Unit 5: Torque and Rotational Dynamics

Rotational Motion

Rotational motion occurs when an object spins around an axis or a fixed point. Unlike linear motion, where an object translates in a straight path, rotational motion describes turning or spinning around a central axis. Many real-world systems—wheels, planets, gears, and turbines—are governed by rotational dynamics. To study this motion, we use angular quantities such as angular displacement, angular velocity, and angular acceleration, which are rotational counterparts to linear displacement, velocity, and acceleration.

Key Angular Quantities

• Angular Displacement (θ): The angle through which an object rotates, measured in radians. One full revolution corresponds to \(2\pi\) radians. Angular displacement helps describe how much "turning" has occurred, with positive for counterclockwise and negative for clockwise directions.
• Angular Velocity (ω): The rate of change of angular displacement over time. Measured in radians per second, it shows how fast an object rotates. For example, a ceiling fan’s speed is its angular velocity, which remains constant when running steadily.
• Angular Acceleration (α): The rate of change of angular velocity. Measured in radians per second squared, it indicates whether an object is speeding up, slowing down, or changing its rotational direction. For instance, when a record player starts spinning, the angular acceleration determines how quickly it reaches its steady speed.
• Moment of Inertia (I): The rotational analog of mass. It measures an object’s resistance to changes in rotational motion and depends on both mass and how that mass is distributed relative to the axis. A figure skater pulling in her arms decreases her moment of inertia, causing her to spin faster due to conservation of angular momentum.
• Torque (τ): The measure of a force’s ability to cause rotation. Torque depends on both the magnitude of the force and its perpendicular distance from the axis of rotation. For example, pushing a door near its handle produces more torque than pushing close to its hinges.

Rotational Kinematics

Rotational kinematics is the study of rotational motion without considering the forces that cause it. Just as in linear kinematics, there are four primary equations of motion (often called the “SUVAT” equations in linear motion). These rotational equations mirror the linear ones, with angular variables replacing linear variables.

Rotational Kinematics Equations

• \( \theta = \theta_0 + \omega_0 t + \tfrac{1}{2}\alpha t^2 \) Describes angular displacement after time \(t\), given initial angular velocity and angular acceleration.
• \( \omega = \omega_0 + \alpha t \) Gives final angular velocity after time \(t\), based on initial velocity and angular acceleration.
• \( \omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0) \) Relates angular velocity to angular displacement without requiring time explicitly.
• \( \theta - \theta_0 = \tfrac{1}{2}(\omega + \omega_0)t \) Provides angular displacement using average angular velocity over time \(t\).

Relationship Between Linear and Angular Motion

• Linear velocity of a point on a rotating object: \[ v = r\omega \] where \(r\) is the distance from the axis. This means that outer points on a rotating wheel move faster than inner points.
• Linear acceleration of a point on a rotating object: \[ a = r\alpha \] connecting angular acceleration with linear acceleration.
• This relationship allows us to solve problems that combine rotational and translational motion, such as rolling objects or planetary orbits.

Center of Mass

The center of mass (COM) is the point in an object or system where its mass can be considered to be concentrated for analyzing motion. In physics problems, we often treat an entire object as if all its mass were located at the center of mass. This simplifies the study of both linear and rotational motion, especially when objects are irregularly shaped. The COM may lie inside, outside, or even in an empty region of an object depending on how mass is distributed.

Properties of the Center of Mass

• The center of mass of a symmetrical object with uniform density (such as a sphere or cube) is located at its geometric center. This makes predicting its motion straightforward.
• If the object has an uneven mass distribution, the COM shifts toward the heavier portion. For example, in a hammer, the COM is closer to the metal head than the wooden handle.
• The motion of an object can be analyzed as if all forces act at the COM. For example, when a ball is thrown, the COM follows a smooth parabolic trajectory even though the ball may spin or wobble in the air.
• The COM of a system of particles is calculated by weighting each particle’s position by its mass. This ensures heavier particles contribute more strongly to the overall COM position.
• The COM is conserved in the absence of external forces. In space, for instance, a system of interacting objects (like an astronaut and a tool) has a COM that moves in a straight line at constant velocity unless acted on by an external force.

