Unit 5: Rotation

Angular Variables and Rotational Kinematics

Rotation is the language used to describe the motion of a rigid object (one that does not deform) turning about an axis. Rotational motion has direct analogs to linear motion: position becomes angle, velocity becomes angular speed, and acceleration becomes angular acceleration. Many AP Physics C systems combine translation and rotation (wheels, pulleys, gears, planets, motors), so success often depends on switching between angular and linear descriptions without mixing variables.

Angular position, displacement, rigid bodies, and sign conventions

The fundamental rotational “position” variable is angular position, measured in radians. A change in angular position is angular displacement.

$\Delta\theta=\theta_f-\theta_i$

A useful way to visualize angular variables is to mark several dots along a radius on a disk and use that radius as a reference line. If the disk rotates as a rigid body, all points on that radial line share the same angular displacement in any time interval. This is the defining feature of rigid-body rotation: different points can have different linear motion, but the same angular motion.

Sign conventions matter. In planar problems, the standard convention is counterclockwise positive when looking along the positive axis direction. In AP-style fixed-axis rotation, quantities like angular position, angular velocity, angular acceleration, torque, and angular momentum are often treated as signed scalars, but conceptually their directions are along the axis via the right-hand rule.

Radians and the arc-length connection

Radians are defined so that arc length relates simply to radius and angle:

$s=r\theta$

This relationship is the bridge between angular and linear descriptions. It only has this simple form when the angle is in radians.

Angular velocity and angular acceleration (average and instantaneous)

Average angular velocity is defined by

$\omega_{\text{avg}}=\frac{\Delta\theta}{\Delta t}$

Instantaneous angular velocity is

$\omega=\frac{d\theta}{dt}$

Average angular acceleration is

$\alpha_{\text{avg}}=\frac{\Delta\omega}{\Delta t}$

Instantaneous angular acceleration is

$\alpha=\frac{d\omega}{dt}=\frac{d^2\theta}{dt^2}$

A rigid body rotating about a fixed axis has a single shared $\omega$ and $\alpha$, even though different points on the object have different linear speeds and linear accelerations.

Relating angular and linear kinematics for a point on the object

From the radian definition, dividing $s=r\theta$ by time gives

$v=r\omega$

So points farther from the axis move faster linearly.

Differentiating $v=r\omega$ (with constant $r$) gives the tangential acceleration relationship:

$a_t=r\alpha$

A point also has centripetal (radial) acceleration due to changing direction:

$a_c=r\omega^2$

A common confusion is to mix up $a_t$ and $a_c$. Tangential acceleration changes speed along the path; centripetal acceleration points toward the axis and changes direction.

Constant angular acceleration and the “Big Five” rotational kinematics equations

When angular acceleration is constant, rotational kinematics mirrors constant-acceleration linear kinematics, with $x$ replaced by $\theta$, $v$ by $\omega$, and $a$ by $\alpha$.

The core equations are:

$\omega=\omega_0+\alpha t$

$\theta=\theta_0+\omega_0 t+\frac{1}{2}\alpha t^2$

$\omega^2=\omega_0^2+2\alpha\left(\theta-\theta_0\right)$

Two additional commonly used constant-$\alpha$ forms are:

$\theta-\theta_0=\frac{1}{2}\left(\omega+\omega_0\right)t$

$\theta-\theta_0=\omega t-\frac{1}{2}\alpha t^2$

These five are often called the “Big Five” for rotational motion. A helpful way to remember them is that each equation is missing one of the five variables $\theta$, $\omega$, $\omega_0$, $\alpha$, or $t$.

Worked example: turntable point acceleration

A turntable starts from rest and reaches angular speed 6 rad/s in 3 s with constant angular acceleration. Find angular acceleration. Then find the tangential and centripetal accelerations of a point 0.20 m from the center at that moment.

