Unit 5: Torque and Rotational Dynamics

Torque: The Rotational Equivalent of Force

Torque (\( \tau \)) is a measure of how effectively a force causes an object to rotate about an axis. It depends not just on the magnitude of the force, but also on the distance from the axis and the angle at which the force is applied. Mathematically, torque is defined as:

\[ \tau = rF\sin(\theta) \]

Here, \( r \) is the distance from the axis of rotation to the point where the force is applied, \( F \) is the magnitude of the force, and \( \theta \) is the angle between the position vector and the force vector. When the force is perpendicular to the radius vector, \( \theta = 90^\circ \), and the torque is maximized.

Torque Direction and Sign Convention

• Torque is a vector quantity; its direction is determined by the right-hand rule. Curl your fingers in the direction of rotation caused by the force, and your thumb points in the direction of the torque vector. This is crucial for analyzing systems in equilibrium and understanding rotational dynamics correctly.
• Clockwise torques are typically considered negative, while counterclockwise torques are positive. This sign convention helps maintain consistency when setting up equations for net torque, especially when multiple forces act at various angles or positions around a pivot.
• Multiple torques can act on a system simultaneously, and the net torque is the algebraic sum of all individual torques. If net torque is zero, the object may still be rotating but it won’t have any angular acceleration; this condition is essential in rotational equilibrium problems.

Right-Hand Rule for Torque

• The right-hand rule helps determine the direction of torque as a vector, which is crucial for understanding rotation in three dimensions. To apply it, point your right-hand fingers in the direction of the position vector \( \vec{r} \) (from the pivot to the point of force application), then curl them toward the direction of the force vector \( \vec{F} \). Your thumb will point in the direction of the torque vector \( \vec{\tau} \).
• If your thumb points out of the page, the torque is considered **positive** (counterclockwise). If your thumb points into the page, the torque is **negative** (clockwise). This is consistent with the sign conventions used in rotational kinematics and dynamics. It’s important to establish a clear axis of rotation (usually the z-axis) before applying the rule.
• The right-hand rule is especially useful when forces act at angles or when torque problems are framed in 3D, like rotating disks or inclined planes. It connects the vector cross product formula for torque: \[ \vec{\tau} = \vec{r} \times \vec{F} \] The direction of the torque vector is always perpendicular to the plane formed by \( \vec{r} \) and \( \vec{F} \), which is what the right-hand rule helps visualize.
• A common student mistake is using the left hand or pointing fingers in the wrong order (e.g., force before position). Always go from pivot to point of force (that's \( \vec{r} \)), then curl to the force vector. Practicing this rule physically with your hand helps solidify the spatial understanding required for AP problems on torque and angular momentum.

Translating Newton's Second Law into Rotational Form

Just as \( F = ma \) governs linear motion, its rotational equivalent is \( \tau = I\alpha \), where \( I \) is the moment of inertia (rotational equivalent of mass) and \( \alpha \) is angular acceleration. This law connects torque with how fast an object spins up or slows down, forming the backbone of rotational dynamics.

Conceptually, torque is like a “twist force,” and the moment of inertia resists that twist. An object with a large moment of inertia requires more torque to achieve the same angular acceleration as an object with a smaller moment of inertia. This is why it’s harder to rotate heavy or extended objects.

Angular acceleration \( \alpha \) has units of \( \text{rad/s}^2 \), and moment of inertia \( I \) depends on both mass and geometry. The equation \( \tau = I\alpha \) allows you to analyze any rotating system as long as the correct moment of inertia is used for its shape.

Moment of Inertia

• The moment of inertia (\( I \)) measures an object’s resistance to changes in its rotational motion, just as mass measures resistance to linear acceleration. It depends on both the total mass and how that mass is distributed relative to the axis of rotation. The farther the mass is from the axis, the greater the moment of inertia, and the harder it is to spin the object. This explains why figure skaters spin faster when they pull their arms in — they reduce their moment of inertia.
• Moment of inertia has units of \( \text{kg} \cdot \text{m}^2 \), and for different shapes, it must be computed using geometry. For example, a solid disk has \( I = \frac{1}{2}MR^2 \), while a hoop has \( I = MR^2 \). These values come from integrating mass over distance squared: \[ I = \int r^2 \, dm \] This calculus definition applies to irregular shapes and continuous bodies.
• When a system contains multiple components, you calculate total moment of inertia by summing each part’s contribution. This is similar to adding masses in linear systems but uses each part’s distance from the axis squared. If a point mass \( m \) is a distance \( r \) from the axis, then \( I = mr^2 \). Recognizing when to use point-mass vs. extended-body formulas is critical on AP problems.
• Moment of inertia is not constant for a given mass — it depends on the axis of rotation. For example, a rod spinning about its center has less inertia than the same rod spinning about one end. This leads to the parallel axis theorem: \[ I = I_{\text{COM}} + Md^2 \] where \( d \) is the distance between the new axis and the center-of-mass axis.