Formula for the Center of Mass

For discrete masses along a line, the position of the center of mass is given by:

\[ x_{COM} = \frac{m_1x_1 + m_2x_2 + \cdots + m_nx_n}{m_1 + m_2 + \cdots + m_n} \]

where \(m_i\) are the masses and \(x_i\) their positions. Similar formulas apply in two and three dimensions using coordinates \(y_{COM}\) and \(z_{COM}\).

Example

A 3 kg mass is placed at \(x = 0 \, m\), and a 5 kg mass at \(x = 4 \, m\). The center of mass is:

\[ x_{COM} = \frac{(3)(0) + (5)(4)}{3 + 5} = \frac{20}{8} = 2.5 \, m \]

Thus, the system behaves as if all 8 kg were concentrated at 2.5 m from the origin.

Moment of Inertia

The moment of inertia (I) is the rotational analog of mass in linear motion. It measures an object’s resistance to changes in its rotational motion about a specific axis. While mass alone determines resistance to linear acceleration, the moment of inertia depends on both mass and how that mass is distributed relative to the axis of rotation. Objects with more mass located farther from the axis have larger moments of inertia and are harder to spin or stop spinning.

Key Ideas

• The formula for a system of point masses is: \[ I = \sum m_i r_i^2 \] where \(m_i\) is each mass and \(r_i\) is its perpendicular distance from the axis of rotation. This weighting by the square of the radius shows why mass farther from the axis has a much greater effect.
• For continuous bodies, calculus is used to integrate over the object’s mass distribution. AP Physics 1 focuses on simpler shapes with known formulas (like rods, disks, hoops, and spheres).
• Moment of inertia depends strongly on the axis of rotation. For example, spinning a pencil around its length is easy (small \(I\)), but spinning it around its end is much harder (large \(I\)).
• Reducing the moment of inertia increases angular acceleration for the same torque. This is why figure skaters spin faster when pulling their arms in: mass moves closer to the axis, reducing \(I\).
• The SI unit of moment of inertia is \( \text{kg·m}^2 \).

Common Moments of Inertia (about central axes)

• Thin hoop or cylinder: \[ I = MR^2 \]
• Solid disk or cylinder: \[ I = \tfrac{1}{2}MR^2 \]
• Solid sphere: \[ I = \tfrac{2}{5}MR^2 \]
• Thin rod about center: \[ I = \tfrac{1}{12}ML^2 \]
• Thin rod about end: \[ I = \tfrac{1}{3}ML^2 \]

Parallel-Axis Theorem

If the axis of rotation is shifted parallel to a known axis by a distance \(d\), the moment of inertia is:

\[ I = I_{cm} + Md^2 \]

where \(I_{cm}\) is the moment of inertia through the center of mass and \(M\) is the total mass. This is often used for rods, plates, or disks rotating around points other than their center.

Example

A solid disk of mass 2.0 kg and radius 0.30 m spins about its center. Its moment of inertia is:

\[ I = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(2.0)(0.30^2) = 0.090 \, \text{kg·m}^2 \]

Torque

Torque is the measure of how much a force causes an object to rotate about an axis. In linear motion, force causes acceleration. In rotational motion, torque causes angular acceleration. The larger the torque, the greater the tendency for rotation. Torque depends not just on how much force is applied, but also on where and in what direction it is applied relative to the axis of rotation.

Torque Formula

The magnitude of torque \( \tau \) is calculated by: \[ \tau = rF\sin\theta \] where:

• \(r\) is the distance from the axis of rotation to the point where the force is applied (the lever arm).
• \(F\) is the magnitude of the force.
• \(\theta\) is the angle between the force vector and the position vector (lever arm).

