}\times\vec{F} = 10\,\hat{k}\ \text{N}\cdot\text{m}

places the force at the cm; the spin piece uses

\vec{r}_{\text{cm},i}

, from the cm to where the force acts.

❌ That is the whole torque about $S$ .

\vec{r}_{S,i}\times\vec{F} = 15\,\hat{k}\ \text{N}\cdot\text{m}

is orbital plus spin; subtract the orbital

10\,\hat{k}

❌ Not quite. The spin piece vanishes only when the force's line of action passes through the cm; this force acts

1.0\ \text{m}

to the side of it.

❌ Not quite. Use

\vec{r}_{\text{cm},i} = (3.0 - 2.0,\ 1.0 - 1.0) = (1.0,\ 0)\ \text{m}

Show solution

The vector from the cm to the point of application is

\vec{r}_{\text{cm},i} = (3.0,\ 1.0) - (2.0,\ 1.0) = (1.0,\ 0)\ \text{m}.

With $(\vec{r}\times\vec{F})_z = r_xF_y - r_yF_x$ :

\vec{\tau}_{\text{cm}}^{\,\text{ext}} = \big[(1.0)(5.0) - (0)(0)\big]\,\hat{k} = 5.0\,\hat{k}\ \text{N}\cdot\text{m}.

Check with the split: the orbital piece is $\vec{r}_{S,\text{cm}}\times\vec{F} = \big[(2.0)(5.0) - (1.0)(0)\big]\,\hat{k} = 10\,\hat{k}\ \text{N}\cdot\text{m}$ , and the torque about $S$ computed directly is $\big[(3.0)(5.0) - (1.0)(0)\big]\,\hat{k} = 15\,\hat{k}\ \text{N}\cdot\text{m} = 10\,\hat{k} + 5.0\,\hat{k}$ .

Problem 2 · The Baton in Flight

Given: a baton is thrown end over end and air resistance is negligible. Which statement is correct?

✅ Correct! Uniform gravity gives zero torque about the cm but a torque

\vec{r}_{S,\text{cm}}\times m_T\vec{g}

about a ground point, which changes only the orbital piece.

❌ Not quite.

\sum_i\vec{r}_{\text{cm},i}\times m_i\vec{g} = \left(\sum_i m_i\vec{r}_{\text{cm},i}\right)\times\vec{g} = \vec{0}

: the mass-weighted positions about the cm sum to zero.

❌ That is the trap. The spin relation holds about the cm even when

\vec{A}_{\text{cm}} \neq \vec{0}

; the

-m_i\vec{A}_{\text{cm}}

terms add up to

-\left(\sum_i m_i\vec{r}_{\text{cm},i}\right)\times\vec{A}_{\text{cm}} = \vec{0}

❌ Not quite. About

S

the weight acts at the cm with lever arm

\vec{r}_{S,\text{cm}}

, so

\vec{\tau}_S^{\,\text{ext}} = \vec{r}_{S,\text{cm}}\times m_T\vec{g} \neq \vec{0}

whenever the cm is not directly above

S

Show solution

Spin: the only external force is $m_i\vec{g}$ on each particle, and $\vec{g}$ is the same for all of them:

\vec{\tau}_{\text{cm}}^{\,\text{ext}} = \sum_i\vec{r}_{\text{cm},i}\times m_i\vec{g} = \left(\sum_i m_i\vec{r}_{\text{cm},i}\right)\times\vec{g} = \vec{0}.

The spin relation holds about the cm even though the cm accelerates at $\vec{g}$ , so $\vec{L}_{\text{cm}}^{\,\text{spin}} = I_{\text{cm}}\vec{\omega}_{\text{cm}}$ is constant and the baton turns at a steady rate.

Orbital: about a fixed ground point $S$ ,

\frac{d\vec{L}_S^{\,\text{orbital}}}{dt} = \vec{r}_{S,\text{cm}}\times m_T\vec{g},

which is not zero while the cm is to the side of $S$ . So $\vec{L}_S^{\,\text{sys}} = \vec{L}_S^{\,\text{orbital}} + \vec{L}_{\text{cm}}^{\,\text{spin}}$ changes, entirely through its orbital piece.

Problem 3 · Orbital Angular Momentum of the Baton

Given: a baton of mass $m_T = 0.40\ \text{kg}$ is in flight. At one instant its cm is at $\vec{r}_{S,\text{cm}} = (2.0,\ 1.5)\ \text{m}$ from a point $S$ on the ground and moves with $\vec{V}_{\text{cm}} = (3.0,\ -1.0)\ \text{m/s}$ ; $\vec{g} = (0,\ -9.8)\ \text{m/s}^2$ , $\hat{k}$ out of the page. Find $d\vec{L}_S^{\,\text{orbital}}/dt$ and $\vec{L}_S^{\,\text{orbital}}$ at that instant.

What is $d\vec{L}_S^{\,\text{orbital}}/dt$ ?

What is $\vec{L}_S^{\,\text{orbital}}$ ?