Rotational Kinematics

• Rotational kinematics describes how an object’s angular position, angular velocity, and angular acceleration change over time. These are the rotational equivalents of displacement, velocity, and acceleration. The three key rotational quantities are angular displacement (\( \theta \), in radians), angular velocity (\( \omega \), in rad/s), and angular acceleration (\( \alpha \), in rad/s²). The motion of spinning objects can be analyzed using similar equations to linear kinematics — but in angular form.
• The core rotational kinematic equations mirror their linear counterparts:
  • \( \omega = \omega_0 + \alpha t \)
  • \( \theta = \omega_0 t + \frac{1}{2}\alpha t^2 \)
  • \( \omega^2 = \omega_0^2 + 2\alpha\theta \)
  These equations apply when angular acceleration is constant. They help predict final angular speed, rotation angle, and time intervals for rotating systems.
• Angular displacement is measured in **radians**, not degrees. One full rotation is \( 2\pi \) radians. This unit is chosen because radians make angular relationships with arc length and rotational inertia simpler — for instance, \( s = r\theta \), where \( s \) is arc length. Always convert to radians in calculations unless otherwise specified.
• There are strong parallels between linear and rotational motion. Just like linear acceleration results from net force, angular acceleration results from net torque (\( \tau = I\alpha \)). Linear velocity \( v \) is related to angular velocity \( \omega \) through: \[ v = r\omega \quad \text{and} \quad a = r\alpha \] These relationships allow us to translate between spinning and linear systems — for example, analyzing rolling wheels or gears.
• In problems involving **non-uniform rotation** (where \( \alpha \) is not constant), you must use calculus instead of these equations. Angular acceleration is the time derivative of angular velocity: \[ \alpha = \frac{d\omega}{dt} \quad \text{and} \quad \omega = \frac{d\theta}{dt} \] To analyze such motion, you often integrate to find angle or angular velocity as a function of time. The AP exam may include graphical or conceptual interpretations of these relationships.

Rotational Kinetic Energy

• Just as moving objects have linear kinetic energy, rotating objects have rotational kinetic energy given by: \[ K_{\text{rot}} = \frac{1}{2} I \omega^2 \] This formula mirrors the translational expression \( \frac{1}{2}mv^2 \), but replaces mass with moment of inertia \( I \), and velocity with angular velocity \( \omega \). This energy exists in any spinning object, from gears to spinning planets. Since \( I \) depends on mass distribution, two objects with the same mass and spin rate can have different rotational energy depending on how their mass is arranged.
• When an object both rotates and translates (e.g., a rolling sphere), its total kinetic energy includes both forms: \[ K_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \] This is especially important in problems involving rolling without slipping, where \( v = r\omega \). This condition connects linear and angular motion, allowing us to solve for unknowns using conservation of energy. Rotational kinetic energy helps explain why it’s harder to start a solid disk spinning than a hollow ring of the same mass — the distribution of mass matters.

Static Equilibrium and Torque Balance

• Static equilibrium occurs when an object is at rest and remains at rest. This means both the **net force** and the **net torque** on the object must be zero. The condition for force is: \[ \sum F = 0 \quad \text{(translational equilibrium)} \] and the condition for torque is: \[ \sum \tau = 0 \quad \text{(rotational equilibrium)} \] Both must be satisfied to ensure the object doesn’t move or rotate unexpectedly.
• To analyze torque balance, we choose a pivot point (often where an unknown force acts) and calculate the torques caused by all other forces about that point. If clockwise and counterclockwise torques are equal, the object is in rotational equilibrium. This strategy is especially useful in problems involving seesaws, bridges, ladders, or beams. Choosing your pivot wisely can eliminate forces from your equations, simplifying your math.
• In many AP Physics C problems, you’ll analyze systems with multiple forces, angles, and distances — so setting the correct sign for torque (positive for counterclockwise, negative for clockwise) is essential. Additionally, using components of angled forces is key. For example, the torque from a diagonal force \( F \) at angle \( \theta \) is: \[ \tau = rF\sin(\theta) \] Mistakes often come from skipping the angle or using the wrong trigonometric function.