1) Angular acceleration:

$\alpha=\frac{\omega-\omega_0}{t}=\frac{6-0}{3}=2\ \text{rad/s}^2$

2) Tangential acceleration:

$a_t=r\alpha=0.20(2)=0.40\ \text{m/s}^2$

3) Centripetal acceleration:

$a_c=r\omega^2=0.20\left(6^2\right)=7.2\ \text{m/s}^2$

This illustrates why $a_c$ can be much larger than $a_t$: $a_c$ scales with the square of angular speed.

Sample problem: average angular acceleration

The angular velocity of a rotating disk increases from 2 rad/s to 5 rad/s in 0.5 s. The average angular acceleration is

$\alpha_{\text{avg}}=\frac{\Delta\omega}{\Delta t}=\frac{5-2}{0.5}=6\ \text{rad/s}^2$

Exam Focus

• Typical question patterns:
  • Convert between angular and linear variables using $s=r\theta$ and $v=r\omega$.
  • Use constant-$\alpha$ kinematics to connect $\omega$, $\alpha$, $\theta$, and $t$.
  • Compute accelerations of points on a rotating rigid body and separate tangential vs centripetal components.
• Common mistakes:
  • Using degrees in formulas like $s=r\theta$ without converting to radians.
  • Treating centripetal acceleration as $a_c=r\alpha^2$ (incorrect); it is $a_c=r\omega^2$.
  • Forgetting that different points have different $v$ and $a$ even though they share the same $\omega$ and $\alpha$.

Torque: The Rotational Cause of Acceleration

If force causes linear acceleration, torque causes angular acceleration. Torque measures how effective a force is at producing rotation about a chosen axis, so the axis choice is part of the problem: the same force can produce different torques about different pivots.

What torque is (vector form, magnitude form, and lever arm)

Torque about an origin is defined by the cross product

$\boldsymbol{\tau}=\mathbf{r}\times\mathbf{F}$

The magnitude is

$\tau=rF\sin\phi$

where $\phi$ is the angle between $\mathbf{r}$ (from pivot to point of application) and $\mathbf{F}$.

An equivalent and often faster method uses the moment arm (lever arm), the perpendicular distance from the axis to the force’s line of action:

$\tau=Fr_\perp$

This encodes two key ideas: increasing force increases torque, and applying the force farther from the axis increases torque.

Units, direction, and the right-hand rule

Torque units are newton-meter (N m). In some engineering contexts you may also see pound-feet (lb-ft).

Torque is a vector quantity. Its direction is perpendicular to the plane of rotation and follows the right-hand rule. In many AP planar problems, you simplify this by treating torque as a signed scalar: counterclockwise positive and clockwise negative.

Torque is central in many applications (engines, motors, machines). For example, engine performance discussions often emphasize torque output and relate it to power (and sometimes report power in horsepower). Torque can be increased by increasing the applied force or by increasing the moment arm.

Choosing signs in planar problems

A common misconception is to assign the torque sign based on the force direction alone. The torque direction depends on both $\mathbf{r}$ and $\mathbf{F}$ (i.e., the tendency to rotate about the pivot).

Net torque and equilibrium preview

Net torque is the sum of individual torques:

$\tau_{\text{net}}=\sum\tau_i$

If net torque about a fixed axis is zero, angular acceleration is zero. That situation corresponds to either static equilibrium (at rest) or dynamic equilibrium (rotating with constant angular velocity).

Worked example: torque from an angled push

You push on a door at its handle, 0.80 m from the hinges, with force magnitude 40 N at an angle 30 degrees to the door (in the plane of the door). The angle between $\mathbf{r}$ and $\mathbf{F}$ is 30 degrees, so

$\tau=rF\sin\phi=0.80(40)\sin 30^\circ=16\ \text{N m}$

The sign depends on whether the push tends to rotate the door counterclockwise (positive) or clockwise (negative).

