AP Physics 1 Torque and Rotational Motion Basics
Defining Torque and Lever Arm
Torque (τ) is the rotational equivalent of force, representing the tendency of a force to cause an object to rotate about a specific axis. On the AP Physics 1 exam, torque is defined mathematically as $\tau = r F sin\theta$, where $r$ is the distance from the pivot point to the point of force application, and $\theta$ is the angle between the force vector and the radial line. A critical concept here is the lever arm (or moment arm), which is the perpendicular distance from the axis of rotation to the line of action of the force. Many students lose points by using the total length of a rod rather than the perpendicular component. If the force is applied parallel to the radial line, the torque is zero because the lever arm is zero. Understanding that torque is a vector quantity—typically assigned a direction of clockwise or counterclockwise—is the first step in analyzing rotational systems.
Calculating Net Torque
To determine the behavior of an extended object, one must calculate the net torque ($Sigma\tau$). This involves identifying all external forces acting on the body and their respective distances from the chosen pivot. A common exam scenario involves a beam supported by two pillars or a see-saw with multiple masses. When calculating $Sigma\tau$, it is vital to establish a consistent sign convention; usually, counterclockwise (CCW) is positive and clockwise (CW) is negative. The choice of pivot point is arbitrary in static systems, but strategic selection can simplify calculations. For instance, placing the pivot at the location of an unknown force eliminates that force from the torque equation because its lever arm becomes zero. This technique is frequently required in Unit 7 problems to isolate variables like tension or normal force.
Conditions for Rotational Equilibrium
An object is in rotational equilibrium if the net torque acting upon it is zero ($Sigma\tau = 0$). This state implies that the object’s angular acceleration is zero, meaning it is either at rest or rotating at a constant angular velocity. However, for a rigid body to be in total mechanical equilibrium, it must satisfy two conditions: the net force must be zero ($Sigma F = 0$) and the net torque must be zero. The AP exam often tests this through "ladder problems" or balanced beams. In these cases, the sum of upward forces must equal downward forces, and the sum of CCW torques must equal CW torques. If a system is in static equilibrium, it is completely stationary. If it is in dynamic equilibrium, it may be moving at a constant speed, but the lack of net torque ensures no change in its rotational state.
Rotational Kinematics
Angular Displacement, Velocity, and Acceleration
Rotational kinematics AP Physics problems require students to describe motion using angular variables. Angular displacement ($Delta\theta$) is measured in radians and represents the change in angular position. Angular velocity ($omega$) is the rate of change of displacement ($Delta\theta / Delta t$), and angular acceleration ($alpha$) is the rate of change of angular velocity ($Deltaomega / Delta t$). Unlike linear variables, these angular quantities apply to the entire rigid body; every point on a spinning record has the same $omega$ and $alpha$, regardless of its distance from the center. Exam questions often test the distinction between these angular rates and the tangential speeds of specific points on the object. Radians are the standard unit; failing to convert revolutions to radians is a common source of calculation error on the AP exam.
The Rotational Kinematics Equations
For systems with constant angular acceleration, the AP Physics 1 equation sheet provides a set of formulas that mirror the standard linear kinematics equations. These include $omega_f = omega_i + alpha t$ and $\theta_f = \theta_i + omega_i t + \frac{1}{2} alpha t^2$. These equations allow students to predict the future state of a rotating object. For example, if a flywheel slows down due to a constant frictional torque, these equations can determine the total number of rotations before it stops. It is essential to recognize the "starting from rest" or "comes to a stop" cues in word problems, which set $omega_i$ or $omega_f$ to zero. Mastery of these equations is often assessed in conjunction with graphing, where the slope of an angular position-time graph represents angular velocity, and the slope of an angular velocity-time graph represents angular acceleration.
Relating Linear and Angular Quantities
Connecting the rotation of a body to the linear motion of a point on that body is a frequent requirement. The bridge between these two worlds is the radius ($r$). The tangential displacement is $s = r\theta$, the tangential velocity is $v = romega$, and the tangential acceleration is $a_t = ralpha$. It is important to distinguish tangential acceleration (which changes the speed) from centripetal acceleration ($a_c = v^2/r = omega^2 r$), which changes the direction of motion. An object can have zero $alpha$ but still have $a_c$ if it is spinning at a constant rate. In AP problems involving strings unwinding from a pulley, the linear acceleration of the falling mass is equal to the tangential acceleration of the pulley's rim, provided the string does not slip.
