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6.1 Rotational Kinetic Energy

Connect rotation to stored motion energy

Rotational kinetic energy is the energy a rigid system has because it rotates. It depends on the system's rotational inertia and angular speed about the same axis.

$K_{\mathrm{rot}}=\frac12 I\omega^2$

Match quantities to one axis

Symbol	Meaning	SI unit
$I$	Rotational inertia about the chosen axis	$\text{kg}\,\text{m}^2$
$\omega$	Angular velocity; its magnitude is angular speed	$\text{rad}\,\text{s}^{-1}$
$K_{\mathrm{rot}}$	Rotational kinetic energy	$\text{J}$

Recover particle kinetic energy

For one object of mass $m$ rotating a distance $r$ from a fixed axis, $I=mr^2$ and $v=r\omega$ . Therefore

$K_{\mathrm{rot}}=\tfrac12(mr^2)\omega^2=\tfrac12m(r\omega)^2=\tfrac12mv^2$ .

The rotational expression adds the translational kinetic energies of the moving parts.

Add center-of-mass and rotational motion

$K_{\mathrm{total}}=\frac12 Mv_{\mathrm{cm}}^2+\frac12 I_{\mathrm{cm}}\omega^2$

Calculate and scale the energy

Rotating system: if $I=0.80\,\text{kg}\,\text{m}^2$ and $\omega=5.0\,\text{rad}\,\text{s}^{-1}$ ,

$K_{\mathrm{rot}}=\tfrac12(0.80)(5.0)^2=10\,\text{J}$ .

Doubling angular speed would make the energy four times as large because $K_{\mathrm{rot}}\propto\omega^2$ .

Keep energy scalar

Rotational kinetic energy is a scalar: reversing the rotation changes the sign of $\omega$ but not $K_{\mathrm{rot}}$ . A stationary center of mass means the translational term is zero; the system can still have rotational kinetic energy because its parts are moving.

6.2 Torque and Work

Connect torque to energy transfer

A torque transfers energy into or out of a rigid system only while the system undergoes an angular displacement. Positive rotational work adds energy; negative rotational work removes energy.

Use the constant-torque relationship

$W=\tau\,\Delta\theta$

Quantity or condition	Meaning
$\tau$	Constant signed torque during the interval; use $\tau_{\mathrm{net}}$ for net work
$\Delta\theta$	Signed angular displacement in radians
Same rotational sense	$W>0$ : energy transferred into the system
Opposite rotational senses	$W<0$ : energy transferred out

Calculate constant-torque work

Constant torque: a net torque of $+4.0\,\text{N}\,\text{m}$ acts while a wheel turns through $+3.0\,\text{rad}$ .

$W=\tau_{\mathrm{net}}\Delta\theta=(+4.0)(+3.0)=+12\,\text{J}$ .

The positive sign means $12\,\text{J}$ of energy is transferred into the wheel.

Read work as signed graph area

Torque–angular-position graph feature	Rotational meaning
Horizontal axis	Angular position $\theta$ in radians
Vertical axis	Signed torque $\tau$ in $\text{N}\,\text{m}$
Area above the axis	Positive work
Area below the axis	Negative work
Total signed area	Net work over the interval

Calculate varying-torque work

Varying torque: suppose torque increases linearly from $0$ to $6.0\,\text{N}\,\text{m}$ while angular position changes by $4.0\,\text{rad}$ . The graph region is a triangle, so

$W=\tfrac12(4.0\,\text{rad})(6.0\,\text{N}\,\text{m})=12\,\text{J}$ .

Choose product or area correctly

Do not multiply one instantaneous torque value by the entire angular displacement when torque varies. Use signed graph area instead. In $W=\tau\Delta\theta$ , the angle is in radians; degrees must be converted first.

6.3 Angular Momentum and Angular Impulse

Choose the reference first

Angular momentum describes rotational motion relative to a specified axis or point. Changing that reference can change the rotational inertia, perpendicular geometry, and therefore the measured angular momentum.

