Unit 5 Rotation: Understanding Torque, Rotational Dynamics, and Rolling (AP Physics C: Mechanics)

Torque

What torque is (and why it’s the real “cause” of rotation)

Torque is the rotational “push” that tends to change an object’s rotational motion. In linear motion, forces cause changes in translational motion. In rotational motion, forces only matter in the way they are applied relative to an axis of rotation—this is what torque captures.

A key idea: the same force can produce very different rotational effects depending on where it’s applied and which direction it points. Pushing on a door near the hinges is frustratingly ineffective; pushing at the handle is easy. Torque is the quantity that explains this.

How torque depends on force, distance, and angle

For a force F applied at a position vector r measured from the axis (or pivot point) to the point of application, the magnitude of the torque is

\tau = rF\sin\theta

where \theta is the angle between r and F.

• The factor \sin\theta is what makes “perpendicular push” most effective. If you push directly toward or away from the pivot, \sin\theta=0 and you get no torque.
• The factor r is the lever-arm effect—farther from the axis means a larger turning effect.

You will often see torque written using the lever arm (also called the moment arm) r_\perp, the perpendicular distance from the axis to the line of action of the force:

\tau = r_\perp F

This is often the fastest way to compute torque on AP problems: draw the line of action of F, then measure the perpendicular distance from the axis.

Direction/sign of torque: In AP Physics C, torque is commonly treated as positive or negative depending on whether it tends to cause counterclockwise (CCW) or clockwise (CW) rotation about the chosen axis (right-hand rule). Your sign convention must be consistent.

Net torque and the idea of choosing an axis wisely

Real objects usually have multiple forces acting. The net torque about a chosen axis is the sum of the individual torques (with signs):

\sum \tau = \tau_1 + \tau_2 + \cdots

A critical strategic skill is choosing an axis that simplifies the torque sum. If you choose the axis through a point where an unknown force acts, that force produces zero torque about that axis (because r=0), so it disappears from the torque equation. This is especially powerful in equilibrium and rigid-body problems.

Torque from gravity: why the center of mass matters

For uniform gravity, the weight force effectively acts at the center of mass. So the gravitational torque about a pivot can be found by treating mg as applied at the COM.

This is why a meterstick balanced on a pivot behaves like all its mass is concentrated at its center.

Worked example 1: torque on a door

A door is 0.90 m wide. You push with 25 N at the handle, perpendicular to the door.

Use \tau = rF\sin\theta with r=0.90\text{ m} and \theta=90^\circ:

\tau = (0.90)(25)\sin 90^\circ

\tau = 22.5\text{ N·m}

If instead you push at 0.10 m from the hinge with the same force and direction:

\tau = (0.10)(25)\sin 90^\circ = 2.5\text{ N·m}

Same force, 9 times less torque.

Worked example 2: torque from an angled force

A wrench of length 0.20 m experiences a 60 N force applied at its end at an angle of 30° above the wrench (so the force is not perpendicular to the handle).

Here r=0.20 m and \theta=30^\circ:

\tau = (0.20)(60)\sin 30^\circ

\tau = (0.20)(60)(0.5) = 6.0\text{ N·m}

A common mistake is using \cos\theta here—torque depends on the perpendicular component, which corresponds to \sin\theta when \theta is defined between r and F.

Exam Focus

• Typical question patterns
  • Compute torques about a pivot using lever arms and determine the direction (CW vs CCW).
  • Use torque balance to find unknown forces/angles, often by choosing a smart axis.
  • Determine where to apply a force to maximize torque or achieve equilibrium.
• Common mistakes
  • Using the wrong angle in \tau = rF\sin\theta (angle must be between r and F).
  • Forgetting sign conventions (adding magnitudes instead of signed torques).
  • Measuring r to the point of application rather than using the perpendicular distance to the line of action.

Rotational Inertia (Moment of Inertia)

What rotational inertia is

Rotational inertia (also called moment of inertia) measures how strongly an object resists changes in rotational motion about a particular axis. It is the rotational analog of mass.

Mass tells you how hard it is to change linear motion. Moment of inertia tells you how hard it is to change angular motion—and it depends not just on “how much mass,” but how that mass is distributed relative to the axis.

If more mass lies far from the axis, the object is harder to spin up or slow down.