Rotational Dynamics and Moment of Inertia
Moment of Inertia for Common Shapes
Moment of inertia ($I$) is the rotational analog of mass, measuring an object's resistance to changes in its rotational motion. Unlike mass, $I$ depends on how the mass is distributed relative to the axis of rotation. The general formula $I = Sigma mr^2$ shows that mass located further from the pivot contributes significantly more to the inertia. The AP Physics 1 curriculum expects students to understand qualitatively how shapes like hoops, disks, and spheres differ. For example, a hoop ($I = MR^2$) has a higher moment of inertia than a solid disk ($I = \frac{1}{2}MR^2$) of the same mass and radius because its mass is concentrated at the edge. On the exam, you may be asked to rank objects based on their inertia or explain why a figure skater's $I$ decreases when they pull their arms inward.
Newton's Second Law for Rotation
Rotational dynamics is centered on the angular version of Newton’s Second Law: $\tau_{net} = Ialpha$. This relationship dictates that the net torque applied to a system is directly proportional to the resulting angular acceleration and inversely proportional to the moment of inertia. This is a high-frequency concept on the AP exam, often appearing in Free Response Questions (FRQs) where students must derive the acceleration of a system. If multiple torques act on an object, they must be summed before being set equal to $Ialpha$. This law explains why it is harder to swing a long bat than a short one; the increased length increases $r$ for the mass distribution, raising $I$ and requiring more torque to achieve the same $alpha$.
Applying τ_net = Iα to Problems
When solving complex problems involving both translation and rotation (like a mass hanging from a pulley), students must use a system of equations. One equation describes the linear motion of the mass ($Sigma F = ma$) and another describes the rotation of the pulley ($Sigma\tau = Ialpha$). The link between these equations is usually the constraint $a = ralpha$. For instance, in a system where a block falls and pulls a string wrapped around a disk, the tension in the string creates the torque that accelerates the disk. Solving these problems requires careful free-body diagrams (FBDs) for the block and extended free-body diagrams for the rotating object to show exactly where forces are applied. Correctly identifying the direction of the friction or tension force relative to the pivot is crucial for determining the sign of the torque.
Rotational Kinetic Energy and Work
An object spinning about an axis possesses rotational kinetic energy ($K_{rot} = \frac{1}{2}Iomega^2$). This energy is distinct from translational kinetic energy ($K_{trans} = \frac{1}{2}mv^2$). For a rigid body undergoing pure rotation, the total kinetic energy is simply $K_{rot}$. However, for objects that are both moving through space and spinning, the total mechanical energy includes both components. The AP exam often tests this through energy conservation problems. It is vital to remember that $I$ must be calculated about the axis of rotation. Because $I$ involves the square of the distance from the axis, small changes in mass distribution lead to significant changes in the energy required to reach a certain angular velocity.
Work Done by a Torque
Just as work in linear mechanics is force times displacement ($W = Fd$), the work done by a torque is the product of the torque and the angular displacement: $W = \tauDelta\theta$. This relationship is governed by the Work-Energy Theorem, which states that the net work done on a system equals its change in kinetic energy. If a constant torque is applied to a stationary wheel, the work done will manifest as an increase in $K_{rot}$. AP questions may ask for the power delivered by a motor, where Power ($P$) is the rate of doing work, or $P = \tauomega$. Understanding $W = \tauDelta\theta$ is essential for analyzing systems where external torques, such as friction in a bearing, remove energy from a rotating system over time.
Conservation of Energy with Rotation
In an isolated system where only conservative forces do work, the total mechanical energy ($E_{tot} = U + K_{trans} + K_{rot}$) remains constant. A classic AP Physics 1 scenario involves an object rolling down an incline. At the top, the object has gravitational potential energy ($mgh$). At the bottom, this energy has been converted into both translational and rotational kinetic energy. Consequently, a rolling object will reach the bottom with a lower translational velocity than a sliding object because some of the potential energy had to be "spent" on making the object rotate. Students must be able to set up the equation $mgh = \frac{1}{2}mv^2 + \frac{1}{2}Iomega^2$ and use the rolling constraint $v = romega$ to solve for the final speed.