Match the model to the motion

Physical model	Angular-momentum magnitude	Geometry
Rigid system rotating about an axis	$L=I\omega$	$I$ and $\omega$ use the same axis
Object moving relative to a point	$L=rmv\sin\theta$	$\theta$ is between $\vec r$ and $\vec v$

Calculate rigid-system momentum

Rigid system: $I=1.2\,\text{kg}\,\text{m}^2$ and $\omega=4.0\,\text{rad}\,\text{s}^{-1}$ .

$L=I\omega=(1.2)(4.0)=4.8\,\text{kg}\,\text{m}^2\,\text{s}^{-1}$ .

Calculate moving-object momentum

Moving object: $m=2.0\,\text{kg}$ , $v=3.0\,\text{m}\,\text{s}^{-1}$ , $r=0.50\,\text{m}$ , and $\theta=90^\circ$ .

$L=rmv\sin\theta=(0.50)(2.0)(3.0)\sin90^\circ=3.0\,\text{kg}\,\text{m}^2\,\text{s}^{-1}$ .

Keep the perpendicular geometry

The distance $r$ alone does not determine a moving object's angular momentum. Only the velocity component perpendicular to $\vec r$ contributes: radial motion has $\theta=0^\circ$ and therefore $L=0$ about that point.

Connect torque and duration

Angular impulse measures the effect of a torque acting over a time interval. Within a one-dimensional sign convention, it has the same signed rotational sense as the torque.

$J_{\mathrm{ang}}=\tau\,\Delta t$

Calculate constant-torque impulse

Constant torque: a signed torque of $+5.0\,\text{N}\,\text{m}$ acts for $0.40\,\text{s}$ .

$J_{\mathrm{ang}}=(+5.0)(0.40)=+2.0\,\text{N}\,\text{m}\,\text{s}$ .

Read signed torque–time area

Torque–time graph feature	Angular impulse
Area above the time axis	Positive
Area below the time axis	Negative
Total signed area	Angular impulse over the interval

Calculate varying-torque impulse

Varying torque: torque rises linearly from $0$ to $8.0\,\text{N}\,\text{m}$ over $0.50\,\text{s}$ . The triangular graph area is

$J_{\mathrm{ang}}=\tfrac12(0.50)(8.0)=2.0\,\text{N}\,\text{m}\,\text{s}$ .

Use the correct graph and method

Angular impulse uses area on a torque–time graph. Area on a torque–angular-position graph represents work instead. Use $\tau\Delta t$ only when torque is constant over the interval.

Measure final minus initial momentum

$\Delta L=L_f-L_i$

Equate impulse and momentum change

$\Delta L=J_{\mathrm{ang}}=\tau_{\mathrm{net}}\Delta t$

When rotational inertia is constant, the rotational second law produces the impulse–momentum theorem:

$\tau_{\mathrm{net}}=\dfrac{\Delta L}{\Delta t}=I\dfrac{\Delta\omega}{\Delta t}=I\alpha$ .

Multiplying by $\Delta t$ gives $\tau_{\mathrm{net}}\Delta t=\Delta L$ .

Connect slope, area, torque, and change

Graph	Operation	Result
Angular momentum $L$ vs. time $t$	Slope	Net torque $\tau_{\mathrm{net}}$
Net external torque $\tau$ vs. time $t$	Signed area	Change in angular momentum $\Delta L$

Calculate a momentum change

Constant inertia: $I=0.50\,\text{kg}\,\text{m}^2$ and angular velocity changes from $2.0$ to $6.0\,\text{rad}\,\text{s}^{-1}$ .

$\Delta L=I(\omega_f-\omega_i)=(0.50)(6.0-2.0)=2.0\,\text{kg}\,\text{m}^2\,\text{s}^{-1}$ .

The delivered angular impulse is therefore $2.0\,\text{N}\,\text{m}\,\text{s}$ .

Use one-dimensional signs only

AP Physics 1 uses one-dimensional signed conventions to manipulate angular-momentum and angular-impulse magnitudes. Full vector directions for these quantities are beyond scope. Keep one sign convention consistent when subtracting $L_i$ from $L_f$ .

6.4 Conservation of Angular Momentum

Add the parts about one axis

For a chosen rotational axis, a system's total angular momentum is the signed sum of the angular momenta of all its parts:

$L_{\mathrm{total}}=L_1+L_2+\cdots$ .

