Classical-Mechanics · Unit 21 · Video 3 · Interactive Practice

One Equation for the Whole Body: Newton's Second Law Summed into Torque Equals I Alpha

IKey Formulas

Formula	Name	What it says
$(\vec{\tau}_{S,i})_z = r_i\,F_{\theta,i}$	Axial torque on one element	Only the tangential component turns; $F_{r,i}$ , $F_{z,i}$ and the height $z_i$ drop out
$F_{\theta,i} = \Delta m_i\, r_i\,\alpha_z$	Newton's second law, tangential	From $a_{\theta,i} = r_i\,\alpha_z$ , so $(\vec{\tau}_{S,i})_z = \Delta m_i\, r_i^2\,\alpha_z$
$\vec{\tau}_{S,j,i} + \vec{\tau}_{S,i,j} = (\vec{r}_{S,i} - \vec{r}_{S,j}) \times \vec{F}_{j,i}$	Torque of one internal pair	Independent of $S$ ; zero when $\vec{F}_{j,i}$ lies along the line joining the pair
$(\tau_S^{\,\text{ext}})_z = I_S\,\alpha_z$ with $I_S = \sum_i \Delta m_i\, r_i^2$	Rotational equation of motion	Rigid body; internal forces along the lines joining each pair

Key Insight: $(\tau_S^{\,\text{ext}})_z = I_S\,\alpha_z$ has the shape of $\vec{F} = m\vec{a}$ — torque about the axis for force, moment of inertia for mass, angular acceleration for acceleration — and, with internal forces along the lines joining each pair, only forces from outside the body appear on the left.

IIOnly the Tangential Component Turns

Of the net force on an element, only $F_{\theta,i}$ has a torque along the axis: $(\vec{\tau}_{S,i})_z = r_i F_{\theta,i}$ .

radial F_r (N) 2.0 tangential F_θ (N) 3.0 axial F_z (N) 1.0 height z_i (m) 0.60

IIISumming Over the Body

One shared $\alpha_z$ makes each element need $\Delta m_i\, r_i^2\,\alpha_z$ of torque, so the whole body needs $I_S\,\alpha_z$ .

α_z (rad/s²) 2

💡 Each $r_i$ is the perpendicular distance from the axis, not the distance from $S$ : the height $z_i$ left with the term $z_i\hat{k}\times\vec{F}_i$ , so $I_S$ is the same for every $S$ on the axis.

IVWhen Internal Torques Cancel

A third-law pair exerts $(\vec{r}_{S,i} - \vec{r}_{S,j}) \times \vec{F}_{j,i}$ : the same about every $S$ , and zero only when the forces lie along the joining line.

F_j,i along i–j, pull (N) 3 F_j,i sideways (N) 0

💡 Summed over every pair, the strong form of the third law gives $\vec{\tau}_S^{\,\text{int}} = \vec{0}$ , so $\vec{\tau}_S = \vec{\tau}_S^{\,\text{ext}}$ and the torque in $\tau_{S,z} = I_S\,\alpha_z$ is the external torque alone.

VQuiz Questions

Problem 1 · The Axial Torque on One Element

Given: An element of a rigid body turning about the fixed $z$ -axis sits $r_i = 0.40\ \text{m}$ from the axis, at height $z_i = 0.30\ \text{m}$ above the reference point $S$ on the axis. The net force on it has components $F_{r,i} = 2.0\ \text{N}$ , $F_{\theta,i} = 3.0\ \text{N}$ and $F_{z,i} = 6.0\ \text{N}$ (magnitude $7.0\ \text{N}$ ). Find $(\vec{\tau}_{S,i})_z$ .

✅ Correct! Only the tangential component has a lever arm about the axis:

r_iF_{\theta,i} = (0.40)(3.0) = 1.2\ \text{N}\cdot\text{m}

❌ That is $r_iF_{r,i}$ . The radial component points straight away from the axis, and

\hat{r}\times\hat{r} = \vec{0}

: it has no torque about the axis however large it is.

❌ That uses the distance from $S$ .

|\vec{r}_{S,i}| = \sqrt{0.30^2 + 0.40^2} = 0.50\ \text{m}

, but the height term

z_i\hat{k}\times\vec{F}_i

has no

z

-component; the lever arm is the perpendicular distance

r_i = 0.40\ \text{m}

❌ That multiplies $r_i$ by the whole force. Of the

7.0\ \text{N}

, only the

3.0\ \text{N}

tangential part turns the element about the axis; the radial and axial parts give horizontal torque only.

