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Angular acceleration

Angular acceleration is the rate of change of angular velocity with respect to time, characterizing the rotational equivalent of linear acceleration in objects undergoing rotational motion around a fixed axis.^[1] It is typically denoted by the Greek letter α and quantified using the formula α = Δω / Δt, where Δω represents the change in angular velocity and Δt is the corresponding change in time.^[1] The standard unit for angular acceleration is radians per second squared (rad/s²), reflecting its dimensional basis in angle per unit time squared.^[1] In rotational kinematics, angular acceleration connects directly to linear motion through the tangential acceleration a_t = r α, where r is the radius from the axis of rotation, enabling the analysis of curved paths in systems like wheels or gears.^[1] For constant angular acceleration, kinematic equations analogous to those for linear motion apply, such as ω = ω_0 + α t and θ = θ_0 + ω_0 t + (1/2) α t², facilitating predictions of rotational displacement and final velocity.^[2] Within rotational dynamics, angular acceleration arises from net torque according to Newton's second law for rotation: Στ = I α, where I is the moment of inertia quantifying an object's resistance to angular change based on its mass distribution.^[3] This relationship underscores angular acceleration's role in engineering applications, such as vehicle propulsion where torque from engines produces wheel acceleration, or in machinery like grindstones where varying loads alter rotational speedup.^[1] Understanding angular acceleration is fundamental to fields including mechanics, robotics, and astrophysics, where it models phenomena from planetary orbits to spinning spacecraft.

Fundamentals

Definition

Angular acceleration, denoted by the symbol α, is defined as the rate of change of angular velocity with respect to time.^[4] It is mathematically expressed as

\alpha = \frac{d\omega}{dt}

where ω represents angular velocity.^[5] This quantity quantifies how quickly the rotational speed of an object changes and applies to both point particles undergoing curvilinear motion and extended bodies in rotation. The average angular acceleration over a time interval Δt is calculated as

\bar{\alpha} = \frac{\Delta\omega}{\Delta t}

where Δω is the change in angular velocity during that interval.^[4] The instantaneous angular acceleration corresponds to the limit of this average as Δt approaches zero, yielding the derivative form.^[4]

Mathematical representation

Angular acceleration is mathematically represented in scalar form for rotations about a fixed axis, typically in two dimensions, as the second derivative of the angular displacement \theta with respect to time t: \alpha = \frac{d^2\theta}{dt^2}.^[6] This expression captures the instantaneous rate of change of angular velocity \omega = \frac{d\theta}{dt}, where \alpha has units of radians per second squared.^[1] In three dimensions, angular acceleration is treated as a vector quantity \vec{\alpha}, defined as the time derivative of the angular velocity vector \vec{\omega}: \vec{\alpha} = \frac{d\vec{\omega}}{dt}.^[7] The vector \vec{\omega} points along the axis of rotation following the right-hand rule, and \vec{\alpha} similarly aligns with the instantaneous axis, indicating both the magnitude and direction of the change in rotational speed.^[8] Notation conventions for vectors include boldface (\mathbf{\alpha}) or arrows (\vec{\alpha}) to distinguish them from scalar quantities.^[9] For rotations about a constant axis, the general time-dependent expression simplifies to the scalar form \alpha(t) = \frac{d^2\theta(t)}{dt^2}, where \theta(t) describes the angular displacement as a function of time.^[6] This formulation assumes the axis direction remains fixed, allowing the vector \vec{\alpha} to be parallel or antiparallel to \vec{\omega} depending on whether the rotation is speeding up or slowing down.^[8]