Key Concepts of Torque

• Lever arm (perpendicular distance): Torque depends on the perpendicular distance from the axis to the line of action of the force. A longer lever arm increases torque, which is why doors open more easily when pushed at the edge rather than near the hinge.
• Direction matters: Torque is a vector and can be clockwise or counterclockwise. Conventionally, counterclockwise torque is taken as positive and clockwise as negative. Always watch signs carefully in problems involving equilibrium or rotational motion.
• Maximum torque: When the force is applied perpendicular to the lever arm (\(\theta = 90^\circ\)), torque is maximized because \(\sin(90^\circ) = 1\). If the force is parallel to the lever arm (\(\theta = 0^\circ\)), no torque is produced.
• Torque and angular acceleration: Newton’s Second Law for rotation relates torque to moment of inertia: \[ \tau = I\alpha \] where \(I\) is the moment of inertia and \(\alpha\) is angular acceleration. This is the rotational version of \(F = ma\).
• Units: The SI unit of torque is Newton-meter (\( \text{N·m} \)). It is not the same as energy (also measured in Joules), because torque is a vector that causes rotation, not scalar energy.

Right-Hand Rule

Use the right-hand rule to determine the direction of torque: Point your fingers in the direction of the position vector (\( \vec{r} \)), then curl them toward the force vector (\( \vec{F} \)). Your thumb points in the direction of the torque vector. If your thumb points out of the page, torque is positive (counterclockwise); if it points into the page, torque is negative (clockwise).

Equilibrium Condition

For an object to be in rotational equilibrium, the net torque must be zero: \[ \sum \tau = 0 \] This means the object is not accelerating rotationally — it may be at rest or rotating at constant angular velocity.

Example

A 20 N force is applied at the end of a 0.6 m long wrench at an angle of \(60^\circ\) from the handle. Find the torque produced about the bolt. \[ \tau = rF\sin\theta = (0.6)(20)\sin(60^\circ) \approx (0.6)(20)(0.866) = 10.392 \, \text{N·m} \] The torque is approximately 10.4 N·m, and it will cause counterclockwise rotation.

Rotational Equilibrium

An object is in rotational equilibrium when the net torque acting on it is zero. This means the object has no angular acceleration — it may either remain at rest or continue rotating at a constant angular velocity. Rotational equilibrium is the rotational counterpart of translational equilibrium, where the net force is zero. In many AP Physics 1 problems, both linear and rotational equilibrium conditions must be satisfied simultaneously.

Conditions for Equilibrium

• First condition (translational equilibrium): \[ \sum \vec{F} = 0 \] The vector sum of all external forces on the object must be zero. This ensures the center of mass does not accelerate linearly.
• Second condition (rotational equilibrium): \[ \sum \tau = 0 \] The algebraic sum of all torques about any axis must be zero. This ensures the object does not angularly accelerate.
• Both conditions must be satisfied at the same time for static equilibrium, which means the object is completely at rest.
• Choosing a convenient pivot point often simplifies torque calculations because forces passing through the pivot exert no torque. This strategy is essential in solving AP-level equilibrium problems efficiently.
• Rotational equilibrium problems are common in AP exams, especially with beams, levers, or see-saws where multiple forces act at different points.

Example

A uniform 4 m long beam of mass 20 kg is supported at its center. A 40 N weight hangs 1 m from the right end. Find the upward force the support must provide.

Step 1: Place pivot at the beam’s center. Net torque = 0.
Torque due to the weight: \( \tau = rF = (1.0)(40) = 40 \, \text{N·m} \).
Torque due to beam’s weight cancels (acts at pivot).
The support provides an equal and opposite torque to balance.
So the support force = 40 N downward + beam’s own 196 N = 236 N upward.

Rotational Dynamics

Rotational Dynamics is the study of how torques affect the rotational motion of objects. Just as Newton’s Second Law describes how forces change linear motion, there’s a rotational equivalent: \[ \tau_{\text{net}} = I\alpha \] where \( \tau_{\text{net}} \) is the net torque, \( I \) is the moment of inertia (rotational mass), and \( \alpha \) is the angular acceleration. This law helps us predict how an object’s rotation will change in response to external torques.