Check external angular impulse

$J_{\mathrm{ext}}=0\quad\Longrightarrow\quad L_{\mathrm{total},i}=L_{\mathrm{total},f}$

Calculate a nonrigid shape change

Shape change: a rotating person has $I_i=3.0\,\text{kg}\,\text{m}^2$ and $\omega_i=2.0\,\text{rad}\,\text{s}^{-1}$ . Pulling mass inward changes the inertia to $I_f=1.5\,\text{kg}\,\text{m}^2$ , with negligible external angular impulse.

$L_i=I_i\omega_i=(3.0)(2.0)=6.0\,\text{kg}\,\text{m}^2\,\text{s}^{-1}$

$\omega_f=\dfrac{L_i}{I_f}=\dfrac{6.0}{1.5}=4.0\,\text{rad}\,\text{s}^{-1}$ .

Track internal exchanges

Parts of the system can exchange angular momentum. Their interaction gives equal-and-opposite angular impulses, so those internal changes cancel in the system total even though each part's angular momentum may change.

Know when the total changes

Conservation does not mean angular momentum can never change. If the surroundings exert a net angular impulse, then $\Delta L_{\mathrm{system}}=J_{\mathrm{ext}}$ . Always name the system and axis before applying conservation.

Classify torque using the boundary

A torque is external only when the object exerting it lies outside the selected system. The same interaction can therefore be internal for one system choice and external for another.

Change only the selected system

Selected system	Torque from the interaction	Angular-momentum result
Both interacting objects together	Internal	Total $L$ is constant if other external torques are zero
Only one interacting object	External	That object's $L$ changes and momentum crosses the boundary

Apply both choices to coupled disks

Two disks couple on a low-friction axle. For the system containing both disks, their mutual frictional torques are internal and the total angular momentum is constant when the axle exerts negligible external torque. For disk A alone, disk B's frictional torque is external, so $L_A$ changes.

Account for boundary transfer

$\Delta L_{\mathrm{selected\ system}}=J_{\mathrm{external}}$

Do not conserve each object separately

“Angular momentum is conserved in interactions” refers to a sufficiently inclusive system. It does not mean every individual object keeps the same angular momentum. State the boundary, then test net external torque or angular impulse.

6.5 Rolling

Recognize both motions

A rolling rigid system can translate through the motion of its center of mass and rotate about its center of mass at the same time. Its total kinetic energy includes both motions.

Add the energy contributions

$K_{\mathrm{tot}}=K_{\mathrm{trans}}+K_{\mathrm{rot}}=\frac12 Mv_{\mathrm{cm}}^2+\frac12 I_{\mathrm{cm}}\omega^2$

Contribution	Quantities used
Translation	Total mass $M$ and center-of-mass speed $v_{\mathrm{cm}}$
Rotation	Inertia $I_{\mathrm{cm}}$ about the center of mass and angular speed $\omega$

Calculate each part before totaling

Example: $M=2.0\,\text{kg}$ , $v_{\mathrm{cm}}=3.0\,\text{m}\,\text{s}^{-1}$ , $I_{\mathrm{cm}}=0.18\,\text{kg}\,\text{m}^2$ , and $\omega=10\,\text{rad}\,\text{s}^{-1}$ .

$K_{\mathrm{trans}}=\tfrac12(2.0)(3.0)^2=9.0\,\text{J}$

$K_{\mathrm{rot}}=\tfrac12(0.18)(10)^2=9.0\,\text{J}$

$K_{\mathrm{tot}}=9.0+9.0=18\,\text{J}$ . The energy is split equally in this case; that is not a general rule.

Keep the decomposition consistent

Do not count only $\tfrac12Mv^2$ or only $\tfrac12I\omega^2$ when the system both translates and rotates. Use the center-of-mass axis consistently so the two terms form one valid decomposition.

Start from the contact condition

Rolling without slipping means the point touching the surface is instantaneously at rest relative to that surface. Translation and rotation are then locked by the rolling radius $r$ .