Sample problem: torque from a rope on a pulley

A student pulls down with a force of 40 N on a rope that winds around a pulley of radius 5 cm. Assuming the tension is tangential (perpendicular to the radius), the torque magnitude is

$\tau=rF=0.050(40)=2.0\ \text{N m}$

Exam Focus

• Typical question patterns:
  • Compute torque using $\tau=rF\sin\phi$ or $\tau=Fr_\perp$.
  • Determine whether a force produces zero torque (line of action through the pivot).
  • Combine multiple torques with correct sign to find $\tau_{\text{net}}$.
• Common mistakes:
  • Using $\tau=rF$ without the sine factor when the force is not perpendicular.
  • Measuring $r$ from the wrong point (it must be from the axis/pivot).
  • Confusing force components: the component perpendicular to $\mathbf{r}$ is what matters.

Moment of Inertia: Rotational “Mass” and How to Compute It

In linear motion, mass tells you how strongly an object resists acceleration for a given force. In rotation, the analog is moment of inertia (rotational inertia) $I$, which measures resistance to angular acceleration for a given torque. It depends not just on how much mass there is, but on how that mass is distributed relative to the axis.

Definition for point masses and continuous mass distributions

For discrete point masses,

$I=\sum m_i r_i^2$

For a continuous object,

$I=\int r^2\,dm$

The units of moment of inertia are kg m^2. Axis choice matters dramatically because of the square on distance.

Standard moments of inertia you should know

Common AP results (each about the stated axis):

Thin hoop (ring), radius $R$, about central axis perpendicular to plane:

$I=MR^2$

Solid disk (or solid cylinder), radius $R$, about central axis:

$I=\frac{1}{2}MR^2$

Solid sphere, radius $R$, about a diameter:

$I=\frac{2}{5}MR^2$

Thin rod, length $L$, about its center (axis perpendicular to rod):

$I=\frac{1}{12}ML^2$

Thin rod, length $L$, about one end (axis perpendicular to rod):

$I=\frac{1}{3}ML^2$

An important comparison intuition is that “more mass farther out” means larger $I$, so a hoop has larger $I$ than a disk with the same $M$ and $R$.

Parallel-axis theorem (shifting the axis)

If you know the moment of inertia about an axis through the center of mass, you can shift to a parallel axis using

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

where $d$ is the perpendicular distance between the two parallel axes.

This theorem is exact for rigid bodies regardless of whether the mass distribution is uniform, and it does not require $d$ to be “small.” External forces like friction and air resistance do not change the geometric relationship that the theorem expresses, though they may matter for the motion in a dynamics problem.

Parallel-axis theorem is widely used in physics and engineering: it helps compute moments of inertia for complex objects, informs machine and structure design (stability and resistance to rotation), appears in celestial mechanics (planets and other bodies), and is used in contexts like molecular rotational motion.

Perpendicular-axis theorem (planar objects)

For a flat lamina in the $xy$-plane, about perpendicular axes through the same point,

$I_z=I_x+I_y$

Worked example: rod about an end via parallel-axis theorem

A uniform thin rod has length $L$ and mass $M$. Given

$I_{\text{cm}}=\frac{1}{12}ML^2$

find the moment of inertia about an axis through one end, perpendicular to the rod. The center of mass is at distance $d=L/2$ from the end, so

$I=I_{\text{cm}}+Md^2=\frac{1}{12}ML^2+M\left(\frac{L}{2}\right)^2=\frac{1}{3}ML^2$

Sample problem: three beads on a rod

Three beads, each of mass $m$, are arranged along a rod of negligible mass and length $L$. Find the rotational inertia about (i) an axis through the center bead and (ii) an axis through one end bead.

(i) Axis through the center bead: the left and right beads are each at distance $L/2$, and the center bead is at zero distance.

$I=m\left(\frac{L}{2}\right)^2+m\left(\frac{L}{2}\right)^2=\frac{1}{2}mL^2$

(ii) Axis through an end bead: distances are 0, $L/2$, and $L$.