Newton’s Second Law for Rotation

• Formula: \( \tau_{\text{net}} = I\alpha \) tells us that angular acceleration is directly proportional to net torque and inversely proportional to moment of inertia. This mirrors the linear relationship \( F = ma \).
• Moment of inertia \(I\): The rotational equivalent of mass. Objects with more mass farther from the axis of rotation have a higher moment of inertia and are harder to spin. For example, a ring is harder to spin than a solid disc of the same mass and radius.
• Angular acceleration \( \alpha \): The rate of change of angular velocity. It’s measured in radians per second squared (\( \text{rad/s}^2 \)) and only occurs when the net torque is nonzero.
• Signs matter: Be careful with direction! Positive torque (counterclockwise) and negative torque (clockwise) follow the right-hand rule. Consistently track signs when summing torques or forces.
• Rotational analogs: You can think of torque, moment of inertia, and angular acceleration as analogs to force, mass, and linear acceleration. They’re all connected through Newton’s laws in rotational motion.

Common Shapes and Moments of Inertia

• Solid disk or cylinder: \( I = \frac{1}{2}MR^2 \)
• Solid sphere: \( I = \frac{2}{5}MR^2 \)
• Thin hoop or ring (about center): \( I = MR^2 \)
• Rod about center: \( I = \frac{1}{12}ML^2 \)
• Rod about end: \( I = \frac{1}{3}ML^2 \)

These formulas will be provided on the AP equation sheet, but you must recognize which to use depending on where the axis is located.

Example Problem

A 0.5 kg solid disc with radius 0.2 m is acted on by a net torque of 0.1 N·m. What is its angular acceleration?

Step 1: Use moment of inertia of solid disc: \( I = \frac{1}{2}MR^2 = \frac{1}{2}(0.5)(0.2)^2 = 0.01 \, \text{kg·m}^2 \)
Step 2: Apply \( \tau = I\alpha \): \( 0.1 = 0.01 \alpha \Rightarrow \alpha = 10 \, \text{rad/s}^2 \)
The angular acceleration is \( 10 \, \text{rad/s}^2 \).

Connecting Linear and Rotational Motion

Linear and rotational motion are deeply connected, and many equations from linear kinematics have rotational counterparts. When an object rotates around an axis, every point on the object also undergoes linear motion along a circular path. These connections allow us to translate between linear and angular variables, which is crucial in AP Physics 1 when solving rolling motion problems or analyzing rotating systems.

Key Relationships

• \( v = r\omega \) — The linear speed of a point at a distance \(r\) from the axis is proportional to angular velocity \( \omega \). Outer points on a wheel move faster than inner points.
• \( a = r\alpha \) — The linear acceleration of a point at radius \(r\) is linked to angular acceleration \( \alpha \). This ensures that changes in rotational speed affect the linear speed of all points.
• Arc length: \( s = r\theta \) — Linear displacement along the edge of a circle relates directly to angular displacement \( \theta \).
• These relationships are valid only when \( \theta \) is measured in radians. Using degrees will give incorrect results because radians are defined directly from arc length.
• This connection allows us to analyze problems where objects both rotate and translate, such as a rolling ball or a yo-yo unwinding.

Rotational Inertia

Rotational inertia, another name for moment of inertia, describes how resistant an object is to changes in its rotational motion. Just as heavier objects resist changes in linear motion, objects with larger rotational inertia resist changes in angular motion. The distribution of mass is equally important: mass located farther from the axis increases rotational inertia dramatically.

Key Points

• The formula for discrete point masses is: \[ I = \sum m_i r_i^2 \] showing that distance from the axis is squared, making far-away mass much more impactful.
• For continuous bodies, formulas are derived through integration, but the AP exam provides the most common ones for rods, disks, hoops, and spheres.
• Objects with the same mass and radius can have very different rotational inertias. A hoop is harder to spin than a solid disk because more of its mass is located farther from the axis.
• Rotational inertia is axis-dependent: rotating a rod about its end requires more torque than about its center, even though mass and length are unchanged.
• Because torque equals \( I\alpha \), reducing rotational inertia increases angular acceleration. This principle explains why figure skaters spin faster when pulling their arms in.