Planar rotation

Point particle in two dimensions

In the context of a point particle undergoing circular motion in a plane with constant radius r, the angular acceleration \alpha is defined as the time rate of change of the angular velocity \omega, and it relates directly to the tangential component of the linear acceleration a_t by the expression \alpha = \frac{a_t}{r}. This relation holds because the tangential acceleration arises solely from changes in the magnitude of the velocity, while the radial (centripetal) acceleration a_c = \frac{v^2}{r} = r \omega^2 does not contribute to \alpha.^[10] Assuming constant angular acceleration \alpha, the kinematic equations for the angular motion of the particle mirror those of linear kinematics and are given by \omega = \omega_0 + \alpha t and \theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2, where \omega_0 and \theta_0 are the initial angular velocity and position, respectively.^[10] These equations describe the evolution of the angular position \theta and velocity over time t for motion about a fixed axis in the plane. The sign convention follows the right-hand rule, with positive \alpha corresponding to counterclockwise acceleration when viewing the plane such that the rotation axis points toward the observer.^[10] For more general planar motion where the radius r may vary, such as in polar coordinates, the theta-component of acceleration is a_\theta = r \alpha + 2 \dot{r} \omega, where \dot{r} = \frac{dr}{dt} is the radial velocity, leading to \alpha = \frac{a_\theta - 2 \dot{r} \omega}{r}.^[11] If \dot{r} = 0, this simplifies to the constant-radius case. An illustrative example is a point particle of mass m attached to a string being reeled in toward the center at a constant radial speed |\dot{r}|, with no tangential forces acting (so a_\theta = 0); here, \alpha = -\frac{2 \dot{r} \omega}{r}, meaning that as the radius decreases (\dot{r} < 0), the angular acceleration becomes positive for positive \omega, causing the angular speed to increase to conserve angular momentum in the absence of torque. This demonstrates how radial motion induces angular acceleration even without direct tangential influences.

Rigid body in two dimensions

In the context of planar rotation, angular acceleration describes the rate of change of angular velocity for a rigid body, where the body undergoes rotation about a fixed axis perpendicular to the plane of motion. Unlike a point particle, which may follow an orbital path with position-dependent angular acceleration, a rigid body maintains a fixed shape and relative distances between its points, ensuring that the angular acceleration \alpha is uniform and identical for every point within the body. This uniformity arises because the body's rigidity constrains all parts to rotate together as a single unit, with \alpha defined as the second derivative of the angular displacement \theta with respect to time: \alpha = \frac{d^2 \theta}{dt^2}.^[12]^[9] The kinematics of a rigid body in two dimensions relate the angular acceleration to the linear acceleration of any point on the body at a perpendicular distance r from the axis of rotation. The total linear acceleration \mathbf{a} of a point is given by \mathbf{a} = \boldsymbol{\alpha} \times \mathbf{r} + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}), where \boldsymbol{\omega} is the angular velocity vector (directed along the axis) and \mathbf{r} is the position vector from the axis to the point. In the planar case, this decomposes into a tangential component a_t = r \alpha, which is perpendicular to \mathbf{r} and responsible for changes in speed along the circular path, and a centripetal (normal) component a_n = r \omega^2, directed toward the axis. The tangential acceleration a_t varies linearly with r but shares the same \alpha across the body, highlighting how the overall rotational dynamics propagate uniformly.^[13]^[14] For dynamics, the angular acceleration of a rigid body is linked to the net torque \tau about the axis through the rotational form of Newton's second law: \tau = I \alpha, where I is the moment of inertia, a scalar measure of the body's mass distribution relative to the axis that quantifies its resistance to angular acceleration. This equation sets up the framework for analyzing how external torques induce uniform \alpha, with full derivations deferred to torque relations. Consider a uniformly accelerating disk, such as a wheel with constant \alpha: points farther from the center experience greater tangential acceleration (e.g., a_t = \alpha r), yet all share the same \alpha, demonstrating the body's cohesive rotational behavior independent of specific point locations.^[15]^[14]