Parallel Axis Theorem

• The Parallel Axis Theorem allows us to calculate the moment of inertia of an object about any axis that is parallel to, but not through, the center of mass. This is useful when analyzing rotational motion around a point other than the center, such as the end of a rod or an offset pivot. The theorem states: \[ I = I_{\text{COM}} + Md^2 \] where \( I \) is the new moment of inertia, \( I_{\text{COM}} \) is the moment of inertia through the center of mass, \( M \) is the object’s total mass, and \( d \) is the distance between the axes.
• This theorem is especially helpful in problems involving physical pendulums or objects that are not spinning about their geometric center. For example, a uniform rod has \( I_{\text{COM}} = \frac{1}{12}ML^2 \), but if it rotates about one end, we use the parallel axis theorem to find: \[ I = \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{3}ML^2 \] This is critical for understanding how axis location affects rotational inertia and dynamics.

Angular Momentum

• Angular momentum is the rotational analog of linear momentum. For a rotating object, it is defined as: \[ L = I\omega \] where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. Like linear momentum, angular momentum is a vector quantity and obeys a conservation law — meaning that if no net external torque acts on a system, its total angular momentum remains constant.
• For point particles moving in a circle or orbit, angular momentum is given by: \[ L = r \times p = mvr \] where \( r \) is the radius and \( p = mv \) is linear momentum. This expression shows up in systems like planets orbiting the Sun, ice skaters pulling in their arms, or collapsing stars increasing spin rate. As \( r \) decreases, \( \omega \) increases to conserve \( L \), assuming no external torque.
• The time rate of change of angular momentum equals net external torque: \[ \tau = \frac{dL}{dt} \] This is the rotational version of Newton’s Second Law. It helps explain why a spinning wheel resists changes in orientation (gyroscopic stability) and how rotational systems evolve dynamically when torques act. If \( \tau = 0 \), then angular momentum is conserved — a crucial principle on the AP exam.

Equilibrium in Rotational Motion

• A system is in rotational equilibrium when the net torque acting on it is zero. This is a special case of dynamic balance where there is no angular acceleration, and thus the object spins at constant angular velocity or remains at rest. The condition is expressed mathematically as: \[ \sum \tau = 0 \] This must be true about **any pivot point**, not just the center of mass, and is used heavily in statics problems.
• To fully ensure equilibrium, both translational and rotational conditions must be met: \[ \sum F_x = 0, \quad \sum F_y = 0, \quad \sum \tau = 0 \] This guarantees the object doesn’t translate or rotate unexpectedly. In ladder problems, bridge balancing, or scaffolding support systems, all three must be checked to solve for unknowns. A frequent AP trick is to give you a hidden unbalanced torque and ask if the system is really in equilibrium.
• Even rotating objects can be in equilibrium if there’s no change in angular velocity — for example, a merry-go-round spinning at a constant rate. In that case, there’s motion but no net torque or angular acceleration. This distinction between equilibrium and rest is a common misconception and is essential for conceptual understanding.

Example Problem 1: Hanging Rod in Static Equilibrium

A uniform rod of length \( L = 2.0 \, \text{m} \) and mass \( M = 5.0 \, \text{kg} \) is suspended horizontally by two vertical strings: one at its left end and one located \( 0.50 \, \text{m} \) from the right end. The rod remains at rest.

Question: What is the tension in each string?

Solution:

• Step 1: Choose a pivot point. We’ll place the pivot at the left end so the torque due to the left tension becomes zero. Let \( T_1 \) be the tension at the left and \( T_2 \) the tension at the right.
• Step 2: The torque due to the rod’s weight acts at its center of mass, which is at \( 1.0 \, \text{m} \) from the left. The torque from the right tension acts at \( 1.5 \, \text{m} \) from the left.
• Step 3: Apply torque balance: \[ T_2(1.5) - Mg(1.0) = 0 \Rightarrow T_2 = \frac{(5.0)(9.8)(1.0)}{1.5} = 32.67 \, \text{N} \]
• Step 4: Use vertical force balance: \[ T_1 + T_2 = Mg = 49.0 \Rightarrow T_1 = 49.0 - 32.67 = 16.33 \, \text{N} \]
• Final Answer: \( T_1 = 16.33 \, \text{N}, \quad T_2 = 32.67 \, \text{N} \)

Example Problem 2: Angular Acceleration of a Pulley System

A light pulley of radius \( R = 0.20 \, \text{m} \) and moment of inertia \( I = 0.04 \, \text{kg} \cdot \text{m}^2 \) has a rope wrapped around it. A block of mass \( 3.0 \, \text{kg} \) is attached to the rope and released from rest.