The definition: point masses and continuous bodies

For discrete point masses m_i at distances r_i from the axis:

I = \sum m_i r_i^2

For a continuous mass distribution, you generalize to an integral:

I = \int r^2\,dm

Here, r is the perpendicular distance from the axis of rotation to the mass element dm.

This definition explains two high-yield facts:

• Moment of inertia depends on the axis you choose.
• Units: \text{kg·m}^2.

Standard moments of inertia you should know

AP Physics C expects comfort with common shapes (and the ability to use theorems to shift axes). Common results include:

• Thin hoop/ring about center axis perpendicular to plane:

I = MR^2

• Solid disk/cylinder about center axis perpendicular to face:

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

• Thin rod of length L about center (axis perpendicular to rod):

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

• Thin rod about one end (axis perpendicular to rod):

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

• Solid sphere about diameter:

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

• Thin spherical shell about diameter:

I = \frac{2}{3}MR^2

The pattern is important: distributing mass farther out increases I.

Parallel-axis theorem (moving the axis)

Often you know the moment of inertia about an axis through the center of mass, but the problem uses a parallel axis a distance d away. The parallel-axis theorem is

I = I_{cm} + Md^2

• I_{cm} is the moment of inertia about the center-of-mass axis.
• M is total mass.
• d is the perpendicular distance between the axes.

This is especially common for rods pivoting about an end. You can get the end result from the center result:

I_{end} = \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{3}ML^2

Perpendicular-axis theorem (planar objects)

For a flat object lying in the xy plane, with moments about perpendicular axes through the same point:

I_z = I_x + I_y

This can be useful for some lamina shapes, though many AP problems give you what you need directly.

Why moment of inertia matters in dynamics and energy

Moment of inertia is what connects torque to angular acceleration (next section) and also sets the rotational kinetic energy:

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

This becomes essential in rolling motion, where objects have both translational and rotational kinetic energy.

Worked example 1: comparing rotational inertia (qualitative)

Two objects have the same mass M and radius R: a hoop and a solid disk.

• Hoop: I = MR^2
• Disk: I = \frac{1}{2}MR^2

The hoop has double the moment of inertia, meaning for the same applied torque it will have half the angular acceleration. Physically, more mass is located at the rim, farther from the axis.

Worked example 2: using the parallel-axis theorem

A uniform rod of mass 2.0 kg and length 1.5 m rotates about an axis perpendicular to the rod through one end. Find I.

Start from the center result:

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

Shift by d=L/2:

I = I_{cm} + Md^2

Substitute:

I = \frac{1}{12}(2.0)(1.5^2) + (2.0)(0.75^2)

Compute:

I = \frac{1}{12}(2.0)(2.25) + (2.0)(0.5625)

I = 0.375 + 1.125 = 1.50\text{ kg·m}^2

A common error is forgetting to square d in Md^2.

Notation reference (common equivalent forms)

Exam Focus

• Typical question patterns
  • Compute I for composite objects (sum of parts) and/or shifted axes using the parallel-axis theorem.
  • Compare accelerations or energies of different rolling objects using different I values.
  • Use a given I to connect torque, angular acceleration, and rotational kinetic energy.
• Common mistakes
  • Treating I as an intrinsic property like mass (it depends on axis).
  • Using the wrong radius (distance must be perpendicular to the rotation axis).
  • Mixing up formulas for hoop vs disk vs sphere (learn the physical “mass farther out means larger I” idea to self-check).

Newton’s Second Law for Rotation

The rotational version of F=ma

In translation, Newton’s second law says net force causes translational acceleration:

\sum F = ma

For rotation about a fixed axis, the analogous statement is: net torque causes angular acceleration:

\sum \tau = I\alpha

This is one of the central equations of rotational dynamics.

• \sum\tau is the net torque about the chosen axis.
• I is the moment of inertia about that same axis.
• \alpha is the angular acceleration.

The analogy is deep and useful:

• Force corresponds to torque.
• Mass corresponds to moment of inertia.
• Linear acceleration corresponds to angular acceleration.

When you can (and cannot) use \sum\tau = I\alpha

For AP Physics C, \sum\tau = I\alpha is used most cleanly for a rigid body rotating about a fixed axis (or about its center of mass with a consistent axis) where I is constant.

A more general statement that connects torque to angular momentum is

\sum \tau = \frac{dL}{dt}

If L = I\omega with constant I about that axis, then

\sum \tau = I\alpha

This perspective helps you avoid a common misconception: \sum\tau = I\alpha is not “always true” in every possible situation unless the conditions for using a constant I about a fixed axis are met.