$I=m(0)^2+m\left(\frac{L}{2}\right)^2+m(L)^2=\frac{5}{4}mL^2$

Exam Focus

• Typical question patterns:
  • Use a standard $I$ and apply the parallel-axis theorem to a new pivot.
  • Compare rotational behavior of different shapes (hoop vs disk) using $I$.
  • Identify how changing axis location changes $I$ and therefore angular acceleration.
• Common mistakes:
  • Plugging the wrong $d$ into $I=I_{\text{cm}}+Md^2$ (it is axis-to-axis distance).
  • Using a formula for the wrong axis (for a rod, “about center” vs “about end” is a classic trap).
  • Thinking $I$ depends only on mass; it depends strongly on geometry and axis choice.

Rotational Dynamics: Newton’s Second Law for Rotation

Rotational dynamics connects torque, inertia, and angular acceleration the way linear dynamics connects force, mass, and linear acceleration. This is where many multi-step AP problems live, especially pulley systems, rotating platforms, and systems with strings.

The rotational form of Newton’s second law

For a rigid body rotating about a fixed axis,

$\tau_{\text{net}}=I\alpha$

The analogy to translation is direct:

How forces produce torque in dynamics problems

In many systems, you do not “apply a torque” directly; instead, forces like tension, friction, the normal force, and weight act at lever arms and create torques. A reliable workflow is:

1) Draw a free-body diagram for translating parts.
2) Draw a torque diagram (rotational free-body diagram) for rotating parts.
3) Use kinematic constraints to connect linear and angular variables.

When a tension force acts tangentially at radius $R$ (as in many pulley problems), the torque magnitude is simply $\tau=TR$.

Constraint relations: when translation and rotation are linked

If a string does not slip on a pulley of radius $R$, then the string’s linear acceleration equals the tangential acceleration of the rim:

$a=\alpha R$

Similarly,

$v=\omega R$

These are kinematic constraints from geometry, not extra force laws.

Worked example: hanging mass and rotating disk pulley

A mass $m$ hangs from a light string wrapped around a solid disk pulley (mass $M$, radius $R$). The string does not slip. Find the acceleration of the falling mass.

Translation of the mass (downward positive):

$mg-T=ma$

Rotation of the pulley: tension applies torque $\tau=TR$. For a solid disk,

$I=\frac{1}{2}MR^2$

Apply rotational dynamics:

$TR=I\alpha$

No slip gives $\alpha=a/R$, so

$TR=I\frac{a}{R}$

$T=\frac{Ia}{R^2}$

Substitute $I=\frac{1}{2}MR^2$:

$T=\frac{1}{2}Ma$

Combine with $mg-T=ma$:

$mg-\frac{1}{2}Ma=ma$

$a=\frac{mg}{m+\frac{1}{2}M}$

The pulley’s rotational inertia effectively adds inertia to the system, reducing the acceleration below $g$.

Exam Focus

• Typical question patterns:
  • Pulley systems where you must write both $\sum F=ma$ and $\sum\tau=I\alpha$ with a no-slip constraint.
  • Determine tension differences on either side of a massive pulley.
  • Compare accelerations for different pulley shapes using different $I$.
• Common mistakes:
  • Treating tension as the same on both sides of a massive pulley (it is only equal for a massless pulley).
  • Forgetting to apply $a=\alpha R$ (or applying it with the wrong radius).
  • Mixing up which forces create torque about the axis (forces through the axis create zero torque).

Work, Energy, and Power in Rotational Motion

Energy methods are often the fastest route in rotation problems, especially when forces are conservative or when you care about speeds rather than forces. Rotational energy also reveals the meaning of moment of inertia: it determines how much energy is stored for a given angular speed.

Rotational kinetic energy and total kinetic energy of a rigid body

A rigid body rotating about a fixed axis has rotational kinetic energy

$K_{\text{rot}}=\frac{1}{2}I\omega^2$

A rigid body that both translates and rotates can have its total kinetic energy written as

$K=\frac{1}{2}Mv_{\text{cm}}^2+\frac{1}{2}I_{\text{cm}}\omega^2$

This decomposition is essential for rolling motion.