Show solution

Write both vectors in the cylindrical basis, $\vec{r}_{S,i} = z_i\hat{k} + r_i\hat{r}$ and $\vec{F}_i = F_{r,i}\hat{r} + F_{\theta,i}\hat{\theta} + F_{z,i}\hat{k}$ , and use $\hat{r}\times\hat{\theta} = \hat{k}$ , $\hat{\theta}\times\hat{k} = \hat{r}$ , $\hat{k}\times\hat{r} = \hat{\theta}$ :

\vec{\tau}_{S,i} = -z_iF_{\theta,i}\,\hat{r} + \left(z_iF_{r,i} - r_iF_{z,i}\right)\hat{\theta} + r_iF_{\theta,i}\,\hat{k}

\vec{\tau}_{S,i} = \left(-0.90\,\hat{r} - 1.8\,\hat{\theta} + 1.2\,\hat{k}\right)\ \text{N}\cdot\text{m}

The height $z_i$ and the components $F_{r,i}$ and $F_{z,i}$ appear only in the horizontal parts. The axial torque is

(\vec{\tau}_{S,i})_z = r_iF_{\theta,i} = (0.40\ \text{m})(3.0\ \text{N}) = 1.2\ \text{N}\cdot\text{m}

Problem 2 · Equal and Opposite Is Not Enough

Given: Inside a rigid body, elements $i$ and $j$ are $0.20\ \text{m}$ apart. Suppose the force on $i$ due to $j$ had magnitude $5.0\ \text{N}$ and pointed perpendicular to the line joining them, with the force on $j$ due to $i$ equal and opposite, $\vec{F}_{i,j} = -\vec{F}_{j,i}$ . Find the magnitude of the pair's total torque about a reference point $S$ .

✅ Correct! The pair's torque is

(\vec{r}_{S,i} - \vec{r}_{S,j})\times\vec{F}_{j,i}

, of magnitude

(0.20)(5.0) = 1.0\ \text{N}\cdot\text{m}

about every

S

: a couple, although the two forces cancel.

❌ The forces cancel; their torques do not. They act at different points, and

(\vec{r}_{S,i} - \vec{r}_{S,j})\times\vec{F}_{j,i}

vanishes only when

\vec{F}_{j,i}

is parallel to the joining line. Here it is perpendicular.

❌ That is one force's torque about the midpoint. Each force has a

0.10\ \text{m}

lever arm about the midpoint and the two turn the same way, so they add:

2(0.10)(5.0) = 1.0\ \text{N}\cdot\text{m}

❌ $S$ drops out. With

\vec{F}_{i,j} = -\vec{F}_{j,i}

the pair's torque is

(\vec{r}_{S,i} - \vec{r}_{S,j})\times\vec{F}_{j,i}

, and

\vec{r}_{S,i} - \vec{r}_{S,j}

is the vector from

j

i

whatever

S

is.

Show solution

Add the two torques and replace $\vec{F}_{i,j}$ by $-\vec{F}_{j,i}$ :

\vec{\tau}_{S,j,i} + \vec{\tau}_{S,i,j} = \vec{r}_{S,i}\times\vec{F}_{j,i} + \vec{r}_{S,j}\times\left(-\vec{F}_{j,i}\right) = \left(\vec{r}_{S,i} - \vec{r}_{S,j}\right)\times\vec{F}_{j,i}

$\vec{r}_{S,i} - \vec{r}_{S,j}$ is the $0.20\ \text{m}$ vector from $j$ to $i$ , the same for every $S$ , and the force is perpendicular to it:

\left|\vec{\tau}_{S,j,i} + \vec{\tau}_{S,i,j}\right| = (0.20\ \text{m})(5.0\ \text{N})\sin 90\degree = 1.0\ \text{N}\cdot\text{m}

This couple is why the derivation needs the strong form of the third law: with $\vec{F}_{j,i}$ along the joining line the angle is $0\degree$ or $180\degree$ , and every pair's torque vanishes.

Problem 3 · From the Elements to the Angular Acceleration

Given: A rigid rotor turns about a fixed vertical axis. Three small masses sit on its light frame: $2.0\ \text{kg}$ at $0.30\ \text{m}$ from the axis and $0.40\ \text{m}$ above $S$ ; $1.0\ \text{kg}$ at $0.50\ \text{m}$ from the axis, level with $S$ ; $3.0\ \text{kg}$ at $0.20\ \text{m}$ from the axis and $1.20\ \text{m}$ below $S$ . A belt pulls tangentially with $5.5\ \text{N}$ on the rim of a light pulley of radius $0.30\ \text{m}$ on the same axis; the bearings push on the shaft only toward the axis and along it, and gravity is parallel to the axis.

What is the moment of inertia $I_S$ ?

What is the angular acceleration $\alpha_z$ ?

✅ Correct!

I_S = 0.18 + 0.25 + 0.12 = 0.55\ \text{kg}\cdot\text{m}^2

, only the belt has an external

z

-torque,

1.65\ \text{N}\cdot\text{m}

, and

\alpha_z = 1.65/0.55 = 3.0\ \text{rad/s}^2

❌ The radius must be squared.