Angular Momentum
Defining and Calculating Angular Momentum
Angular momentum ($L$) is a measure of the amount of rotation an object has, taking into account its mass, shape, and speed. For a rigid body rotating about a fixed axis, $L = Iomega$. For a point mass moving in a straight line relative to a reference point, $L = mvrsin\theta$, where $r$ is the distance from the reference point to the mass. This second definition is frequently tested in "collision" problems where a ball hits a rod. Angular momentum is a vector, and its direction is determined by the right-hand rule. On the AP exam, the units for $L$ are $kg cdot m^2/s$. It is important to note that an object does not need to be moving in a circle to have angular momentum; any object moving linearly has $L$ relative to any point not on its line of motion.
The Law of Conservation of Angular Momentum
One of the most fundamental laws in physics is the Law of Conservation of Angular Momentum: if the net external torque on a system is zero, the total angular momentum of the system remains constant ($L_i = L_f$). This principle is the rotational equivalent of the conservation of linear momentum. The most famous example is the figure skater: as they pull their arms in, their moment of inertia ($I$) decreases. To keep $L$ constant, their angular velocity ($omega$) must increase. In AP Physics 1, this conservation law is applied to planetary orbits (Kepler’s Second Law) and systems where the internal distribution of mass changes. Note that while $L$ is conserved in these cases, kinetic energy often is not, as internal work is done to move the mass.
Collisions and Angular Momentum
Angular momentum is particularly useful for analyzing collisions involving rotation. A common AP FRQ involves a projectile hitting a pivoted rod. Since the pivot exerts an external force, linear momentum is not conserved. However, if the pivot is the reference point, the force from the pivot exerts no torque (lever arm = 0), meaning angular momentum AP Physics 1 principles can be used to find the post-collision angular velocity. The equation would be $L_{initial, projectile} = L_{final, system}$. Students must be careful to include both the projectile and the rod in the final moment of inertia calculation ($I_{total} = I_{rod} + m_{projectile}r^2$) if the projectile sticks to the rod. This demonstrates the transition from a point mass's $L$ to a rigid body's $L$.
Rolling Motion Without Slipping
Kinetic Energy of a Rolling Object
Rolling without slipping is a special case where an object’s point of contact with the ground is instantaneously at rest. This motion is a combination of translation of the center of mass and rotation about the center of mass. The total kinetic energy is $K = rac{1}{2}Mv_{cm}^2 + rac{1}{2}I_{cm}omega^2$. Because the object is rolling without slipping, there is a direct relationship between its linear and angular speeds. On the AP exam, students are often asked to compare different shapes (like a sphere vs. a cylinder) rolling down a hill. The object with the smallest moment of inertia coefficient (like a solid sphere with $I = 0.4MR^2$) will have the smallest portion of its energy in rotation, leaving more for translation, and thus will reach the bottom first.
The Relationship Between Linear and Angular Speed
The "no-slip condition" is mathematically expressed as $v = Romega$ and $a = Ralpha$. This condition implies that for every full rotation ($2pi$ radians), the object travels a linear distance equal to its circumference ($2pi R$). In the context of the AP Physics 1 exam, this relationship is a critical substitution tool. When asked to find the acceleration of a rolling cylinder, one replaces $omega$ with $v/R$ in the energy equations or $alpha$ with $a/R$ in the dynamics equations. If an object does slip, these equalities no longer hold, and kinetic friction must be accounted for as an external torque and force that changes the motion until the no-slip condition is restored.
Problem-Solving with Rolling Constraints
Solving rolling problems often requires a multi-step approach involving forces and torques. Consider a ball rolling down an incline: static friction is the force responsible for the torque that causes the ball to rotate. Without friction, the ball would simply slide. In an FBD, the friction force acts at the point of contact, pointing up the incline, creating a torque $\tau = f_s R$. By setting $f_s R = Ialpha$ and $Sigma F = mgsin\theta - f_s = ma$, and using the constraint $alpha = a/R$, students can solve for the linear acceleration $a$. This type of derivation is a hallmark of the AP Physics 1 curriculum, requiring a synthesis of rotational dynamics and linear mechanics to describe the motion of real-world objects.