✅ Correct! Both are clockwise:

\vec{L}_S^{\,\text{orbital}} = -2.6\,\hat{k}\ \text{kg}\cdot\text{m}^2/\text{s}

, growing more negative at

7.84\ \text{N}\cdot\text{m}

❌ Check the sign. The weight points down and the cm is to the right of

S

, so the torque about

S

is clockwise:

(2.0)(-3.92) - (1.5)(0) = -7.84

❌ Wrong lever arm. For a vertical force only the horizontal component of

\vec{r}_{S,\text{cm}}

contributes:

r_xF_y

with

r_x = 2.0\ \text{m}

, not the height

1.5\ \text{m}

❌ That is the spin piece. Gravity has no torque about the cm, but about

S

the weight acts at the cm with lever arm

2.0\ \text{m}

❌ Check the torque. Use

\vec{r}_{S,\text{cm}}\times m_T\vec{g}

with

m_T\vec{g} = (0,\ -3.92)\ \text{N}

❌ Check the sign.

(\vec{r}\times\vec{p})_z = xp_y - yp_x

, not

yp_x - xp_y

❌ Missing the mass.

\vec{r}_{S,\text{cm}}\times\vec{V}_{\text{cm}} = -6.5\,\hat{k}\ \text{m}^2/\text{s}

; multiply by

m_T = 0.40\ \text{kg}

❌ Check the cross product. It is a difference,

xV_y - yV_x

, not a sum

xV_y + yV_x

❌ Check the cross product.

\vec{L}_S^{\,\text{orbital}} = \vec{r}_{S,\text{cm}}\times m_T\vec{V}_{\text{cm}}

Show solution

Rate of change. By the orbital relation the weight $m_T\vec{g} = (0,\ -3.92)\ \text{N}$ acts at the cm:

\frac{d\vec{L}_S^{\,\text{orbital}}}{dt} = \vec{r}_{S,\text{cm}}\times m_T\vec{g} = \big[(2.0)(-3.92) - (1.5)(0)\big]\,\hat{k} = -7.84\,\hat{k}\ \text{N}\cdot\text{m}.

Orbital angular momentum. $m_T\vec{V}_{\text{cm}} = (1.2,\ -0.40)\ \text{kg}\cdot\text{m/s}$ , so

\vec{L}_S^{\,\text{orbital}} = \big[(2.0)(-0.40) - (1.5)(1.2)\big]\,\hat{k} = (-0.80 - 1.8)\,\hat{k} = -2.6\,\hat{k}\ \text{kg}\cdot\text{m}^2/\text{s}.

Both are clockwise. The spin of the baton plays no part: it adds a constant $\vec{L}_{\text{cm}}^{\,\text{spin}}$ to $\vec{L}_S^{\,\text{sys}}$ .

Problem 4 · Torque About an Accelerating Point

Given: a uniform rod, $M = 2.0\ \text{kg}$ and $1.2\ \text{m}$ long, is released from rest while horizontal and falls without turning ( $g = 9.8\ \text{m/s}^2$ ). Take $P$ at the rod's left end, falling with it, and measure $\vec{L}_P$ with velocities relative to $P$ . Find $d\vec{L}_P/dt$ ( $\hat{k}$ out of the page).

✅ Correct! The extra term

-M\,\vec{r}_{P,\text{cm}}\times\vec{a}_P = +11.76\,\hat{k}\ \text{N}\cdot\text{m}

cancels the torque of the weight, consistent with a rod that never starts turning.

❌ That is the trap.

-11.76\,\hat{k}\ \text{N}\cdot\text{m}

\vec{\tau}_P^{\,\text{ext}}

alone.

P

accelerates, so the extra term

-M\,\vec{r}_{P,\text{cm}}\times\vec{a}_P

must be added; only about the cm is it absent.

❌ Check the sign of the extra term. With

\vec{a}_P = \vec{g}

it is

-M\,\vec{r}_{P,\text{cm}}\times\vec{g}

, the opposite of the weight's torque, not a copy of it.

❌ Not quite.

+11.76\,\hat{k}

is the extra term by itself; it still has to be added to

\vec{\tau}_P^{\,\text{ext}} = -11.76\,\hat{k}\ \text{N}\cdot\text{m}

Show solution

About an accelerating point $P$ the same derivation as for the spin relation leaves an extra term:

\frac{d\vec{L}_P}{dt} = \vec{\tau}_P^{\,\text{ext}} - M\,\vec{r}_{P,\text{cm}}\times\vec{a}_P.

The cm is $0.60\ \text{m}$ to the right of $P$ : $\vec{r}_{P,\text{cm}} = (0.60,\ 0)\ \text{m}$ , and $M\vec{g} = (0,\ -19.6)\ \text{N}$ .

\vec{\tau}_P^{\,\text{ext}} = \vec{r}_{P,\text{cm}}\times M\vec{g} = (0.60)(-19.6)\,\hat{k} = -11.76\,\hat{k}\ \text{N}\cdot\text{m}

$P$ falls with the rod, so $\vec{a}_P = \vec{g}$ and

-M\,\vec{r}_{P,\text{cm}}\times\vec{a}_P = -(0.60)(-19.6)\,\hat{k} = +11.76\,\hat{k}\ \text{N}\cdot\text{m}.

\frac{d\vec{L}_P}{dt} = -11.76\,\hat{k} + 11.76\,\hat{k} = \vec{0}.

Applying $\vec{\tau}_P^{\,\text{ext}} = d\vec{L}_P/dt$ about the falling end would predict a rod that starts turning; about the cm, $\vec{\tau}_{\text{cm}}^{\,\text{ext}} = \vec{0}$ gives the right answer with no correction.

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