Newton’s First Law in Rotational Form

Newton’s First Law applies to rotational motion as well: An object will continue rotating at a constant angular velocity (or remain at rest) unless acted upon by a net external torque. This law emphasizes inertia in rotational systems — objects resist changes in rotational motion just as they resist changes in linear motion.

Applications

• If the net torque on an object is zero, the object will not angularly accelerate. It may remain at rest or continue rotating steadily, similar to how an object moves at constant velocity in linear motion without a net force.
• For example, a spinning top on a frictionless surface would spin indefinitely without slowing down, because no net torque acts on it.
• In reality, friction and air resistance act as external torques, gradually slowing down rotating objects. These forces break the condition of zero net torque.
• Newton’s First Law in rotational form highlights the importance of analyzing external torques. If no torque acts, angular momentum remains constant (a principle expanded in the conservation of angular momentum).
• This law sets the foundation for equilibrium analysis: to keep a system stable, the net torque must equal zero, preventing angular acceleration.

Rolling Motion Without Slipping

Rolling motion without slipping occurs when a circular object, like a wheel or ball, rolls along a surface such that its point of contact is instantaneously at rest relative to the surface. This means there is no sliding or skidding. Rolling combines both translational and rotational motion, and friction plays a critical role in preventing slipping.

Condition for Rolling Without Slipping

The key condition is: \[ v = r\omega \] where \( v \) is the linear velocity of the object’s center of mass, \( r \) is the radius, and \( \omega \) is the angular velocity.

• When rolling without slipping, every point on the rim of the wheel traces a cycloidal path. The bottommost point has zero velocity relative to the ground at any instant, while the topmost point moves twice as fast as the center of mass.
• Static friction (not kinetic) is responsible for rolling without slipping. It prevents the object from sliding, but it does not do work if there is no slipping. This is why rolling motion can occur without continual energy loss to friction.
• The condition \( v = r\omega \) ensures that the rotational and translational motions are linked. If this condition fails, the object either slips forward (if \( v > r\omega \)) or skids backward (if \( v < r\omega \)).
• Acceleration also follows a similar relation: \[ a = r\alpha \] where \( a \) is linear acceleration and \( \alpha \) is angular acceleration.
• In AP Physics, rolling motion often appears in problems with ramps, spheres, and cylinders. Moment of inertia is crucial here, since it affects how energy splits between rotational and translational forms.

Energy in Rolling Motion

The total kinetic energy of a rolling object is the sum of translational and rotational kinetic energy: \[ KE_{total} = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2 \] Using the rolling condition, both terms can be expressed in terms of either \(v\) or \(\omega\).

Example

A solid sphere (mass 2.0 kg, radius 0.10 m) rolls without slipping at 3.0 m/s. Find its total kinetic energy.

Step 1: Find moment of inertia for a solid sphere: \( I = \tfrac{2}{5}MR^2 = \tfrac{2}{5}(2.0)(0.10^2) = 0.008 \, \text{kg·m}^2 \).
Step 2: Use rolling condition to find \(\omega\): \( \omega = v/r = 3.0 / 0.10 = 30 \, \text{rad/s} \).
Step 3: Compute energy: \( KE = \tfrac{1}{2}(2.0)(3.0^2) + \tfrac{1}{2}(0.008)(30^2) = 9 + 3.6 = 12.6 \, J \).
Total kinetic energy = 12.6 J.

Multi-Step Challenge Problems for Unit 5: Torque & Rotational Dynamics

Problem 1: Torque and Angular Acceleration

A uniform solid disk of mass 5.0 kg and radius 0.20 m is free to rotate about its central axis. A 10 N force is applied tangentially at its rim. Find: (a) the torque on the disk, (b) its angular acceleration, and (c) the angular speed after 4.0 s, assuming it starts from rest.