Work done by a torque

The rotational analog of work is

$W=\int\tau\,d\theta$

If torque is constant,

$W=\tau\Delta\theta$

Radians are crucial here so units come out in joules.

Rotational power

Rotational power is

$P=\tau\omega$

This is why torque is discussed alongside engine power output.

Energy conservation with rotation

When only conservative forces do work,

$K_i+U_i=K_f+U_f$

but remember that $K$ may include translational and rotational parts.

A common misconception is that a rolling object “has less energy because some is rotational.” The total mechanical energy change is still set by gravity (or other conservative forces); rotation changes how the energy is partitioned and therefore changes the speed.

Worked example: spinning disk from a constant torque

A disk with moment of inertia $I$ starts from rest. A constant torque $\tau$ acts through an angular displacement $\Delta\theta$. Work done becomes rotational kinetic energy:

$\tau\Delta\theta=\frac{1}{2}I\omega^2$

So the final angular speed is

$\omega=\sqrt{\frac{2\tau\Delta\theta}{I}}$

This approach avoids solving for time.

Exam Focus

• Typical question patterns:
  • Use energy to find speed after falling or rolling without solving for tensions or friction forces.
  • Compute work from torque and relate it to $\Delta K_{\text{rot}}$.
  • Use $P=\tau\omega$ to connect power, torque, and rotation rate.
• Common mistakes:
  • Forgetting rotational kinetic energy when an object rotates (using only translational kinetic energy).
  • Using degrees in $W=\tau\Delta\theta$ instead of radians.
  • Assuming friction always dissipates energy; static friction in pure rolling often does zero work.

Angular Momentum and Its Conservation

Angular momentum is the rotational analog of linear momentum, but it depends on the choice of origin. It becomes a powerful tool whenever external torque about a point is zero or negligible, especially in collisions, collapsing systems (like a skater pulling in arms), or rotating platforms.

Angular momentum for a particle

For a particle with momentum $\mathbf{p}=m\mathbf{v}$, angular momentum about an origin is

$\mathbf{L}=\mathbf{r}\times\mathbf{p}$

Magnitude:

$L=rp\sin\phi$

where $\phi$ is the angle between $\mathbf{r}$ and $\mathbf{p}$.

Angular momentum for a rigid body (fixed axis)

For a rigid body rotating about a fixed axis,

$L=I\omega$

Torque and angular momentum (rotational analog of Newton’s second law for momentum)

The rotational analog of

$F_{\text{net}}=\frac{dp}{dt}$

is

$\boldsymbol{\tau}_{\text{net}}=\frac{d\mathbf{L}}{dt}$

So if net external torque about a point is zero, angular momentum about that point is constant.

Conservation of angular momentum

If

$\tau_{\text{ext}}=0$

then

$L_i=L_f$

For fixed-axis rigid bodies where $L=I\omega$,

$I_i\omega_i=I_f\omega_f$

This explains why reducing $I$ (pulling mass inward) increases $\omega$.

Worked example: skater pulling in arms

With negligible external torque,

$I_i\omega_i=I_f\omega_f$

so

$\omega_f=\frac{I_i}{I_f}\omega_i$

Sample problem: child walking inward on a merry-go-round

A child of mass 30 kg stands at the edge of a small merry-go-round rotating at 1 rad/s. The merry-go-round is a disk of radius 2.5 m and mass 100 kg. The child walks toward the center and stops 0.5 m from the center. Ignoring friction, angular momentum is conserved.

