---
title: Rotational Dynamics
module: Rotation
moduleNumber: 5
lessonNumber: 2
order: 502
summary: >
  A force applied to a wheel does nothing unless it acts off the axis: what turns
  a rigid body is torque, force times lever arm. This lesson makes that precise
  and turns it into the rotational Newton's second law, $\sum\tau=I\alpha$ about a
  fixed axis, the exact analogue of $\sum F=ma$. From there we get rotational work
  $W=\int\tau\,\d\theta$ and power $P=\tau\omega$, size a motor to a load, and
  solve pulleys and Atwood machines where the pulley's own inertia can no longer
  be ignored — always insisting that every torque be measured about the same axis.
topics: [Rotation]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 9 — Rotation; §§9-4–9-5"
---

## Torque, force systems, and safety

Torque measures the rotational effect of a force about a selected axis. Its vector
definition is $\vec\tau=\vec r\times\vec F$; in a planar fixed-axis
problem, its signed magnitude is

$$
\tau=rF\sin\phi=F\ell,
$$

where $\ell$ is the perpendicular distance from the axis to the force line. A
radial force has zero lever arm and therefore zero torque, regardless of its
magnitude. Torque depends on the chosen axis; the same force can have different
torques about different points.

For particles rigidly constrained at fixed radii, tangential Newton's law is
$F_t=mr\alpha$. Multiplication by $r$ and summation give

$$
\sum\tau=\left(\sum_i m_ir_i^2\right)\alpha=I\alpha.
$$

This is the fixed-axis rotational form of Newton's second law. The inertia and
all torques must refer to the same axis. A zero net torque gives zero angular
acceleration, but does not require zero angular velocity.

$$
% caption: The lever arm $\ell$ is the perpendicular distance from the axis $O$ to the force's line of action, giving $\tau=F\ell=rF\sin\phi$. A force pointed along $\vec r$ has $\ell=0$ and no torque.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\coordinate (O) at (0,0);
\coordinate (P) at (4,0);
\draw[black,dashed] (4.55,-0.95)--(2.55,2.51);
\draw[very thick] (O)--(P);
\node at (2,-0.35) {$r$};
\draw[->,black,very thick] (P)--(3.1,1.56) node[above right] {$F$};
\draw[acc,thick] (O)--(3,1.732);
\node[acc] at (1.05,1.28) {arm};
\draw[fill=white,draw=black] (O) circle (2.4pt) node[below left] {$O$};
\draw[fill=white,draw=black] (P) circle (2.2pt);
\end{tikzpicture}
$$

**Force pairs and resultant force systems.**

Newton's third-law force pairs act on different bodies and must not be cancelled
inside one free-body diagram. A force couple, by contrast, contains two forces
acting on the same body and can have zero resultant force with nonzero net torque.
A support force and the load it supports act on different bodies and cannot cancel
within one free-body diagram.

An arbitrary planar set of forces can be reduced at a chosen origin to one
resultant force plus one couple moment. Moving the resultant force along its own
line of action leaves torque unchanged. Moving it to a parallel line changes torque
and requires an added couple to preserve the same external effect. This reduction
separates translational tendency from rotational tendency in a statics calculation
without discarding either.

> **Worked example.** A $100\ \mathrm N$ force acts upward at $0.40\ \mathrm m$
> from an origin, and a $60\ \mathrm N$ force acts downward at $0.20\ \mathrm m$.
> The resultant force is $40\ \mathrm N$ upward and the net torque is
> $100(0.40)-60(0.20)=28\ \mathrm{N\,m}$ counterclockwise, so the equivalent single
> resultant must act at $28/40=0.70\ \mathrm m$ from the origin. That reduced
> system matches both the net force and the net torque; placing the $40\ \mathrm N$
> force anywhere else would require an added couple.

**Torque limits, safety factors, and reporting.**

The torque demanded by a mechanism is rarely the torque that can be applied
continuously. Shafts have allowable shear stress, gears have tooth-contact limits,
bearings have friction and life limits, and motors have thermal current limits. A
design torque budget includes steady load torque, acceleration torque $I\alpha$,
friction, uncertainty, and a safety factor appropriate to the consequences of
failure. Peak torque may be acceptable for a short transient while continuous
torque is restricted by heating.

Shaft twist can become a performance limit before material failure. For a uniform
circular shaft in the elastic range, torque is proportional to twist angle. A
large twist changes alignment, can introduce backlash in a gear train, and makes a
rigid-body fixed-axis model incomplete. The actuator then drives a torsional
oscillator rather than one rigid rotor. A measured torque-speed response with
oscillation or delay often indicates this compliance.

Torque measurements should state the reference axis, sign convention, time or
angle interval, sensor calibration, and whether reported torque is applied, load,
or net torque. Power measurements require simultaneous torque and angular-speed
data on the same shaft. Work calculations require torque against angular
displacement, including the sign of both quantities. These details prevent a
numerically correct value from being assigned an incorrect physical meaning.

**Dimensional and limiting checks.**

Torque has dimensions $\mathrm{kg\,m^2\,s^{-2}}$. Dividing by moment of inertia
gives angular acceleration with dimensions $\mathrm{s^{-2}}$. Multiplying torque
by angle gives energy; multiplying by angular speed gives power. These dimensional
relations expose errors such as using force instead of torque in a rotational
equation or using a linear distance where angular displacement is required.

Several limiting cases test a result without recalculating it. If a lever arm tends
to zero, torque must tend to zero. If pulley inertia tends to zero, its two ideal
cord tensions must become equal. If external torque tends to zero, angular impulse
and angular acceleration must vanish. If a gear ratio tends to one, reflected
inertia and torque-speed relations must reduce to direct coupling. Such checks
expose reversed radii or signs in multistep calculations where algebra can obscure
them.