Solution:

Step 1: Torque: \( \tau = rF = 0.20 \times 10 = 2.0 \, \text{N·m} \).
Step 2: Moment of inertia of solid disk: \( I = \tfrac{1}{2}MR^2 = 0.5(5.0)(0.20^2) = 0.10 \, \text{kg·m}^2 \).
Step 3: Angular acceleration: \( \alpha = \tau/I = 2.0 / 0.10 = 20 \, \text{rad/s}^2 \).
Step 4: Angular speed: \( \omega = \omega_0 + \alpha t = 0 + (20)(4.0) = 80 \, \text{rad/s} \).

Answer: (a) 2.0 N·m, (b) 20 rad/s², (c) 80 rad/s.

Problem 2: Rotational Equilibrium Beam

A uniform 4.0 m long beam of mass 30 kg is supported at its left end. A 600 N student stands 3.0 m from the pivot. Find the upward force the pivot exerts to keep the beam in equilibrium.

Solution:

Step 1: Choose pivot at left end. The beam’s weight acts at its center (2.0 m).
Beam weight: \( W_b = mg = 30(9.8) = 294 \, N \).
Step 2: Calculate torques.
Torque from beam: \( \tau_b = 294(2.0) = 588 \, N·m \).
Torque from student: \( \tau_s = 600(3.0) = 1800 \, N·m \).
Step 3: Total downward force: \( F_{down} = 294 + 600 = 894 \, N \).
Since no vertical acceleration, upward pivot force = 894 N.
Step 4: Net torque check → pivot provides equal upward torque balancing student + beam torque.

Answer: Pivot exerts an upward force of 894 N.

Problem 3: Rolling Motion Without Slipping

A solid sphere of mass 2.0 kg and radius 0.10 m starts from rest and rolls without slipping down a 2.0 m high incline. Find: (a) the linear speed at the bottom, and (b) the angular speed at the bottom.

Solution:

Step 1: Use energy conservation: \( mgh = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2 \).
Step 2: Moment of inertia of solid sphere: \( I = \tfrac{2}{5}MR^2 \). Rolling condition: \( v = r\omega \).
Step 3: Substituting: \( mgh = \tfrac{1}{2}mv^2 + \tfrac{1}{2}(\tfrac{2}{5}MR^2)(v^2/R^2) \).
Simplify: \( mgh = \tfrac{1}{2}mv^2 + \tfrac{1}{5}mv^2 = \tfrac{7}{10}mv^2 \).
Step 4: Solve: \( v = \sqrt{\tfrac{10gh}{7}} = \sqrt{\tfrac{10(9.8)(2.0)}{7}} \approx 5.29 \, m/s \).
Step 5: Angular speed: \( \omega = v/r = 5.29 / 0.10 = 52.9 \, rad/s \).

Answer: (a) 5.29 m/s, (b) 52.9 rad/s.

Common Misconceptions in Unit 5: Torque & Rotational Dynamics

Torque

• Misconception: Torque only depends on the size of the force.
  Correction: Torque also depends on the perpendicular distance from the axis. A small force applied far from the pivot can produce more torque than a large force applied close to it.
• Misconception: Torque and energy are the same because they both use N·m units.
  Correction: Torque is a vector causing rotation, while energy is a scalar quantity. Though units look the same, they represent different physical concepts.

Rotational Inertia

• Misconception: Moment of inertia depends only on mass.
  Correction: It depends on both mass and its distribution. Mass farther from the axis contributes much more strongly due to the \( r^2 \) factor.
• Misconception: Two objects with the same mass and radius always have the same moment of inertia.
  Correction: A hoop and a disk with identical mass and radius have different \(I\) values, because their mass is distributed differently.

Rolling Motion

• Misconception: Rolling motion always requires kinetic friction doing work.
  Correction: In rolling without slipping, static friction prevents sliding but does no work. Energy remains mechanical (translational + rotational kinetic energy).
• Misconception: The bottom point of a rolling wheel is moving.
  Correction: Relative to the ground, the bottom point is momentarily at rest when rolling without slipping.

Newton’s First Law (Rotational)

• Misconception: A rotating object must always have a net torque acting on it.
  Correction: If net torque is zero, an object can rotate indefinitely at constant angular velocity. A torque is only required to change rotational speed, not to maintain it.