Match linear and angular quantities

Center-of-mass quantity	Rotational quantity	No-slip relationship
Displacement $\Delta x_{\mathrm{cm}}$	Angular displacement $\Delta\theta$	$\Delta x_{\mathrm{cm}}=r\Delta\theta$
Speed $v_{\mathrm{cm}}$	Angular speed $\omega$	$v_{\mathrm{cm}}=r\omega$
Acceleration $a_{\mathrm{cm}}$	Angular acceleration $\alpha$	$a_{\mathrm{cm}}=r\alpha$

Calculate center-of-mass speed

Example: a wheel of radius $0.25\,\text{m}$ rolls without slipping at $\omega=8.0\,\text{rad}\,\text{s}^{-1}$ .

$v_{\mathrm{cm}}=r\omega=(0.25)(8.0)=2.0\,\text{m}\,\text{s}^{-1}$ .

Interpret ideal friction

In the ideal no-slip case, the contact point has no displacement relative to the surface. Static friction may still set the required translation and rotation, but it does not dissipate mechanical energy at that contact.

Apply the equations only without slip

These relationships apply only while there is no slipping. Use the same physical radius and one consistent sign convention; the equations shown here express magnitudes when direction is already understood.

Recognize that the coupling has failed

While an object is slipping, its contact point moves relative to the surface. The no-slip coupling fails, so $v_{\mathrm{cm}}$ and $r\omega$ cannot be set equal.

Compare translation-too-fast and spin-too-fast

Forward-rolling state	Contact point slips	Kinetic friction on object	Qualitative change
$v_{\mathrm{cm}}>r\omega$	Forward	Backward	$v_{\mathrm{cm}}$ decreases; spin rate increases
$v_{\mathrm{cm}}<r\omega$	Backward	Forward	$v_{\mathrm{cm}}$ increases; spin rate decreases

Follow force and torque together

Kinetic friction opposes the relative slipping at the contact point. Its force changes the center-of-mass motion, and its torque changes the rotation, tending to reduce the mismatch between $v_{\mathrm{cm}}$ and $r\omega$ .

Account for dissipation

Because the point where kinetic friction acts moves relative to the surface, mechanical energy is dissipated. Momentum and angular-momentum changes must still be analyzed using the chosen system and external forces or torques.

Keep the analysis qualitative

AP Physics 1 expects a qualitative explanation of linear and angular changes while slipping, not a precise general mathematical relationship between them. Rolling friction is also outside this Topic's scope.

6.6 Motion of Orbiting Satellites

Choose the two-body model

Model a satellite of mass $m$ interacting only with a much more massive central body of mass $M$ . The central body's motion is then negligible, while gravity transfers energy between kinetic and gravitational potential forms without changing the system's total mechanical energy.

Compare circular and elliptical orbits

Quantity	Circular orbit	Elliptical orbit
System total mechanical energy	Constant	Constant
Satellite angular momentum	Constant	Constant
Gravitational potential energy	Constant	Changes with radius
Satellite kinetic energy and speed	Constant	Change around the orbit

Use the gravitational energy reference

$\begin{gathered}U_g=-\frac{GMm}{r}\\(U_g=0\text{ at }r\to\infty)\end{gathered}$

Trace energy around an ellipse

In an elliptical orbit, moving closer makes $U_g$ more negative. Because $E=K+U_g$ stays constant, $K$ increases and the satellite moves faster. Moving farther away reverses the exchange. Angular momentum remains constant throughout.

Derive the minimum escape speed

Minimum escape condition: choose the satellite–central-body system. At launch radius $r$ , set its total mechanical energy to zero so that at infinite distance both $U_g$ and the final speed approach zero.

$0=\tfrac12mv_{\mathrm{esc}}^2-\dfrac{GMm}{r}$

$\tfrac12mv_{\mathrm{esc}}^2=\dfrac{GMm}{r}$

$v_{\mathrm{esc}}=\sqrt{\dfrac{2GM}{r}}$ .

The satellite mass cancels; for the same central body, doubling $r$ reduces escape speed by a factor of $\sqrt2$ .

Interpret escape correctly

Escape velocity is the initial speed for zero total mechanical energy under gravity alone. It is not a speed maintained during escape: at the minimum value, the satellite slows toward zero speed only as $r$ approaches infinity. Atmosphere, thrust, and other forces are outside this model.

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