> **Worked example.** A robotic-arm joint has load inertia
> $0.40\ \mathrm{kg\,m^2}$, desired acceleration $8.0\ \mathrm{rad\,s^{-2}}$, and
> estimated friction torque $0.30\ \mathrm{N\,m}$. The acceleration torque is
> $I\alpha=3.2\ \mathrm{N\,m}$, so the actuator must supply at least
> $3.2+0.30=3.5\ \mathrm{N\,m}$ before any safety margin. Inserting a 2:1 speed
> reduction between motor and load cuts the reflected load inertia to one quarter
> of its value but makes the motor turn twice as fast: the choice depends on the
> motor's torque-speed capability and the required joint speed, not on inertia
> alone.

## Signed rotational dynamics and measurement

Rotational calculations are verified by tracing signs through one physical motion.
Take a motor that accelerates a rotor in the declared positive direction. Applied
torque, angular acceleration, and the increasing angular velocity must all have
the positive sign. A braking torque on the same spinning rotor has negative sign
and negative power. If a calculation predicts positive brake power while angular
speed is positive, either the torque direction or the power convention has been
reversed. This direct physical check is often more reliable than checking symbols
in a long algebraic expression.

Torque slopes determine inertia experimentally. With constant net torque,
a graph of angular velocity against time has slope $\alpha=\tau_{net}/I$. Repeating
the measurement for several torque settings gives a graph of torque against
acceleration. Its slope is inertia when the intercept has been corrected for
constant loss torque. A nonzero intercept represents a torque that persists even
when acceleration is zero, commonly bearing friction or sensor bias.

The slope method assumes the rotating assembly is unchanged across trials. Adding
a clamp, changing a chuck, or allowing a cable to wind onto a drum changes inertia
or effective radius and can bend the fitted relation. Speed-dependent drag also
causes curvature because net torque decreases at larger angular speed. A robust
measurement therefore uses an initial low-speed window or explicitly fits the loss
model together with inertia.

Inertia inferred from torque slope should agree with an independent geometric,
torsion-pendulum, or energy measurement within uncertainty. Agreement supports the
fixed-axis model. Disagreement can arise from an incorrect axis, unaccounted rotating
hardware, a hidden compliance mode, or an error in torque calibration. The
cross-check tests whether one mechanical model explains all measurements.

**Work-energy checks for varying torque.**

When a torque measurement is available as a function of angle, rotational work is
obtained by numerical integration even if no analytic torque formula is known.
Trapezoidal summation over samples $(\theta_i,\tau_i)$ gives

$$
W\approx\sum_i\frac{\tau_i+\tau_{i+1}}{2}(\theta_{i+1}-\theta_i).
$$

The resulting work should equal measured change in rotational kinetic energy after
losses have been included. A mismatch can identify a torque-sensor calibration
error, an angle encoder scale error, an unmeasured resisting torque, or a rotor
whose inertia is not constant. This comparison is stronger than checking either
the torque trace or speed trace alone because it joins independent measurements
through energy conservation.

With speed-dependent loss torque $\tau_L(\omega)$, rotor net work is
$\int[\tau_{drive}(\theta)-\tau_L(\omega)]\,\d\theta$. The loss term cannot be
subtracted using one mean speed if speed varies strongly through the motion. A
time-domain power integral $\int\tau_L\omega\,\d t$ is equivalent and can be
more natural when loss has been calibrated against speed.

**Angular-acceleration data fitting.**

Angular acceleration is commonly estimated from sampled angle or speed data. A
linear fit to angular velocity over an interval gives its slope as $\alpha$. A
quadratic fit to angle gives twice its curvature as $\alpha$ when acceleration is
constant. The speed fit is usually less sensitive to position offsets, while the
angle fit can use high-resolution encoder counts directly. Both fits should report
their interval because a changing torque can make a single constant-acceleration
value depend on the selected window.

Given a calibrated inertia, plotting net torque against measured angular acceleration
should give a line through the origin with slope $I$. A nonzero intercept indicates
constant resisting torque or sensor offset. Curvature indicates speed-dependent
drag, changing inertia, compliance, or an unmodelled torque contribution. Residual
plots distinguish these physical effects from random measurement scatter.

**Torque-step identification.**

A torque-step experiment identifies rotational dynamics from a measured angular-
speed trace. A known torque is applied rapidly to a rotor, and the initial slope
of $\omega(t)$ estimates angular acceleration before speed-dependent drag becomes
important. If inertia is known, $I\alpha$ gives net torque. Applied torque follows
after a separately measured loss torque is added. The procedure is valid only when
the rotor and its sensor rotate as one rigid body about the calibrated axis.

Numerical differentiation of angular position is sensitive to noise. A direct
encoder velocity channel, a fit over a short initial interval, or repeated trials
can improve the estimate. Curvature in the speed trace is physical evidence of
speed-dependent drag, drive saturation, or changing load torque; it should not be
discarded as random error without an uncertainty analysis.

> **Worked example.** A rotor with $I=0.050\ \mathrm{kg\,m^2}$ shows a measured
> initial slope $18\ \mathrm{rad\,s^{-2}}$, so the net torque is
> $I\alpha=0.90\ \mathrm{N\,m}$. If spin-down data give a loss torque
> $0.12\ \mathrm{N\,m}$ at that speed, the drive torque is $1.02\ \mathrm{N\,m}$.
> The loss correction must be measured over the same speed range, because drag
> varies with angular velocity.

**Power averaging in cyclic machinery.**

Instantaneous mechanical power is the product $P(t)=\tau(t)\omega(t)$. For
cyclic machinery, mean power is the time average of this product. Multiplying the
mean torque by mean angular speed is valid only when fluctuations are uncorrelated
or negligible. Torque can peak when speed is low or high depending on the machine,
and the correlation changes energy transfer over a cycle.

The energy delivered in one cycle is the area under power-time curve. It is also
the closed integral of torque with respect to angle. A flywheel stores energy during
portions with positive excess torque and releases it during portions with negative
excess torque. Its mean speed can remain nearly constant even though instantaneous
speed varies. Increasing inertia reduces this variation but increases required
startup work and bearing load.