Treat the child as a point mass. Disk inertia:

$I_{\text{disk}}=\frac{1}{2}MR^2$

Initial total inertia:

$I_i=\frac{1}{2}MR^2+mR^2$

Final total inertia:

$I_f=\frac{1}{2}MR^2+m\left(0.5\right)^2$

Numerically,

$I_{\text{disk}}=\frac{1}{2}(100)\left(2.5^2\right)=312.5\ \text{kg m}^2$

$I_i=312.5+30\left(2.5^2\right)=312.5+187.5=500\ \text{kg m}^2$

$I_f=312.5+30\left(0.5^2\right)=312.5+7.5=320\ \text{kg m}^2$

Conservation gives

$I_i\omega_i=I_f\omega_f$

so

$\omega_f=\frac{I_i}{I_f}\omega_i=\frac{500}{320}(1)=1.5625\ \text{rad/s}$

Angular speed increases as moment of inertia decreases.

Rotational kinetic energy is not necessarily conserved

Even when angular momentum is conserved, rotational kinetic energy can change because internal work is done. For fixed-axis rotation,

$K=\frac{L^2}{2I}$

So if $L$ is constant and $I$ decreases, $K$ increases (energy supplied by internal mechanisms, like muscles).

Exam Focus

• Typical question patterns:
  • Use $\tau_{\text{ext}}=0$ to justify angular momentum conservation and solve for final angular speed.
  • Collisions involving rotation (a mass sticks to a rotating disk, a projectile hits a rod pivot).
  • Decide whether to conserve energy, angular momentum, or both (often angular momentum during a brief collision).
• Common mistakes:
  • Conserving angular momentum about the wrong point (choose the pivot or a point with zero external torque).
  • Assuming energy is conserved in inelastic rotational collisions (it is not).
  • Writing $L=I\omega$ in situations that are not fixed-axis rotation about the chosen origin (use particle definition when needed).

Rolling Motion Without Slipping

Rolling is where translation and rotation merge. AP problems often test whether you can correctly relate the motion of the center of mass to the rotation about the center.

What “rolling without slipping” means

For pure rolling, the point of contact with the ground is instantaneously at rest relative to the ground. Equivalently, the motion can be viewed as translation of the center of mass plus rotation about the center of mass, and at each instant the object can also be thought of as rotating about the contact point.

For a rolling object of radius $R$,

$v_{\text{cm}}=\omega R$

and

$a_{\text{cm}}=\alpha R$

A key speed insight: only the center of mass has speed $v_{\text{cm}}$. Rim points can have speeds ranging from 0 (contact point) up to $2v_{\text{cm}}$ (top point) relative to the ground.

Energy of a rolling object

Total kinetic energy for rolling without slipping is

$K=\frac{1}{2}Mv_{\text{cm}}^2+\frac{1}{2}I_{\text{cm}}\omega^2$

Using $v_{\text{cm}}=\omega R$,

$K=\frac{1}{2}Mv_{\text{cm}}^2+\frac{1}{2}I_{\text{cm}}\frac{v_{\text{cm}}^2}{R^2}$

This is why objects with larger $I_{\text{cm}}$ roll more slowly down a ramp from the same height.

Rolling down an incline (acceleration result)

For rolling down an incline of angle $\beta$ without slipping,

$a_{\text{cm}}=\frac{g\sin\beta}{1+\frac{I_{\text{cm}}}{MR^2}}$

For a hoop,

$a_{\text{cm}}=\frac{g\sin\beta}{2}$

For a solid disk (or solid cylinder),

$a_{\text{cm}}=\frac{2}{3}g\sin\beta$

So the disk accelerates faster than the hoop.

The role of friction in rolling

Rolling without slipping usually requires static friction, but static friction does not necessarily do negative work on the rolling object. In ideal rigid-body pure rolling on a rigid surface, the contact point does not slide, so the friction force often does zero work on the object as a whole. Friction is still essential because it provides the torque needed for angular acceleration.

The direction of static friction depends on the tendency to slip. A freely rolling object down a ramp typically has static friction up the ramp to spin it up. A driven wheel can have static friction forward, propelling the vehicle.