Rotational dynamics often couples to translation

Many AP problems involve strings, pulleys, and rolling objects. In those problems, you typically need both:

• Translational dynamics of the center of mass: \sum F = ma
• Rotational dynamics about the center (or axis): \sum\tau = I\alpha

The bridge between them is usually a kinematic constraint such as “no slipping,” which relates linear and angular acceleration.

Relating torque to tangential force

If a tangential force F_t is applied at radius R, then \tau = RF_t and

RF_t = I\alpha

Also, tangential acceleration at the rim is

a_t = \alpha R

This is frequently used for pulleys where a string pulls on a radius.

Rotational work and power (often used alongside dynamics)

Torque also connects to energy. The work done by a constant torque over an angular displacement \Delta\theta is

W = \tau\Delta\theta

Power delivered by torque at angular speed \omega is

P = \tau\omega

These are especially helpful when problems ask about motors, spinning up flywheels, or energy methods.

Worked example 1: angular acceleration of a disk with a tangential force

A solid disk (mass 4.0 kg, radius 0.30 m) is mounted on a frictionless axle. A tangential force of 12 N is applied at the rim.

1) Find the torque:

\tau = RF = (0.30)(12) = 3.6\text{ N·m}

2) Moment of inertia of a solid disk:

I = \frac{1}{2}MR^2 = \frac{1}{2}(4.0)(0.30^2)

I = 2.0(0.09) = 0.18\text{ kg·m}^2

3) Apply \sum\tau = I\alpha:

\alpha = \frac{\tau}{I} = \frac{3.6}{0.18} = 20\text{ rad/s}^2

A common mistake is using I=MR^2 (that would be for a hoop, not a disk).

Worked example 2: pulley with two masses (rotation + translation)

Two masses are connected by a light string over a pulley of radius R and moment of inertia I. Let m_2 > m_1 so the system accelerates with magnitude a, with m_2 moving down.

For the masses (translation):

m_2g - T_2 = m_2a

T_1 - m_1g = m_1a

For the pulley (rotation about axle): the tensions create torques. Taking the net torque in the direction of m_2 descending:

\sum\tau = (T_2 - T_1)R = I\alpha

No slipping between string and pulley gives the constraint:

a = \alpha R

Substitute into the torque equation:

(T_2 - T_1)R = I\frac{a}{R}

T_2 - T_1 = \frac{I}{R^2}a

From here, you solve the three equations for a. The key conceptual takeaway: unlike the massless-pulley case, T_2\neq T_1 because the difference in tensions is what produces the torque needed to spin the pulley.

Exam Focus

• Typical question patterns
  • “Given forces at certain distances, find angular acceleration” using \sum\tau=I\alpha.
  • Pulley/string systems requiring both \sum F = ma and \sum\tau=I\alpha with the constraint a=\alpha R.
  • Energy vs dynamics comparisons (e.g., compute speed after rotating through an angle using work-energy with W=\tau\Delta\theta).
• Common mistakes
  • Using torque about the wrong axis (you must compute \tau and I about the same axis).
  • Assuming tensions on both sides of a massive pulley are equal.
  • Forgetting the no-slip constraint a=\alpha R or misapplying it with the wrong radius.

Rolling Motion

What “rolling” really means: translation + rotation together

Rolling motion combines:

• Translation of the object’s center of mass (COM)
• Rotation of the object around its COM

A wheel moving along the ground is not doing “either/or.” Points on the wheel have velocities that come from both effects simultaneously.

Rolling without slipping: the key constraint

The most important special case is rolling without slipping (also called pure rolling). It means the point of contact with the ground is instantaneously at rest relative to the ground.

This produces the fundamental kinematic relationship:

v_{cm} = \omega R

and similarly for accelerations along the surface:

a_{cm} = \alpha R

These constraints are not optional “extra formulas”—they are what close the system when you have both rotation and translation.

A misconception to avoid: v=\omega R is not a universal truth for rotating objects. It is specifically the rolling-without-slipping condition relating COM speed to angular speed.

Energy of a rolling rigid body

The total kinetic energy of a rolling object is the sum of translational KE of the COM and rotational KE about the COM:

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

If rolling without slipping, substitute \omega = v_{cm}/R:

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

This form is extremely useful for ramps: gravitational potential energy becomes both kinds of kinetic energy.