A brake illustrates the limiting case. With constant brake torque, power is
proportional to speed and falls to zero at rest. The total heat produced is fixed
by initial rotational kinetic energy, but peak thermal loading occurs immediately
after braking begins. A design that meets total-energy capacity can still fail if
its instantaneous power capacity or heat-transfer rate is inadequate.

**Torque control and actuator limits.**

A controlled actuator applies torque according to a command, but the available
torque is bounded by current, voltage, thermal capacity, and mechanical strength.
At low speed, many electric motors operate near a torque limit. At higher speed,
back electromotive force limits current and torque falls. The resulting torque-
speed curve determines angular acceleration through $\alpha=\tau_{net}/I$.
Load torque, bearing torque, and aerodynamic drag must be subtracted before this
relation predicts acceleration.

Acceleration control is not identical to speed control. A speed controller adjusts
torque until net torque approaches zero at the desired speed. A position controller
often uses torque proportional to position error and velocity error. The mechanical
plant still obeys rotational dynamics, so high gain can excite flexible modes or
cause torque saturation. A rigid fixed-axis model is valid only below frequencies
where shaft and support deformation become important.

$$
% caption: Available drive torque falls with angular speed while the load torque stays roughly constant; their intersection is the steady operating speed, where net torque and angular acceleration vanish.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0,0)--(5.8,0) node[right] {speed};
\draw[->,black] (0,0)--(0,3.2) node[above] {torque};
\draw[thick] (0.3,2.7)--(5.3,0.3);
\draw[black,dashed] (0.3,1.1)--(5.3,1.1);
\draw[fill=acc!20,draw=acc] (3.63,1.1) circle (2.2pt);
\node[black] at (1.15,2.55) {drive};
\node[black] at (4.7,0.82) {load};
\end{tikzpicture}
$$

## Equilibrium and force couples

Static equilibrium requires both zero force resultant and zero torque resultant.
Choosing the hinge as torque origin removes the hinge reaction from the torque
equation because its lever arm is zero. The reaction is still found afterward from
the force equations. This is an algebraic convenience, not a physical omission.

In a uniform beam of length $L$, weight $W_b$ acts at $L/2$. A load $W_L$ at
distance $x$ and a vertical cable tension $T$ at the far end obey

$$
TL-W_b\frac L2-W_Lx=0.
$$

The equation is valid only when the cable force is perpendicular to the beam. For
an inclined cable, replace $T$ by its perpendicular component. A shallow cable
therefore requires high tension to supply the same moment.

$$
% caption: A hinged beam balances torques about the hinge: the beam weight $W_b$ at its midpoint and the load $W_L$ are held by the cable tension $T$ at the far end.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[very thick] (0,0)--(5.8,0);
\draw[fill=white,draw=black] (0,0) circle (2.4pt) node[below left] {hinge};
\draw[->,black,very thick] (2.9,0)--(2.9,-1.1) node[below] {$W_b$};
\draw[->,black,very thick] (4.3,0)--(4.3,-1.45) node[below] {$W_L$};
\draw[->,acc,very thick] (5.8,0)--(5.8,1.6) node[above] {$T$};
\end{tikzpicture}
$$

> **Worked example.** A $3.0\ \mathrm m$ uniform beam weighs $150\ \mathrm N$ and
> carries $240\ \mathrm N$ at $2.2\ \mathrm m$ from the hinge, with a vertical
> cable at the far end. Torque balance about the hinge gives
>
> $$
> T=\frac{150(1.5)+240(2.2)}{3.0}=251\ \mathrm N,
> $$
>
> and vertical force balance then gives a hinge reaction of $139\ \mathrm N$
> upward. A negative cable tension would mean the assumed support cannot supply the
> required force, and the support model would have to change.

**Couples and torque signs.**

Torque is signed only after an axis direction is declared. In a planar problem,
counterclockwise torque is commonly positive. The sign records rotational tendency,
not force direction. A force line passing through the axis has zero torque even if
the force is large. The right-hand rule fixes the corresponding vector direction
when the calculation requires three-dimensional components.

Two equal, opposite, parallel forces separated by perpendicular distance $d$ form
a couple. Their resultant force is zero while their torque is $Fd$. The torque of
a couple is independent of origin, since any shift adds equal and opposite moments
to the two forces. A wrench turned by two hands is a direct example: translation
cancels while rotation remains.

$$
% caption: A couple is two equal, opposite forces separated by a perpendicular distance $d$. Their net force is zero but their torque $Fd$ is the same about every point.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black,very thick] (0,1)--(3,1) node[right] {$F$};
\draw[->,black,very thick] (3,0)--(0,0) node[left] {$F$};
\draw[<->,black] (3.6,0)--(3.6,1) node[midway,right] {$d$};
\draw[->,acc] (1.5,0.25) arc (-72:250:0.42);
\end{tikzpicture}
$$

## Variable torque, work, and power

When torque varies with angular position, its work is the signed area under the
torque-angle curve. A torsional spring with $\tau=-\kappa\theta$ stores energy

$$
U(\theta)=\frac12\kappa\theta^2.
$$

If a motor torque decreases linearly from $\tau_0$ to zero over angular interval
$\Theta$, it does work $\tau_0\Theta/2$. The result is the triangular area,
not peak torque times angle. For a rotor of fixed inertia, this work sets the
change in $I\omega^2/2$ regardless of the time profile used to traverse the angle.

$$
% caption: A torque that falls linearly with angle does work equal to the triangular area under its torque-angle curve; the peak torque alone overstates the energy transferred.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0,0)--(5.8,0) node[right] {angle};
\draw[->,black] (0,0)--(0,3.1) node[above] {torque};
\fill[acc!12] (0.35,0)--(0.35,2.6)--(5.1,0)--cycle;
\draw[acc,thick] (0.35,2.6)--(5.1,0);
\node[acc] at (2.2,0.9) {$W$};
\end{tikzpicture}
$$

For torque specified as a function of time, angular impulse gives speed change.
The two descriptions are connected by the actual motion but should not be mixed.
A torque pulse may have zero net work if it occurs when angular displacement is
negligible, while still changing angular momentum; a torque applied over finite
angle can transfer work even when its time integral is small under sign changes.