Question: What is the angular acceleration \( \alpha \) of the pulley when the block is released?

Solution:

• Step 1: The tension in the rope produces a torque on the pulley: \[ \tau = TR = I\alpha \Rightarrow T = \frac{I\alpha}{R} \]
• Step 2: Apply Newton’s second law to the block: \[ Mg - T = Ma \Rightarrow T = Mg - Ma \]
• Step 3: Link the translational acceleration \( a \) and angular acceleration: \[ a = R\alpha \]
• Step 4: Substitute: \[ T = Mg - M(R\alpha), \quad \text{and} \quad T = \frac{I\alpha}{R} \Rightarrow Mg - MR\alpha = \frac{I\alpha}{R} \]
• Solve for \( \alpha \): \[ (3)(9.8) - (3)(0.2)\alpha = \frac{0.04\alpha}{0.2} \Rightarrow 29.4 - 0.6\alpha = 0.2\alpha \Rightarrow 29.4 = 0.8\alpha \Rightarrow \alpha = 36.75 \, \text{rad/s}^2 \]
• Final Answer: \( \alpha = 36.75 \, \text{rad/s}^2 \)

Example Problem 3: Rotational Kinetic Energy in a Rolling Cylinder

A solid cylinder of mass \( M = 4.0 \, \text{kg} \) and radius \( R = 0.3 \, \text{m} \) rolls without slipping down a \( 2.0 \, \text{m} \) high incline. It starts from rest.

Question: What is the final speed of the center of mass at the bottom?

Solution:

• Step 1: Use conservation of energy. Initial potential energy becomes both translational and rotational kinetic energy: \[ Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 \]
• Step 2: For a solid cylinder, \( I = \frac{1}{2}MR^2 \) and \( \omega = \frac{v}{R} \).
• Step 3: Substitute: \[ Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}\left(\frac{1}{2}MR^2\right)\left(\frac{v^2}{R^2}\right) \Rightarrow Mgh = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2 = \frac{3}{4}Mv^2 \]
• Solve: \[ v^2 = \frac{4gh}{3} = \frac{4(9.8)(2.0)}{3} = \frac{78.4}{3} \Rightarrow v \approx \sqrt{26.13} \approx 5.11 \, \text{m/s} \]
• Final Answer: \( v \approx 5.11 \, \text{m/s} \)

Common Misconceptions: Torque and Rotational Dynamics

• Torque is not the same as force. Many students mistakenly think torque and force are interchangeable because both can cause motion. However, torque depends not only on the magnitude of the force but also on the distance from the pivot and the angle at which the force is applied. For instance, a large force applied close to the axis might produce less torque than a smaller force applied farther away. Torque is the rotational equivalent of force and involves rotational motion around an axis, not just linear displacement.
• The largest torque is not always produced by the largest force. Students often assume that a greater force automatically means more torque. But torque is calculated using \( \tau = rF\sin(\theta) \), and if the angle is not perpendicular or the lever arm is short, even a strong force may result in weak torque. It's the combination of force, angle, and lever arm that determines torque, not the force alone. Misunderstanding this leads to errors when comparing torques acting at different positions or angles.
• In rotational equilibrium, the net torque must be zero—not just the net force. Some students only consider the net force and forget that torques must balance for an object to remain rotationally at rest. An object can have zero net force (so it doesn’t translate), but if the torques aren’t balanced, it will still rotate. True equilibrium means both \( \sum F = 0 \) and \( \sum \tau = 0 \), and both must be considered when solving static problems with extended bodies.
• Direction of torque is a vector concept, not just clockwise or counterclockwise. Students often treat torque as a scalar and forget about the importance of its vector direction. Using the right-hand rule is critical in 3D problems and ensures consistency when adding torques algebraically. Ignoring this can lead to incorrect assumptions about the net effect of multiple torques, especially when dealing with angular momentum or rotational motion about different axes.
• Moment of inertia is not the same for all objects. Many students incorrectly think moment of inertia depends only on mass, but it also depends on how the mass is distributed. A solid disk and a hoop with the same mass and radius will have different moments of inertia. This difference drastically changes how objects accelerate rotationally under the same torque. Understanding this helps avoid mistakes in comparing rolling speeds, angular accelerations, and rotational energies.