Worked example: comparing speeds after rolling down a height

A hoop and a solid disk, both starting from rest, roll without slipping down the same vertical height $h$. Use energy:

$Mgh=\frac{1}{2}Mv^2+\frac{1}{2}I_{\text{cm}}\omega^2$

With $\omega=v/R$,

$Mgh=\frac{1}{2}Mv^2+\frac{1}{2}I_{\text{cm}}\frac{v^2}{R^2}$

Solve:

$v^2=\frac{2gh}{1+\frac{I_{\text{cm}}}{MR^2}}$

Hoop: $I_{\text{cm}}=MR^2$ gives

$v^2=gh$

Solid disk: $I_{\text{cm}}=\frac{1}{2}MR^2$ gives

$v^2=\frac{4}{3}gh$

So the disk reaches a higher speed.

Sample problem: cylinder rolling down an incline (energy + kinematics check)

A cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane of height $h$ and length $L$, with incline angle $\theta$ (relative to the horizontal). Starting from rest, find the speed of the center of mass at the bottom.

Energy conservation gives

$Mgh=\frac{1}{2}Mv^2+\frac{1}{2}I_{\text{cm}}\omega^2$

For a solid cylinder,

$I_{\text{cm}}=\frac{1}{2}MR^2$

and $\omega=v/R$, so

$Mgh=\frac{1}{2}Mv^2+\frac{1}{4}Mv^2=\frac{3}{4}Mv^2$

Thus,

$v=\sqrt{\frac{4}{3}gh}$

You can verify the same result using dynamics plus kinematics: for a solid cylinder rolling without slipping,

$a=\frac{2}{3}g\sin\theta$

Then using a constant-acceleration kinematics relation along the ramp,

$v^2=2aL=2\left(\frac{2}{3}g\sin\theta\right)L=\frac{4}{3}gL\sin\theta$

Since $h=L\sin\theta$, this becomes $v^2=\frac{4}{3}gh$, matching the energy approach.

Exam Focus

• Typical question patterns:
  • Use $v_{\text{cm}}=\omega R$ to connect translation and rotation for rolling.
  • Use energy conservation down a ramp including both translational and rotational kinetic energy.
  • Compare accelerations and speeds for different shapes via different $I_{\text{cm}}$.
• Common mistakes:
  • Treating the contact point as moving at speed $v_{\text{cm}}$ (it is instantaneously at rest for pure rolling).
  • Ignoring rotational kinetic energy and getting the same final speed as sliding.
  • Assuming friction always dissipates energy; static friction can be non-dissipative in pure rolling.

Rotational Statics and Torque Equilibrium (Rigid Bodies at Rest)

Not all rotation problems involve spinning. Many involve objects that are not rotating but could rotate if torques were unbalanced: beams, ladders, signs, cranes, and pivoted rods. These are equilibrium problems and require both force balance and torque balance.

Static vs dynamic rotational equilibrium

Equilibrium in rotational motion occurs when an object is not rotating or is rotating at a constant angular velocity, which means net torque is zero.

• Static equilibrium: the object is at rest and net torque is zero.
• Dynamic equilibrium: the object rotates with constant angular velocity and net torque is zero.

Conditions for static equilibrium

For a rigid body in static equilibrium (no linear acceleration and no angular acceleration):

$\sum F_x=0$

$\sum F_y=0$

and, about any point in the plane you choose,

$\sum\tau=0$

You can take torques about any point, and choosing the pivot strategically to eliminate unknown forces is one of the most important skills in these problems.

Why torque equilibrium is necessary

An object can have zero net force but still rotate (a couple: equal and opposite forces separated by a distance). Torque balance is what prevents rotation.

Center of mass and weight location

For a uniform object, weight acts at its geometric center. For a non-uniform object, the total weight acts at the center of mass. In statics, you typically model the entire weight $Mg$ as applied at that location.

Worked example: uniform beam with a hanging load

A uniform horizontal beam of length $L$ and mass $M$ is supported at its left end by a hinge and at its right end by a vertical cable. A mass $m$ hangs from the beam at distance $x$ from the hinge. Find the cable tension $T$.