**Distributed-force applications.**

A distributed load requires a resultant force and its line of action.
With load density $w(x)$, both force and first moment must be integrated.
The centre-of-pressure calculation for hydrostatic force follows the same rule:
pressure is a force per area, and the torque integral weights deeper surface
elements more strongly. The same integral structure applies to beams, dams, and
nonuniform traction problems.

> **Worked example.** A rotor with $I=0.20\ \mathrm{kg\,m^2}$ receives torque
> $\tau(\theta)=2.0-0.10\theta$ from $0$ to $10\ \mathrm{rad}$. The work is
>
> $$
> W=\int_0^{10}(2.0-0.10\theta)\,\d\theta=15\ \mathrm J.
> $$
>
> Starting from rest, the final speed is
> $\sqrt{2W/I}=12.2\ \mathrm{rad\,s^{-1}}$. The average torque is
> $1.5\ \mathrm{N\,m}$, so average torque times angle gives the same work for this
> linear profile.

**Power averaging.**

Instantaneous mechanical power is $P=\tau\omega$. The average of a product is not
usually the product of averages. Torque ripple can be correlated with speed ripple,
so measuring mean torque and mean speed separately can misstate mean power. Direct
time-resolved multiplication or a physical energy measurement over a known interval
avoids this error.

During a constant-torque acceleration, angular speed rises linearly and power rises
linearly. During constant-power drive, torque must decrease inversely with speed;
this behavior appears in motor control above a base speed. Mechanical limits such
as maximum torque, maximum speed, and thermal loss determine which region is
available in practice.

**Force couples and static torque balance.**

Two equal, opposite, parallel forces separated by perpendicular distance $d$ form
a couple. Their net force is zero, but their torque magnitude is $Fd$. Unlike the
torque of a single force, a couple has the same torque about every origin. Moving
the origin changes the individual force torques by equal and opposite amounts, so
their sum remains unchanged. A steering wheel, screwdriver, and jar lid are common
examples of applied couples.

Static balance of a rigid body requires the force sums and torque sum to vanish.
Choosing a torque origin through an unknown support reaction removes it from the
moment equation, but it remains in the force equations. A sign convention must be
fixed before calculation. In a plane, counterclockwise positive is common; a
negative solved torque then represents clockwise tendency rather than an error.

A beam of length $L$ with a point load $W$ at position $x$ and a vertical
cable tension $T$ at the far end, torque balance about the left hinge is

$$
TL-Wx=0,
\qquad T=\frac{Wx}{L}.
$$

The cable force is smaller than the load only when its lever arm is larger. If the
cable is inclined, replace $T$ by its component perpendicular to the beam. Force
balance then determines hinge components. The geometry of the force line, not the
distance to its point of application, controls torque.

## Torque integrals and pulley systems

Torque sign can be checked by imagining the initial rotation caused by each force.
A force line through the axis has no rotational tendency. A force that tends to
turn a body counterclockwise has positive scalar torque in the stated convention.
The vector direction is perpendicular to the plane by the right-hand rule. In
three-dimensional problems, scalar signs are insufficient and torque components
must be resolved along coordinate axes.

> **Worked example.** A uniform $2.0\ \mathrm m$ beam weighs $120\ \mathrm N$ and
> carries a $180\ \mathrm N$ load $1.50\ \mathrm m$ from a hinge, with a vertical
> cable at the far end. Torque balance about the hinge gives
>
> $$
> T(2.0)-120(1.0)-180(1.50)=0,
> \qquad T=195\ \mathrm N,
> $$
>
> and the hinge vertical reaction is $105\ \mathrm N$ upward. The beam's own weight
> at its centre must be included; omitting it would give a cable tension too small
> to hold equilibrium.

**Torque integrals, braking, and massive pulleys.**

Distributed forces require a moment integral. If a force density $w(x)$ acts
perpendicular to a beam, its resultant and torque about the left end are

$$
F=\int_0^L w(x)\,\d x,
\qquad \tau=\int_0^Lxw(x)\,\d x.
$$

The resultant line of action is $x_{res}=\tau/F$. For a triangular load
$w(x)=w_0x/L$, direct integration gives $F=w_0L/2$ and $x_{res}=2L/3$ from the
zero-load end. The location shifts toward the larger load because that region
contributes more moment. Replacing the load by a force at the geometric midpoint
would satisfy neither the correct torque nor the correct support reactions.

$$
% caption: A load rising linearly from zero to $w_0$ has a triangular intensity; its single resultant acts at $2L/3$ from the zero-load end, closer to the heavier side, not at the span midpoint.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
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\draw[black,thick] (0,0)--(5.8,0);
\draw[thick] (0,0)--(5.2,2);
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\draw[->,acc,very thick] (3.47,2.55)--(3.47,0.1);
\node[acc] at (4.05,2.5) {resultant};
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$$

The same integral describes rotational work when torque varies with angle:

$$
W=\int_{\theta_i}^{\theta_f}\tau(\theta)\,\d\theta.
$$

Under constant brake torque magnitude $\tau_b$, the stopping angle is
$I\omega_i^2/(2\tau_b)$. The work done by the brake equals the initial rotational
energy. A constant torque produces a linearly decreasing angular speed, but power
$|P|=\tau_b\omega$ is largest at the beginning of the stop.

