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5.1 Rotational Kinematics

Define signed angular quantities

Angular displacement measures rotation about a specified axis in radians. Choose either clockwise or counterclockwise as positive; the opposite direction is negative.

$\begin{aligned}\Delta\theta&=\theta-\theta_0\\[3pt]\omega_{\text{avg}}&=\frac{\Delta\theta}{\Delta t}\\[3pt]\alpha_{\text{avg}}&=\frac{\Delta\omega}{\Delta t}\end{aligned}$

Units are rad, rad/s, and rad/s².

Reuse the linear-kinematics structure

One-dimensional linear motion	Rotation about one axis
Position $x$	Angular position $\theta$
Velocity $v$	Angular velocity $\omega$
Acceleration $a$	Angular acceleration $\alpha$

$\begin{aligned}\omega&=\omega_0+\alpha t\\[3pt]\theta&=\theta_0+\omega_0t+\frac12\alpha t^2\\[3pt]\omega^2&=\omega_0^2+2\alpha(\theta-\theta_0)\end{aligned}$

Read slopes and areas

Graph	Slope gives	Signed area gives
$\theta$ versus $t$	$\omega$	—
$\omega$ versus $t$	$\alpha$	$\Delta\theta$
$\alpha$ versus $t$	Rate of change of $\alpha$	$\Delta\omega$

Calculate constant angular acceleration

Constant- $\alpha$ example: choose counterclockwise as positive. A wheel has $\omega_0=+2.0\,\text{rad/s}$ and $\alpha=-0.50\,\text{rad/s}^2$ for $4.0\,\text{s}$ .

$\omega=\omega_0+\alpha t=2.0+(-0.50)(4.0)=0\,\text{rad/s}$ ,

$\Delta\theta=\omega_0t+\tfrac12\alpha t^2=(2.0)(4.0)+\tfrac12(-0.50)(4.0)^2=4.0\,\text{rad}$ .

The wheel slows to rest but still turns $4.0$ rad counterclockwise during the interval.

Keep the rigid-system model bounded

A rigid system keeps its shape, but different points move in different directions during rotation, so it is not generally a point object. Treat it as a single object only when its rotation is negligible for the motion being analyzed. Direction descriptions here are limited to clockwise and counterclockwise about the stated axis.

5.2 Connecting Linear and Rotational Motion

Connect radius to the path

For a point a perpendicular distance $r$ from a fixed rotation axis, radius converts angular motion into motion along the circular path. Angular displacement must be in radians.

$\begin{aligned}\Delta s&=r\Delta\theta\\[3pt]v_t&=r\omega\\[3pt]a_t&=r\alpha\end{aligned}$

Separate shared and radius-dependent quantities

For points on one rigid system	Same for every point	Increases with radius $r$
Position change	$\Delta\theta$	Arc length $\Delta s$
Rate of motion	$\omega$	Tangential speed $v_t$
Change of rate	$\alpha$	Tangential acceleration $a_t$

Compare two radii

Two points on one wheel: at one instant, $\omega=3.0\,\text{rad/s}$ and $\alpha=2.0\,\text{rad/s}^2$ . Point A is at $r_A=0.20\,\text{m}$ ; point B is at $r_B=0.50\,\text{m}$ .

$v_A=r_A\omega=(0.20)(3.0)=0.60\,\text{m/s},\quad v_B=(0.50)(3.0)=1.5\,\text{m/s}$ ,

$a_{t,A}=r_A\alpha=(0.20)(2.0)=0.40\,\text{m/s}^2,\quad a_{t,B}=(0.50)(2.0)=1.0\,\text{m/s}^2$ .

Both points share $\omega$ and $\alpha$ , but the farther point has larger tangential speed and acceleration.

Keep radians and acceleration components clear

Use radians in $\Delta s=r\Delta\theta$ . The relation $a_t=r\alpha$ gives only the tangential component of acceleration; it does not by itself describe every acceleration component of a point moving in a circle. Rotation direction here is described only as clockwise or counterclockwise about the stated axis.

5.3 Torque

Locate the rotational effect

A force exerts torque about a specified axis only through the component perpendicular to the position vector from that axis to the force's application point. Torque describes the force's rotational effect about that axis.