Spatial rotation

Point particle in three dimensions

In three dimensions, the orbital motion of a point particle is characterized by vector quantities that capture both the magnitude and direction of its rotation about a reference point, such as the origin. The angular momentum vector \mathbf{L} of the particle with position \mathbf{r} and linear momentum \mathbf{p} = m \mathbf{v} is defined as \mathbf{L} = \mathbf{r} \times \mathbf{p}, which points perpendicular to the plane spanned by \mathbf{r} and \mathbf{v}.^[16] The instantaneous angular velocity vector \boldsymbol{\omega} is then \boldsymbol{\omega} = \mathbf{L} / (m r^2) = (\mathbf{r} \times \mathbf{v}) / r^2, where r = |\mathbf{r}|, and this vector lies along the instantaneous axis of rotation perpendicular to the plane of motion.^[17] The angular acceleration vector \boldsymbol{\alpha} is the time derivative \boldsymbol{\alpha} = d\boldsymbol{\omega}/dt. For cases where the radial distance r is constant, such as uniform circular motion, the relation simplifies to the scalar angular acceleration \alpha = (d\mathbf{L}/dt) / (m r^2) in magnitude, with \boldsymbol{\alpha} perpendicular to the plane of motion.^[18] In general three-dimensional paths, described using spherical coordinates (r, \theta, \phi), the angular acceleration components relate to the second time derivatives of the angular coordinates, such as d^2\theta/dt^2 for polar motion and d^2\phi/dt^2 for azimuthal motion, reflecting changes in both the magnitude |\boldsymbol{\omega}| and direction of rotation. The magnitude is |\boldsymbol{\alpha}| = |d\boldsymbol{\omega}/dt|, and the vector \boldsymbol{\alpha} can have components parallel to \boldsymbol{\omega} (changing speed) or perpendicular to it (changing the rotation axis).^[17] A representative example is a point particle in an elliptical orbit under a central gravitational force, where the motion lies in a fixed plane. Here, \mathbf{L} remains constant in both magnitude and direction due to the central nature of the force, ensuring \boldsymbol{\omega} and \boldsymbol{\alpha} are perpendicular to the orbital plane. Although |\mathbf{L}| is conserved, r varies along the orbit, causing the angular speed |\boldsymbol{\omega}| = L / (m r^2) to increase near perigee (minimum r) and decrease near apogee (maximum r), as dictated by Kepler's second law of equal areas swept in equal times. This variation in |\boldsymbol{\omega}| results in non-zero \boldsymbol{\alpha}, whose magnitude fluctuates due to the continuously changing direction and speed of the tangential velocity component.^[19]

Rigid body in three dimensions

In three-dimensional space, the angular acceleration \boldsymbol{\alpha} of a rigid body is a vector quantity defined as the time derivative of the angular velocity vector \boldsymbol{\omega}, i.e., \boldsymbol{\alpha} = \dot{\boldsymbol{\omega}}, expressed in a body-fixed frame aligned with the principal axes of inertia.^[20] The components of \boldsymbol{\alpha} along these principal axes, denoted \alpha_x = \dot{\omega}_x, \alpha_y = \dot{\omega}_y, and \alpha_z = \dot{\omega}_z, account for the body's rotational dynamics, where the inertia tensor \mathbf{I} is diagonal with principal moments I_{xx}, I_{yy}, and I_{zz}.^[21] This vectorial representation allows for rotations about multiple axes simultaneously, distinguishing it from planar cases by incorporating the full tensorial nature of the body's inertia.^[22] The relationship between angular acceleration, angular momentum \mathbf{L} = \mathbf{I} \boldsymbol{\omega}, and external torque \boldsymbol{\tau} is governed by Euler's rigid body equations in vector form: \boldsymbol{\tau} = \mathbf{I} \dot{\boldsymbol{\omega}} + \boldsymbol{\omega} \times (\mathbf{I} \boldsymbol{\omega}).^[22] In the principal axes frame, this expands to the component equations:

\tau_x = I_{xx} \dot{\omega}_x - (I_{yy} - I_{zz}) \omega_y \omega_z,

\tau_y = I_{yy} \dot{\omega}_y - (I_{zz} - I_{xx}) \omega_z \omega_x,

\tau_z = I_{zz} \dot{\omega}_z - (I_{xx} - I_{yy}) \omega_x \omega_y.

^[21] These equations reveal that \boldsymbol{\alpha} arises not only from direct torque but also from cross-coupling terms involving \boldsymbol{\omega}, leading to phenomena like nutation or precession when rotation is not about a principal axis. For asymmetric bodies (where I_{xx} \neq I_{yy} \neq I_{zz}), \boldsymbol{\alpha} generally does not align with \boldsymbol{\omega}, as the cross-product term \boldsymbol{\omega} \times \mathbf{L} induces perpendicular components in the acceleration.^[20] A classic example is the spinning top or gyroscope, where gravitational torque causes angular acceleration that alters the direction of \boldsymbol{\omega} rather than its magnitude, resulting in steady precession (constant tilt angle with azimuthal rotation) or nutation (oscillatory wobbling).^[23] In this case, the instantaneous angular acceleration about the symmetry axis combines with transverse components to produce the observed motion, as described by Euler's equations in the body frame using Euler angles for orientation.^[23] For a symmetric top with high spin rate, the precession angular velocity \dot{\phi} approximates \tau / (I \omega_z), where \boldsymbol{\alpha} primarily drives the directional change without significant spin deceleration.^[22]