> **Worked example.** A rotor with $I=0.80\ \mathrm{kg\,m^2}$ and initial speed
> $100\ \mathrm{rad\,s^{-1}}$ meets a $20\ \mathrm{N\,m}$ brake torque, giving
> $\alpha=25\ \mathrm{rad\,s^{-2}}$. It stops in $4.0\ \mathrm s$, turns through
> $200\ \mathrm{rad}$, and releases $\tfrac12 I\omega_i^2=4.0\ \mathrm{kJ}$ as
> heat. The braking power $\tau_b\omega$ is largest at the start, $2.0\ \mathrm{kW}$.

Massive pulleys couple translation and rotation. For an Atwood system,

$$
m_1g-T_1=m_1a,
\quad T_2-m_2g=m_2a,
\quad (T_1-T_2)R=I\frac aR.
$$

Elimination gives

$$
a=\frac{(m_1-m_2)g}{m_1+m_2+I/R^2}.
$$

The term $I/R^2$ is rotational effective mass. It reduces acceleration because
some gravitational energy increases pulley kinetic energy. The tensions differ by
$Ia/R^2$; equal tensions would imply zero pulley torque and cannot describe an
accelerating massive pulley.

> **Worked example.** For $m_1=3.0\ \mathrm{kg}$, $m_2=1.0\ \mathrm{kg}$,
> $R=0.20\ \mathrm m$, and $I=0.080\ \mathrm{kg\,m^2}$, the rotational effective
> mass is $I/R^2=2.0\ \mathrm{kg}$, so the acceleration is
> $2g/(4+2)=3.27\ \mathrm{m\,s^{-2}}$. A massless-pulley model would overestimate
> it. The force-torque solution also yields the tensions; an energy solution
> returns the same acceleration and is shorter when only speed after a distance is
> wanted.

**Torque as a spatial integral.**

The torque of a distributed force is obtained by integrating the moment of each
force element:

$$
\vec\tau=\int\vec r\times\d\vec F.
$$

For pressure on a surface, $\d\vec F=-p\hat n\,\d A$. For a magnetic
or gravitational body force, the differential force comes from the relevant force
density. The cross product must be evaluated before integrating when force direction
or lever arm changes across the body. Replacing a distributed force by one resultant
is valid only after its line of action has been determined from the first moment.

A uniform force density on a symmetric body often acts through the geometric centre
by symmetry. A nonuniform distribution generally does not. Hydrostatic pressure is
a standard case: pressure increases with depth, shifting the resultant below the
area centroid. The same mathematical structure appears in a beam with varying load
and in a charged plate in a nonuniform electric field.

**Massive Atwood machine.**

An Atwood machine with a massive pulley requires unequal cord tensions. With
$m_1$ descending and $m_2$ ascending, the force equations are

$$
m_1g-T_1=m_1a,
\qquad T_2-m_2g=m_2a.
$$

The pulley equation and no-slip constraint are

$$
(T_1-T_2)R=I\alpha,
\qquad a=R\alpha.
$$

Elimination yields

$$
a=\frac{(m_1-m_2)g}{m_1+m_2+I/R^2}.
$$

The term $I/R^2$ has mass units and is sometimes called rotational effective
mass. It does not represent material added to either hanging block; it expresses
the pulley energy required for a given cord acceleration.

> **Worked example.** With $m_1=4.0\ \mathrm{kg}$, $m_2=1.0\ \mathrm{kg}$,
> $R=0.20\ \mathrm m$, and $I=0.10\ \mathrm{kg\,m^2}$, the effective rotational
> mass is $I/R^2=2.5\ \mathrm{kg}$, so the acceleration is
> $3g/(5+2.5)=3.92\ \mathrm{m\,s^{-2}}$ and the tension difference is
> $Ia/R^2=9.81\ \mathrm N$. Treating the pulley as massless would predict
> $5.89\ \mathrm{m\,s^{-2}}$, an appreciable overestimate.

**Energy solution of pulley systems.**

If the blocks move distance $s$, gravitational potential-energy loss is
$(m_1-m_2)gs$. The kinetic energy is

$$
K=\frac12m_1v^2+\frac12m_2v^2+\frac12I\left(\frac vR\right)^2.
$$

Setting the potential-energy loss equal to this total gives the same acceleration
after using $v^2=2as$. Energy is often shorter for a speed-after-distance question,
whereas force and torque equations expose tensions and their directions. The
requested unknown determines the appropriate method.

## Angular impulse, braking, and energy

A force applied by a cam or crank may produce torque that changes sign within one
revolution. The net work is the signed area under torque-angle curve; positive and
negative regions represent energy delivered to and extracted from the rotor. A
flywheel smooths this variation by storing kinetic energy, but it does not remove
the need for a net positive average torque to overcome losses.

Over an interval with periodic torque $\tau(t)$, angular impulse is
$\int\tau\,\d t$. If the interval is a complete cycle and net impulse is zero,
angular momentum returns to its original value, though speed can vary during the
cycle. Instantaneous torque and cycle-average torque answer different questions in
engines and reciprocating machinery.

**Rotational work and power.**

A tangential force through angular displacement $\d\theta$ acts through distance
$\d s=r\,\d\theta$. Its work is $\d W=\tau\,\d\theta$, so

$$
W=\int\tau\,\d\theta=\Delta\left(\frac12I\omega^2\right),
\qquad P=\tau\omega.
$$

The sign of power is negative for a braking torque opposite angular velocity.
The work-energy relation applies to a rigid body about a fixed axis; a body whose
axis translates additionally has centre-of-mass kinetic energy.

> **Worked example.** A disk with $I=0.40\ \mathrm{kg\,m^2}$ carries a constant
> net torque $2.0\ \mathrm{N\,m}$, giving $\alpha=5.0\ \mathrm{rad\,s^{-2}}$.
> Starting from rest, after $3.0\ \mathrm s$ it reaches
> $\omega=15\ \mathrm{rad\,s^{-1}}$ with rotational kinetic energy $45\ \mathrm J$,
> equal to torque times its angular displacement.