\sum_i \Delta m_i r_i = 1.70\ \text{kg}\cdot\text{m}

is not even in the right units;

r_i

enters twice, once as the lever arm and once through

a_{\theta,i} = r_i\alpha_z

❌ That uses the heights $z_i$ .

\sum_i \Delta m_i z_i^2 = 4.64

measures distances along the axis; the height dropped out of the

z

-torque, and only the distance from the axis counts.

❌ That uses the distances from $S$ .

\sum_i \Delta m_i\,(r_i^2 + z_i^2) = 5.19

would change as

S

slides along the axis;

I_S

uses only the perpendicular distances

r_i

❌ That divides by the total mass.

1.65/6.0 \approx 0.28

; in the rotational law the mass is replaced by the moment of inertia,

\alpha_z = (\tau_S^{\,\text{ext}})_z / I_S

❌ That divides by $5.19\ \text{kg}\cdot\text{m}^2$ . That sum uses the distances from

S

; with the perpendicular distances,

I_S = 0.55\ \text{kg}\cdot\text{m}^2

❌ That multiplies torque by moment of inertia.

(\tau_S^{\,\text{ext}})_z = I_S\,\alpha_z

, so

\alpha_z

is the torque divided by

I_S

Show solution

The moment of inertia uses each mass's perpendicular distance from the axis; the heights above or below $S$ never enter:

I_S = \sum_i \Delta m_i\,r_i^2 = (2.0)(0.30)^2 + (1.0)(0.50)^2 + (3.0)(0.20)^2 = 0.18 + 0.25 + 0.12 = 0.55\ \text{kg}\cdot\text{m}^2

External $z$ -torques: the belt gives $(0.30\ \text{m})(5.5\ \text{N}) = 1.65\ \text{N}\cdot\text{m}$ . The bearing forces are radial or axial and gravity is axial, so none of them has a $z$ -torque, and the internal forces act along the lines joining each pair, so their torques cancel in pairs.

\alpha_z = \frac{(\tau_S^{\,\text{ext}})_z}{I_S} = \frac{1.65\ \text{N}\cdot\text{m}}{0.55\ \text{kg}\cdot\text{m}^2} = 3.0\ \text{rad/s}^2

Problem 4 · List the Outside Forces

Given: A turntable platter turns on a vertical spindle, with $I_S = 0.060\ \text{kg}\cdot\text{m}^2$ about the spindle axis. A finger presses the rim, $0.15\ \text{m}$ from the axis, with a $2.0\ \text{N}$ force: $1.6\ \text{N}$ tangential (counterclockwise seen from above) and $1.2\ \text{N}$ radially toward the axis. Bearing friction contributes a $z$ -torque of $-0.048\ \text{N}\cdot\text{m}$ . Apart from that friction, the spindle pushes on the platter only toward the axis and vertically, gravity is vertical, and the platter's parts pull on one another with internal forces of up to $50\ \text{N}$ , each along the line joining the two parts. Find $\alpha_z$ .

✅ Correct! Only the finger's tangential push and the friction have

z

-torques:

(0.15)(1.6) - 0.048 = 0.192\ \text{N}\cdot\text{m}

, and

0.192/0.060 = 3.2\ \text{rad/s}^2

❌ That uses the full $2.0\ \text{N}$ . Only the

1.6\ \text{N}

tangential part has a lever arm about the axis; the

1.2\ \text{N}

radial part points at the axis.

❌ The friction torque is missing. The bearing is outside the platter, so its

-0.048\ \text{N}\cdot\text{m}

is an external torque:

0.240 - 0.048 = 0.192\ \text{N}\cdot\text{m}

❌ That uses the radial component. The

1.2\ \text{N}

toward the axis has no lever arm about it; the tangential

1.6\ \text{N}

is the part that turns the platter.

Show solution

List the forces from outside the platter and take each one's $z$ -torque about $S$ on the axis:

finger, tangential part: $(0.15\ \text{m})(1.6\ \text{N}) = 0.240\ \text{N}\cdot\text{m}$
finger, radial part: $0$ (it points at the axis)
bearing friction: $-0.048\ \text{N}\cdot\text{m}$
spindle push (toward the axis or vertical) and gravity (vertical): $0$

The $50\ \text{N}$ internal forces act along the lines joining the parts, so their torques cancel in pairs and never enter.

(\tau_S^{\,\text{ext}})_z = 0.240 - 0.048 = 0.192\ \text{N}\cdot\text{m}

\alpha_z = \frac{(\tau_S^{\,\text{ext}})_z}{I_S} = \frac{0.192\ \text{N}\cdot\text{m}}{0.060\ \text{kg}\cdot\text{m}^2} = 3.2\ \text{rad/s}^2

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