Why different shapes roll down ramps differently

If two objects have the same M and R but different I_{cm}, they will accelerate differently on an incline when rolling without slipping. The reason is energetic and dynamical: larger I_{cm} means more energy must go into rotation for a given v_{cm}, leaving less available to increase translational speed.

A very common AP result comes from combining dynamics (or energy) with the rolling constraint. For an object rolling down an incline of angle \theta without slipping, the COM acceleration is

a_{cm} = \frac{g\sin\theta}{1 + \frac{I_{cm}}{MR^2}}

You don’t need to memorize this if you understand how to derive it, but it’s a helpful checkpoint.

• Hoop: I_{cm}=MR^2 gives a_{cm} = \frac{g\sin\theta}{2}
• Solid cylinder/disk: I_{cm}=\frac{1}{2}MR^2 gives a_{cm} = \frac{2}{3}g\sin\theta

So the solid disk accelerates faster than the hoop.

The role (and direction) of friction in rolling

For rolling without slipping on a rough surface, the friction is typically static friction. Two important consequences:

1) Static friction can be nonzero even without slipping. It may be required to provide the torque that spins the object appropriately.

2) Static friction often does no mechanical work on the rolling object if the contact point is instantaneously at rest relative to the surface. Energy changes then come from gravity or other external work, not from frictional work (in the ideal pure-rolling case).

Direction of friction is a frequent exam trap. For a wheel rolling down an incline without slipping, static friction commonly points up the plane (it provides a torque that increases rotation in the correct sense). But you should determine direction by the required tendency to slip: ask yourself which way the contact point would slide if there were no friction, then friction opposes that would-be relative motion.

Worked example 1: speed at bottom using energy (rolling without slipping)

A solid sphere of mass M and radius R rolls without slipping from rest down a vertical height h.

Use energy conservation (no slipping, static friction does no net work in the ideal model):

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

For a solid sphere:

I_{cm} = \frac{2}{5}MR^2

Rolling constraint \omega = v/R:

Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}\left(\frac{2}{5}MR^2\right)\frac{v^2}{R^2}

Mgh = \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2

Mgh = \frac{7}{10}Mv^2

Solve:

v^2 = \frac{10}{7}gh

v = \sqrt{\frac{10}{7}gh}

If you did the same for a hoop, you’d get a smaller speed because more energy would be tied up in rotation for the same v.

Worked example 2: acceleration down an incline (dynamics approach)

A solid cylinder rolls without slipping down an incline angle \theta.

Forces along the incline: component of gravity down the plane Mg\sin\theta and static friction f up the plane (for the usual rolling-down case). Translational equation:

Mg\sin\theta - f = Ma

Torque about the center: friction provides the torque that spins it. With radius R:

fR = I_{cm}\alpha

No-slip constraint:

a = \alpha R

Substitute \alpha = a/R into torque equation:

fR = I_{cm}\frac{a}{R}

f = \frac{I_{cm}}{R^2}a

For a solid cylinder, I_{cm} = \frac{1}{2}MR^2, so

f = \frac{1}{2}Ma

Plug into translation equation:

Mg\sin\theta - \frac{1}{2}Ma = Ma

Mg\sin\theta = \frac{3}{2}Ma

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

This matches the general formula and gives you a strong consistency check.

Rolling with slipping (what changes conceptually)

If the surface is too slippery (or the incline too steep) for static friction to maintain the no-slip condition, the object will slip while rolling. Then:

• The constraint v_{cm}=\omega R no longer holds.
• Friction becomes kinetic friction with magnitude f_k = \mu_k N.
• Kinetic friction does negative work, so mechanical energy is not conserved.

AP questions sometimes ask you to identify whether rolling without slipping is possible by checking whether the required static friction exceeds \mu_s N.

Exam Focus

• Typical question patterns
  • Compare speeds/accelerations of different shapes rolling down an incline using energy or a_{cm} = \frac{g\sin\theta}{1 + \frac{I_{cm}}{MR^2}}.
  • Find friction direction and magnitude for rolling without slipping using coupled \sum F = Ma and \sum\tau=I\alpha.
  • Use energy with K = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 and the constraint v=\omega R.
• Common mistakes
  • Assuming friction must point down the incline because the object moves down (friction’s direction depends on impending slip, not motion direction).
  • Using v=\omega R in situations with slipping.
  • Forgetting rotational kinetic energy (treating a rolling object like a sliding block).