## Levers, supports, and distributed loading

The torque magnitude can be found either from the perpendicular lever arm or from
the force component perpendicular to the radius. Decomposing a force into radial
and tangential components gives

$$
F_t=F\sin\phi,
\qquad \tau=rF_t.
$$

The radial component produces no torque about the axis because its line of action
passes through that axis. This is a geometric result, not a statement that radial
forces are dynamically unimportant: they can supply centripetal acceleration or
change support reactions.

$$
% caption: A force at radius $r$ resolves into a radial part $F_r$, whose line passes through the axis and gives zero torque, and a tangential part $F_t$; only $F_t$ turns the body, with $\tau=rF_t$.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\coordinate (O) at (0,0);
\coordinate (P) at (3.6,0);
\draw[fill=white,draw=black] (O) circle (2.2pt) node[below left] {$O$};
\draw[very thick] (O)--(P);
\node at (1.8,-0.32) {$r$};
\draw[black,dashed] (3.6,1.6)--(2.8,1.6)--(2.8,0);
\draw[->,black,very thick] (P)--(2.8,1.6) node[above left] {$F$};
\draw[->,acc,thick] (P)--(3.6,1.6) node[above] {$F_t$};
\draw[->,black,thick] (P)--(2.8,0) node[below] {$F_r$};
\draw[fill=white,draw=black] (P) circle (2.2pt);
\end{tikzpicture}
$$

**Multiple torques and equilibrium.**

Several forces can have zero net force while retaining a nonzero couple. The
rotational equation uses the signed sum of every external torque about one axis:

$$
\tau_{net}=\tau_1+\tau_2+\cdots=I\alpha.
$$

If $\tau_{net}=0$, angular acceleration is zero. A rotating flywheel then keeps
constant angular velocity in the ideal model. If angular velocity is already zero,
the same condition describes rotational equilibrium. Force balance and torque
balance are separate conditions for an extended body.

**Pulley dynamics.**

A cord that does not slip on a pulley imposes $a=R\alpha$. If pulley inertia is
not negligible, tensions on its two sides differ:

$$
(T_1-T_2)R=I\alpha=I\frac aR.
$$

A massless pulley is the limiting case $I=0$, where tensions are equal. The
pulley equation, each mass's translational equation, and the cord constraint form
one coupled system. Assigning equal tensions to a massive pulley removes the very
torque needed to accelerate it.

$$
% caption: A massive pulley carries unequal tensions $T_1\neq T_2$; their difference supplies the torque that angularly accelerates it, while the no-slip cord ties both blocks to one linear acceleration.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[acc,thick,fill=acc!8] (2.8,1.5) circle (0.72);
\draw[fill=white,draw=black] (2.8,1.5) circle (2pt);
\draw[black,thick] (2.08,1.5)--(2.08,-0.6);
\draw[black,thick] (3.52,1.5)--(3.52,-1.4);
\draw[black,thick,fill=black!8] (1.73,-1.15) rectangle (2.43,-0.6);
\draw[black,thick,fill=black!8] (3.17,-1.95) rectangle (3.87,-1.4);
\node[black] at (2.08,-1.45) {$m_1$};
\node[black] at (3.52,-2.25) {$m_2$};
\node[black] at (1.68,0.5) {$T_1$};
\node[black] at (3.92,0.5) {$T_2$};
\end{tikzpicture}
$$

> **Worked example.** Masses $m_1=3.0\ \mathrm{kg}$ and $m_2=1.0\ \mathrm{kg}$
> hang from a pulley of radius $0.20\ \mathrm m$ and inertia
> $0.080\ \mathrm{kg\,m^2}$. Taking $m_1$ downward positive and eliminating the
> tensions,
>
> $$
> a=\frac{(m_1-m_2)g}{m_1+m_2+I/R^2}
> =\frac{2g}{4+2}=3.27\ \mathrm{m\,s^{-2}}.
> $$
>
> The term $I/R^2=2.0\ \mathrm{kg}$ acts like extra mass because some
> gravitational energy accelerates the pulley; omitting it would overpredict $a$.

**Rotational work and braking.**

Power $P=\tau\omega$ connects torque-speed specifications with energy transfer.
A motor delivering positive torque in the direction of rotation supplies positive
power. A brake applies opposite torque and has negative mechanical power; the
lost rotational energy becomes thermal energy or electrical energy in regenerative
braking.

An angular impulse changes angular momentum:

$$
\Delta L=\int\tau\,\d t.
$$

For fixed inertia this gives $I\Delta\omega$. An impact can have large torque over
a short time, so angular impulse captures the event more directly than a detailed
force-time model.

**Rotational equilibrium of extended bodies.**

A rigid body held at rest requires both force and torque equations:

$$
\sum F_x=0,\qquad \sum F_y=0,\qquad \sum\tau_O=0.
$$

The torque origin is selected for convenience. Choosing an origin at an unknown
support reaction removes that reaction from the torque equation because its lever
arm is zero. The remaining force equations recover it after other loads are known.
Changing the origin changes individual torques but not the final physical
equilibrium condition.

In a uniform beam of length $L$, weight acts at $L/2$. If a cable at the far end
makes angle $\beta$ with the beam, only its perpendicular component contributes
torque. Balance about the hinge gives

$$
TL\sin\beta-Mg\frac L2=0,
\qquad T=\frac{Mg}{2\sin\beta}.
$$

The tension diverges as the cable becomes nearly horizontal because a small
perpendicular component must balance the same weight torque. This limiting behavior
is a physical warning about high loads in shallow support cables.

**Distributed forces and torque integrals.**

A distributed load $w(x)$ has resultant force $F=\int w(x)\,\d x$ and torque
about origin $\int xw(x)\,\d x$. Its line of action is therefore

$$
x_{res}=\frac{\int xw(x)\,\d x}{\int w(x)\,\d x}.
$$

A triangular load increasing from zero to $w_0$ over length $L$ has
$w(x)=w_0x/L$. Integration gives total force $w_0L/2$ acting at $2L/3$ from the
zero-load end. Replacing it by a point force at the geometric midpoint would give
the correct force only for a uniform load and the wrong torque.