Trace the geometry

Identify each torque in this order:

Name the axis of rotation.
Mark where the force is applied and its direction.
Extend the force's line of action.
Measure the lever arm: the perpendicular distance from the axis to that line.

Recognize zero and nonzero torque

Force geometry about the chosen axis	Lever arm	Torque?
Line of action passes through the axis	Zero	None
Force is parallel to the position vector	Zero	None
Force has a perpendicular component away from the axis	Nonzero	Yes

Apply the geometry to a door

Door comparison: for an axis through the hinges, a push at the handle perpendicular to the door has a large lever arm and produces torque. A push directed along the door toward the hinges has a line of action through the axis, so its lever arm and torque are zero.

Measure the correct distance

The lever arm is not automatically the distance from the axis to the application point. It is the perpendicular distance to the line of action. Also, only the force component perpendicular to the position vector contributes to torque.

Preserve force location

A torque force diagram must show each force's relative magnitude and direction and where it acts relative to the chosen axis. That location information is what an ordinary free-body diagram may omit.

Use either torque form

$\tau=r_{\perp}F=rF\sin\theta$

Symbol	Meaning
$r$	Distance from axis to application point
$\theta$	Angle between $\vec r$ and $\vec F$
$r_\perp=r\sin\theta$	Perpendicular lever arm
$F$	Force magnitude

Calculate an angled force

Angled force: a $30\,\text{N}$ force acts $0.40\,\text{m}$ from an axis at $60^\circ$ to the position vector.

$\tau=rF\sin\theta=(0.40\,\text{m})(30\,\text{N})\sin60^\circ=10.4\,\text{N}\,\text{m}$ .

Equivalently, $r_\perp=(0.40\,\text{m})\sin60^\circ=0.346\,\text{m}$ and $\tau=r_\perp F$ gives the same result.

Keep angle, direction, and units bounded

The angle is between the position vector and force vector, not automatically the angle drawn against a surface. AP Physics 1 requires manipulation of torque magnitude but not the vector direction of torque. Torque uses N·m; do not relabel it as joules.

5.4 Rotational Inertia

Connect mass distribution to rotational change

Rotational inertia measures a rigid system's resistance to a change in rotation about a specified axis. It depends on both how much mass the system has and how far that mass lies from the axis.

Calculate one object's contribution

$I=mr^2$

Change for the same point mass	New rotational inertia	Effect
Double the mass	$2I$	Doubles
Double the perpendicular distance	$4I$	Quadruples

Add contributions about one axis

$I_{\mathrm{tot}}=\sum_i I_i=\sum_i m_i r_i^2$

Evaluate a two-object system

Two objects about one axis: a $2.0\,\text{kg}$ object is $0.30\,\text{m}$ from the axis and a $1.0\,\text{kg}$ object is $0.60\,\text{m}$ away.

$I_{\mathrm{tot}}=(2.0)(0.30)^2+(1.0)(0.60)^2=0.18+0.36=0.54\,\text{kg}\,\text{m}^2$ .

The lighter object contributes more because its larger distance is squared.

Keep the axis and unit explicit

Rotational inertia is not fixed until the axis is specified. Use the perpendicular distance from each mass to that same axis, not the distance between masses. The SI unit is $\text{kg}\,\text{m}^2$ .

Start from the minimum axis

Among parallel axes in a given plane, a rigid system's rotational inertia is smallest for the axis through its center of mass. Shifting to a parallel axis moves the system's mass farther away overall, so rotational inertia increases.

Relate two parallel axes

$I'=I_{\mathrm{cm}}+Md^2$

Symbol	Meaning
$I_{\mathrm{cm}}$	Inertia about a parallel axis through the center of mass
$I'$	Inertia about the shifted parallel axis
$M$	Total mass of the rigid system
$d$	Perpendicular separation between the two parallel axes

Calculate the shifted-axis value

Shifted axis: a rigid body has $I_{\mathrm{cm}}=0.50\,\text{kg}\,\text{m}^2$ and total mass $M=2.0\,\text{kg}$ . For a parallel axis $0.30\,\text{m}$ away,

$I'=0.50+(2.0)(0.30)^2=0.68\,\text{kg}\,\text{m}^2$ .