Dynamics and relations

Relation to torque

Angular acceleration arises from the application of torque, analogous to how linear acceleration results from force in Newton's second law. For rotation about a fixed axis, the net torque \vec{\tau} equals the moment of inertia I times the angular acceleration \vec{\alpha}, expressed as \vec{\tau} = I \vec{\alpha}.^[24] Here, I is a scalar quantity in two-dimensional planar rotation but becomes the inertia tensor \mathbf{I} in three dimensions, accounting for the distribution of mass relative to the axis of rotation.^[24] For a point particle of mass m at a perpendicular distance r from the axis, the torque is \vec{\tau} = \vec{r} \times \vec{F}, where \vec{F} is the applied force. This leads to the moment of inertia I = m r^2, and thus \vec{\alpha} = \vec{\tau} / (m r^2), derived from the linear relation \vec{F} = m \vec{a} with tangential acceleration a = r \alpha.^[24] In the case of rigid bodies, the general vector equation relates torque to the rate of change of angular momentum \vec{L}: \vec{\tau} = \frac{d\vec{L}}{dt}. For a rigid body, \vec{L} = \mathbf{I} \vec{\omega}, where \vec{\omega} is the angular velocity, so \vec{\tau} = \mathbf{I} \vec{\alpha} + \vec{\omega} \times \vec{L}.^[25] This form reduces to \vec{\tau} = \mathbf{I} \vec{\alpha} when the rotation is about a principal axis or when the cross product term vanishes, such as in planar motion or fixed-axis rotation.^[16] Consider twisting a door handle: the applied torque from the hand's force at a distance from the hinge produces an angular acceleration inversely proportional to the door's moment of inertia, illustrating how larger I (due to mass distribution) requires greater torque for the same \alpha.^[24]

Relation to linear acceleration

In rotational motion, the linear acceleration \mathbf{a} of a point at position vector \mathbf{r} relative to the axis of rotation is given by the vector equation \mathbf{a} = \boldsymbol{\alpha} \times \mathbf{r} + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}), where \boldsymbol{\alpha} is the angular acceleration and \boldsymbol{\omega} is the angular velocity.^[26] The first term, \boldsymbol{\alpha} \times \mathbf{r}, represents the tangential acceleration component arising directly from the angular acceleration, which is perpendicular to \mathbf{r} and drives changes in the linear speed. The second term, \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}), accounts for the centripetal acceleration, directed toward the axis and dependent on the instantaneous angular velocity.^[27] This decomposition highlights how angular acceleration contributes to the overall linear motion of points in a rotating system, independent of the centripetal effect.^[27] In two-dimensional planar rotation, the relation simplifies due to the perpendicular nature of the vectors. The tangential linear acceleration a_t is a_t = r \alpha, where r is the radial distance from the axis and \alpha is the magnitude of the angular acceleration.^[4] The radial (centripetal) component is a_r = -r \omega^2, pointing inward, with \omega as the angular speed.^[27] These components combine to yield the total linear acceleration magnitude a = \sqrt{a_t^2 + a_r^2}, illustrating how angular acceleration influences the path and speed of the point.^[27] For rigid bodies, the angular acceleration \boldsymbol{\alpha} is uniform across all points, as the body rotates as a single unit without deformation. However, the resulting linear acceleration \mathbf{a} varies with the position \mathbf{r} of each point relative to the rotation axis, following the same vector relation.^[26] Points farther from the axis experience larger tangential accelerations for the same \boldsymbol{\alpha}, while the centripetal term scales with r as well. This variation is key to understanding the differential motion within the body.^[27] A practical example occurs in an accelerating car's drive wheel, where the angular acceleration \alpha of the wheel produces a tangential linear acceleration a_t = r \alpha at the tire's contact point with the road, directly contributing to the vehicle's forward linear acceleration.^[4] Here, r is the wheel radius, and this tangential component propels the car, while the centripetal acceleration maintains the wheel's circular path but does not affect the net translational motion of the vehicle.^[4]

References

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