## Power transmission and design limits

During a short collision, integrate the torque law:

$$
\int_{t_i}^{t_f}\tau_{ext}\,\d t=\Delta L.
$$

The selected origin can eliminate an unknown impulsive support force. A projectile
embedding in a pivoted rod conserves angular momentum about the pivot during the
brief impact because pivot impulse has zero lever arm there. Kinetic energy is not
conserved in the embedding, so energy calculations begin only after the collision.

$$
% caption: A projectile of momentum $mv$ strikes the free end of a rod pivoted at $O$. Angular momentum about $O$ is conserved through the brief impact because the pivot impulse has zero lever arm.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[very thick] (0,0)--(0,3);
\node at (-0.35,1.5) {rod};
\draw[->,black,very thick] (-1.9,2.7)--(-0.4,2.7) node[midway,above] {$mv$};
\draw[fill=acc!15,draw=acc] (-0.16,2.7) circle (2.8pt);
\draw[fill=white,draw=black] (0,0) circle (2.4pt) node[below left] {$O$};
\end{tikzpicture}
$$

> **Worked example.** A $0.020\ \mathrm{kg}$ projectile moving at
> $300\ \mathrm{m\,s^{-1}}$ embeds at the end of a $1.0\ \mathrm m$ uniform rod of
> mass $1.0\ \mathrm{kg}$ pivoted at its other end. The initial angular momentum
> about the pivot is $mvL=6.0\ \mathrm{kg\,m^2\,s^{-1}}$, and the final inertia is
>
> $$
> I_f=\frac13(1.0)(1.0)^2+(0.020)(1.0)^2=0.353\ \mathrm{kg\,m^2},
> \qquad \omega_f=\frac{mvL}{I_f}=17.0\ \mathrm{rad\,s^{-1}}.
> $$
>
> The post-impact kinetic energy is far below the projectile's initial value
> because embedding dissipates energy as deformation and heat. Angular momentum is
> conserved only about the pivot, and only through the short collision.

**Power transmission and gearing.**

An ideal gear contact has equal tangential speeds at the pitch circles:

$$
r_1\omega_1=r_2\omega_2.
$$

Power conservation then gives $\tau_1\omega_1=\tau_2\omega_2$. A larger driven
gear rotates more slowly and carries larger torque. This tradeoff is not an energy
gain; it is a conversion between torque and speed.

$$
% caption: Meshed gears share the same tangential speed at their pitch circles, so $r_1\omega_1=r_2\omega_2$; the larger gear turns slower and delivers proportionally more torque.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[acc,thick,fill=acc!8] (1.5,0) circle (0.7);
\draw[black,thick,fill=black!4] (3.45,0) circle (1.25);
\draw[fill=white,draw=black] (1.5,0) circle (1.8pt);
\draw[fill=white,draw=black] (3.45,0) circle (1.8pt);
\draw[->,acc,thick] ([shift={(1.5,0)}]70:0.45) arc (70:340:0.45);
\draw[->,black,thick] ([shift={(3.45,0)}]110:0.8) arc (110:-170:0.8);
\node[acc] at (1.5,-1.05) {fast};
\node[black] at (3.45,1.6) {slow};
\end{tikzpicture}
$$

Real transmissions have friction, tooth deformation, and speed-dependent losses.
Efficiency is the ratio of output to input power. Torque and angular speed must be
measured on the same shaft when computing a power balance; combining torque from
one shaft with speed from another gives a meaningless result.

**Torque equilibrium and support limits.**

Equilibrium of an extended body requires both zero resultant force and zero
resultant torque. A calculation that balances forces while ignoring their points of
application cannot decide whether a ladder rotates, a beam sags, or a bracket
remains fixed. The torque equation is independent of origin only after the force
equation is satisfied; when net force is nonzero, torque sums about different
origins differ by the moment of the resultant force.

In a horizontal beam with several vertical loads, select one support as origin.
Each downward load contributes force times horizontal distance. A cable at angle
$\beta$ contributes only its vertical component to torque about the hinge. The
resulting tension can exceed total weight when the cable is shallow because its
line of action has a small perpendicular moment arm.

Support reactions have physical limits. A cable cannot sustain compression; a
frictionless roller cannot exert force parallel to its surface; a pin can exert
force components but no idealized couple. Negative values in a solved reaction
therefore indicate either a reversed assumed direction or loss of contact. These
checks turn an algebraic result into a mechanically admissible solution.

**Distributed-force integral calculation.**

The torque of a distributed load is found by integrating its differential force.
With load $w(x)=w_0(1-x/L)$ on $0\le x\le L$, the resultant and its torque about
the left end are

$$
F=\int_0^Lw_0\left(1-\frac{x}{L}\right)\,\d x=\frac12w_0L,
$$

$$
\tau=\int_0^Lxw_0\left(1-\frac{x}{L}\right)\,\d x=\frac16w_0L^2.
$$

The resultant acts at $x=\tau/F=L/3$ from the larger-load end. The location is
not guessed from the triangle's geometric centre after drawing alone; it follows
from force and torque integrals. The same procedure applies to hydrostatic wall
loads, gravitational loads on nonuniform beams, and electromagnetic force density.

**Rotational work-energy derivation.**

During an infinitesimal rotation $\d\theta$, tangential displacement is
$\d s=r\,\d\theta$. The work of tangential force is $F_t\d s$, hence

$$
\d W=(rF_t)\,\d\theta=\tau\,\d\theta.
$$

Integration yields rotational work. Combining it with fixed-axis dynamics gives

$$
\int\tau\,\d\theta=\int I\alpha\,\d\theta
=\int I\omega\,\d\omega=\Delta\left(\frac12I\omega^2\right),
$$

where $\alpha\,\d\theta=\omega\,\d\omega$. The calculation requires constant
inertia about the chosen axis. A deforming rotor or a body redistributing mass has
an additional energy accounting problem because $I$ can change.