Take torques about the hinge (counterclockwise positive):

• Cable tension at the right end: $+TL$
• Hanging mass: $-mgx$
• Beam weight at its center: $-Mg\left(\frac{L}{2}\right)$

Torque equilibrium:

$TL-mgx-Mg\left(\frac{L}{2}\right)=0$

Solve:

$T=\frac{mgx+Mg\left(\frac{L}{2}\right)}{L}$

Force balance can then be used to find hinge forces if needed.

Tipping and stability

Many equilibrium questions reduce to whether the line of action of weight falls within the base of support. When an object is about to tip, the normal force effectively shifts to the edge, and torque balance about that edge gives the tipping condition.

Exam Focus

• Typical question patterns:
  • Beams, ladders, and rods with multiple forces: use $\sum F=0$ and $\sum\tau=0$.
  • Choose a pivot to eliminate unknown hinge forces and solve for a tension or normal force.
  • Determine maximum load before tipping or slipping (often involves friction plus torque).
• Common mistakes:
  • Writing torque with the wrong lever arm (must be perpendicular distance to the force’s line of action).
  • Forgetting the beam’s own weight (or placing it at the wrong point).
  • Using torque equilibrium about multiple points incorrectly (each torque equation must include all torques about the chosen pivot).

Putting It Together: Multi-Concept Rotation Problems (How to Think Through Them)

Many Unit 5 questions are not about a single isolated idea. They mix torque, energy, angular momentum, rolling constraints, and sometimes collisions. The key is deciding which principle applies during which time interval.

Continuous motion vs brief events

A reliable organizing strategy is to classify the interaction:

1) Continuous motion under known forces or torques (block unwinding a pulley, rolling down a ramp). Use $\sum F=ma$, $\sum\tau=I\alpha$, constraints like $a=\alpha R$, or energy if conservative forces dominate.

2) Brief interactions (collisions, sticking, grabbing, stepping onto a rotating platform). During the brief event, external impulses or external torques about a chosen axis may be negligible, making angular momentum conservation the best tool. Mechanical energy often is not conserved in sticking (inelastic) events.

Example: inelastic rotational collision followed by rotation

A small mass $m$ moving at speed $v$ hits and sticks to the rim of a disk of radius $R$ and moment of inertia $I_{\text{disk}}$ about its center. The disk is initially at rest and mounted on a frictionless axle. Find the angular speed immediately after impact.

During the short collision, take angular momentum about the axle. Initial angular momentum of the particle about the center is

$L_i=mvR$

Final angular momentum is

$L_f=I_{\text{total}}\omega$

where

$I_{\text{total}}=I_{\text{disk}}+mR^2$

Conservation gives

$mvR=\left(I_{\text{disk}}+mR^2\right)\omega$

So

$\omega=\frac{mvR}{I_{\text{disk}}+mR^2}$

You do not set kinetic energy before equal to kinetic energy after because the sticking collision dissipates energy.

Example: when energy beats forces

If a rolling object goes down a ramp from height $h$, you can solve with forces and torques, but energy is often quicker when you only need speed. If you need the friction force magnitude, dynamics is usually necessary because energy alone does not directly give static friction.

Exam Focus

• Typical question patterns:
  • Two-stage problems: (i) collision where angular momentum is conserved, then (ii) rotation where energy or rotational kinematics applies.
  • Decide between dynamics and energy for rolling and pulley systems depending on what is asked (speed vs force).
  • Combine translation and rotation for systems with both moving centers of mass and spinning parts.
• Common mistakes:
  • Conserving angular momentum when external torque about the chosen axis is not zero.
  • Conserving mechanical energy across an inelastic event (sticking, slipping with kinetic friction).
  • Mixing up which $I$ to use (about the axle vs about the center of mass) or forgetting to add $mR^2$ for a stuck mass.