The positive $Md^2$ term confirms that the shifted-axis value exceeds the center-of-mass value.

Compare mass distributions

For equal mass and outer radius, a hoop has greater rotational inertia than a solid disk because more of the hoop's mass lies far from the axis. This comparison is qualitative unless the object's inertia formula is provided.

Apply the theorem within scope

The parallel axis theorem works only for parallel axes; $d$ is their perpendicular separation and $M$ is the total system mass. AP Physics 1 provides extended-body inertias and limits point-object calculations to five or fewer objects in two dimensions.

5.5 Rotational Equilibrium and Newton’s First Law in Rotational Form

Connect net torque to angular velocity

Newton's first law in rotational form: a system's angular velocity remains constant only when the net torque about the chosen axis is zero. Constant angular velocity can mean no rotation or steady nonzero rotation.

$\sum_i \tau_i=0$

Check torque balance

Use a force/torque diagram to check rotational equilibrium:

Choose one axis and one rotation plane.
Assign opposite signs to opposite rotational senses.
Calculate each torque about that axis.
Add the signed torques. If the sum is zero, angular velocity is constant; otherwise it changes.

Separate rotational and translational equilibrium

Condition	What remains constant?	Required net quantity
Rotational equilibrium	Angular velocity	$\sum\tau=0$
Translational equilibrium	Linear velocity	$\sum\vec F=0$
Both	Both velocities	Both sums are zero

Balance two nonzero torques

Balanced torques: take counterclockwise as positive. A perpendicular $25\,\text{N}$ force at $0.40\,\text{m}$ gives $-10\,\text{N}\,\text{m}$ , while a perpendicular $10\,\text{N}$ force at $1.0\,\text{m}$ gives $+10\,\text{N}\,\text{m}$ .

$\sum\tau=+10-10=0$ , so the angular velocity stays constant even though both torques are nonzero.

Do not confuse equilibrium with rest

Equilibrium does not require rest: zero net torque means no angular acceleration, not necessarily zero angular velocity. A system can be in rotational equilibrium without translational equilibrium, or vice versa. AP Physics 1 does not require simultaneous rotation analysis in multiple planes.

5.6 Newton’s Second Law in Rotational Form

Identify when angular velocity changes

A rigid system's angular velocity changes when the net torque about the chosen axis is nonzero. The resulting angular acceleration has the same signed rotational sense as the net torque.

Connect net torque and rotational inertia

$\alpha_{\mathrm{sys}}=\frac{\sum_i\tau_i}{I_{\mathrm{sys}}}=\frac{\tau_{\mathrm{net}}}{I_{\mathrm{sys}}}$

Quantity	Meaning	SI unit
$\tau_{\mathrm{net}}$	Signed sum of torques about the axis	$\text{N}\,\text{m}$
$I_{\mathrm{sys}}$	Rotational inertia about that axis	$\text{kg}\,\text{m}^2$
$\alpha_{\mathrm{sys}}$	Rate of change of angular velocity	$\text{rad}\,\text{s}^{-2}$

Predict proportional changes

Change	Angular acceleration response	Why
Double $\tau_{\mathrm{net}}$ at fixed $I$	Doubles	$\alpha\propto\tau_{\mathrm{net}}$
Double $I$ at fixed $\tau_{\mathrm{net}}$	Halves	$\alpha\propto1/I$
Make $\tau_{\mathrm{net}}=0$	Becomes zero	Angular velocity stays constant

Calculate angular acceleration

Rigid system: take counterclockwise as positive. If $\tau_{\mathrm{net}}=+6.0\,\text{N}\,\text{m}$ and $I_{\mathrm{sys}}=2.0\,\text{kg}\,\text{m}^2$ ,

$\alpha=\dfrac{+6.0}{2.0}=+3.0\,\text{rad}\,\text{s}^{-2}$ .

The positive result means the angular velocity changes in the counterclockwise sense under this convention.

Separate rotational and linear dynamics

Rotational analysis alone does not fully describe every rigid system. Use $\sum\tau=I\alpha$ for rotation about the chosen axis and, when the center of mass also translates, perform a separate linear analysis with $\sum\vec F=m\vec a_{\mathrm{cm}}$ . One equation does not replace the other.

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