> **Worked example.** A motor torque decreases linearly from $4.0\ \mathrm{N\,m}$
> at $\theta=0$ to $1.0\ \mathrm{N\,m}$ at $\theta=12\ \mathrm{rad}$. The average
> torque is $2.5\ \mathrm{N\,m}$, so the work is $30\ \mathrm J$. A rotor of
> inertia $0.15\ \mathrm{kg\,m^2}$ starting from rest reaches
> $\omega=\sqrt{2W/I}=20\ \mathrm{rad\,s^{-1}}$, and $\tfrac12I\omega^2$ returns
> the same energy. The average-torque shortcut works here only because torque is
> linear in angle.

**Power and thermal limits.**

Rotating equipment converts mechanical power to heat when torque opposes motion.
With approximately constant resisting torque $\tau_b$, brake heat-generation
rate is $\tau_b\omega$. During a coast-down, speed decreases and heat rate falls.
The total heat produced equals the initial rotational kinetic energy when other
losses are negligible. A brake design must therefore consider both peak power at
initial speed and total energy over the stop.

**Signed torque from Cartesian components.**

Planar torque calculations are most reliable when the sign is obtained from the
component form of the cross product rather than from an informal clockwise sketch.
With $x$ to the right, $y$ upward, and positive $z$ out of the page, the torque
component about the origin is $\tau_z=xF_y-yF_x$. The coordinates and force
components must refer to the same origin. This expression automatically includes
both the moment arm and the sign. A force applied through the origin has position
coordinates zero and therefore produces zero torque about that origin even when
its magnitude is large.

> **Worked example.** A force with components $F_x=8.0\ \mathrm N$ and
> $F_y=5.0\ \mathrm N$ acts at $(x,y)=(0.40\ \mathrm m,-0.30\ \mathrm m)$. Its
> torque about the origin is
> $\tau_z=xF_y-yF_x=(0.40)(5.0)-(-0.30)(8.0)=+4.4\ \mathrm{N\,m}$. The positive
> sign means a counterclockwise tendency in the stated view. Neither component may
> be dropped for being horizontal or vertical: the horizontal force has a vertical
> lever arm and the vertical force a horizontal one, and both contribute positively
> in this placement.

The scalar sign convention does not replace the vector definition. The vector
$\vec r\times\vec F$ remains perpendicular to the plane of the diagram,
and changing the viewing side reverses the drawn clockwise sense while retaining
the same physical vector. In three dimensions, a torque about one axis can have
components about the other axes as well. A fixed-shaft model retains only the
component along the shaft, after verifying that the bearing constraints prevent
other rotations. Reporting the axis with every torque value avoids combining
moments that act about different lines.

**Net torque about a chosen pivot.**

A chosen pivot can simplify a force system without removing the need for force
balance. Every force whose line of action passes through the pivot has zero lever
arm in the torque equation. Hinge reactions are therefore absent from a moment
balance about the hinge, even though they remain present in the horizontal and
vertical force equations. This choice is a calculation convenience, not evidence
that the hinge force is physically absent. A different origin would give the hinge
force a nonzero moment and would require its components in the torque sum.

> **Worked example.** A horizontal beam is supported at its left end. A downward
> $20\ \mathrm N$ load acts $1.2\ \mathrm m$ from the hinge and a downward
> $50\ \mathrm N$ load acts $2.8\ \mathrm m$ from it; a cable attaches at
> $3.5\ \mathrm m$ at $35^\circ$ above the beam. With counterclockwise positive,
> equilibrium requires $T(3.5)\sin 35^\circ-(20)(1.2)-(50)(2.8)=0$, giving cable
> tension $81.7\ \mathrm N$. Only the perpendicular cable component enters the
> moment; using the full tension without the sine factor would overstate the
> supporting torque.
>
> The torque equation does not finish the problem. The upward cable component is
> $T\sin 35^\circ=46.9\ \mathrm N$, so the hinge must supply $23.1\ \mathrm N$
> upward to balance the $70\ \mathrm N$ of load, plus a horizontal reaction equal
> and opposite to the cable's horizontal pull. Those reactions do not affect the
> moment about the hinge but matter for support design; a complete check needs zero
> net force and zero net torque.

**Rotational work and shaft power.**

Torque transfers energy only through angular displacement. For a fixed shaft,
the differential work is $dW=\tau\,\d\theta$, so the work over a finite motion is
the signed area under a torque--angle graph. Torque in the same sense as the
angular displacement gives positive work; opposing torque gives negative work.
The instantaneous power is $P=\tau\omega$. A brake acting on a shaft rotating in
the chosen positive direction has negative torque and therefore negative
mechanical power for the shaft. The magnitude of that negative power is the rate
at which rotational mechanical energy becomes heat or another output form.

> **Worked example.** A rotor of inertia $0.90\ \mathrm{kg\,m^2}$ spins initially
> at $10\ \mathrm{rad\,s^{-1}}$. A motor applies $+12\ \mathrm{N\,m}$ while a brake
> applies $-3\ \mathrm{N\,m}$ through an angular displacement of $18\ \mathrm{rad}$.
> The net torque is $+9\ \mathrm{N\,m}$ and the net work is $162\ \mathrm J$. The
> initial rotational energy is $45\ \mathrm J$, so the final energy is
> $207\ \mathrm J$ and the final speed is $\sqrt{2K_f/I}=21.4\ \mathrm{rad\,s^{-1}}$.
> Using the motor work alone would ignore the energy the brake removes at the same
> time.
>
> At the instant the shaft reaches $16\ \mathrm{rad\,s^{-1}}$, the motor power is
> $192\ \mathrm W$, the brake power is $-48\ \mathrm W$, and the net shaft power is
> $144\ \mathrm W$. The power is not constant even though each torque is, because
> the speed changes. Integrating net torque over time gives the change in angular
> momentum, while integrating over angle gives the energy change; the two integrals
> answer different questions and carry different units.