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9.2 · Relate angular position, velocity, and acceleration

Learn to relate angular position, velocity, and acceleration through clear examples and targeted practice.

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Introduction to Rigid-Body Dynamics

Describing rotation about a fixed axis

A rotating object changes its angular position over time. Describing that change requires a reference direction, a positive sense of rotation, and a time coordinate. This lesson focuses on rotation about a fixed axis, as seen by an observer in a stationary reference frame. Angular position tells where the object is around the axis; angular velocity tells how quickly and in which sense that position changes; angular acceleration tells how quickly angular velocity changes. The same ideas describe a point moving around a circle and a planar rigid body rotating about a fixed axis.

What you will learn

  • Define angular position and angular displacement using a stated reference and positive direction.
  • Relate angular velocity and angular acceleration to changes in angular position and velocity.
  • Use calculus or constant-angular-acceleration equations to find rotational motion.
  • Check angular units, signs, initial conditions, and limiting cases.

1. Set the angular reference and describe position

Choose the system and observer before writing equations. For a rotating wheel, the system might be the wheel; the observer is stationary in the room. Use a fixed reference frame with the rotation axis perpendicular to the plane of motion. Choose a reference ray in that plane and define the positive angular direction, commonly counterclockwise when viewed along the axis toward the plane.
Angular position, written as θ\theta, is the signed angle from the reference ray to a line fixed to the rotating object. Angular displacement is the change in angular position, Δθ=θ2−θ1\Delta\theta=\theta_2-\theta_1. It is not necessarily the same as total angle turned: a wheel can make a full revolution and return to its starting position, giving zero net angular displacement but a nonzero amount of rotation.
Use radians in angular kinematics. One complete counterclockwise turn is 2π rad2\pi\ \mathrm{rad}, and a clockwise turn is negative under the usual convention. The angle is dimensionless in SI, but writing radians makes its meaning clear. Angular position depends on the chosen reference ray; changes and rates are consistent when that reference stays fixed.
Δθ=θ2−θ1\Delta\theta=\theta_2-\theta_1
  • State the observer, fixed reference frame, reference ray, and positive rotation direction.
  • Angular position depends on a chosen reference; angular displacement is final position minus initial position.
  • Use radians and preserve the sign convention.

2. Angular velocity and angular acceleration

Average angular velocity is angular displacement divided by elapsed time. Instantaneous angular velocity is the rate of change of angular position. A positive value means motion in the selected positive angular direction; a negative value means motion in the opposite direction. Its SI unit is radians per second.
Average angular acceleration is the change in angular velocity divided by elapsed time. Instantaneous angular acceleration is the rate of change of angular velocity, or the second time derivative of angular position. Its SI unit is radians per second squared.
Angular velocity and angular acceleration are signed quantities. If angular velocity is positive but angular acceleration is negative, the object is initially rotating in the positive sense while its angular speed is decreasing. If both are negative, it is rotating in the negative sense and its angular speed is increasing. The sign of acceleration alone does not tell whether the object is speeding up; compare the signs of angular velocity and angular acceleration.
When a point on a rigid body rotates about a fixed axis, its angular position, velocity, and acceleration describe the body's rotation. The point also travels along a circular path about that axis. Here the focus is the angular relationships, not the point's linear motion.
ω=dθdt,α=dωdt=d2θdt2\omega=\frac{d\theta}{dt},\quad \alpha=\frac{d\omega}{dt}=\frac{d^2\theta}{dt^2}
  • Angular velocity is the time rate of change of angular position.
  • Angular acceleration is the time rate of change of angular velocity.
  • Interpret angular velocity and angular acceleration signs relative to the chosen positive direction.

3. Choose a relationship that matches the information

If angular position is given as a function of time, differentiate it to find angular velocity, then differentiate again to find angular acceleration. If angular acceleration is given as a function of time, integrate to find angular velocity and use the initial angular velocity to determine the integration constant. Integrate again and apply the initial angular position.
When angular acceleration is constant, the familiar constant-acceleration equations apply to angular quantities. They are useful when the knowns and unknowns are among angular position, angular velocity, angular acceleration, and time. Select an equation that contains the desired unknown and the known quantities; do not assume constant acceleration unless the problem states or supports that assumption.
Before calculating, list the initial and final states, including the time interval. Keep angles in radians, time in seconds, angular velocity in radians per second, and angular acceleration in radians per second squared. A negative answer is not an error by itself; check whether it agrees with the stated positive direction.
ω=ω0+αt,θ=θ0+ω0t+12αt2,ω2=ω02+2α(θ−θ0)\omega=\omega_0+\alpha t,\quad \theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2,\quad \omega^2=\omega_0^2+2\alpha(\theta-\theta_0)
  • Differentiate position to obtain velocity and acceleration.
  • Integrate acceleration using the initial conditions to recover velocity and position.
  • Use constant-angular-acceleration equations only when angular acceleration is constant.

4. Check signs, units, and initial conditions

A motion description should agree with the selected sign convention. If counterclockwise is positive, a clockwise angular displacement, angular velocity, or angular acceleration is negative. For a constant angular velocity, angular acceleration must be zero and angular position must change linearly with time.
Check dimensions as well as signs. Differentiating an angle with respect to time gives units of inverse seconds, conventionally written as radians per second. Differentiating again gives radians per second squared. In the constant-acceleration position equation, each term has angular units: the terms involving time must combine to give radians.
Check initial conditions by substituting zero elapsed time. The calculated angular position and angular velocity should equal their stated initial values. For a limiting check, let angular acceleration approach zero in a constant-acceleration equation: the result should reduce to constant-angular-velocity motion.
α=0⟹ω=ω0,θ=θ0+ω0t\alpha=0\Longrightarrow\omega=\omega_0,\quad\theta=\theta_0+\omega_0t
  • A signed result must agree with the chosen positive direction.
  • Check angular units and initial conditions by substitution.
  • With zero angular acceleration, angular velocity is constant.

Worked example

Differentiate a stated angular position

A marked line on a wheel rotates about a fixed axis. A stationary observer chooses the initial marked direction as the reference ray and counterclockwise as positive. From the initial state at t=0t=0, the angular position is θ(t)=0.40+1.20t−0.15t2\theta(t)=0.40+1.20t-0.15t^2, with angle in radians and time in seconds. Find angular velocity and acceleration at t=2.0 st=2.0\ \mathrm{s}, and state the direction of rotation.
Wheel rotation
Wheel rotationReference rayPositive angular dirFixed-axis rotationMarked lineInitial angular velocity: 1.20 rad/s

The circular path is counterclockwise; the labelled arrow indicates the initial positive sense of rotation.

  1. Define the states and sign
    The system is the wheel's marked line, the observer is stationary, and the frame is fixed to the room. The reference ray is its initial direction; counterclockwise is positive. The initial state is at t=0t=0, and the requested state is at t=2.0 st=2.0\ \mathrm{s}. The given function supplies angular position directly.
  2. Differentiate position
    Instantaneous angular velocity is the time derivative of position. Differentiating the stated function gives velocity as a function of time; differentiating once more gives angular acceleration.
    ω(t)=1.20−0.30t,α(t)=−0.30\omega(t)=1.20-0.30t,\quad\alpha(t)=-0.30
  3. Evaluate at the requested time
    At t=2.0 st=2.0\ \mathrm{s}, angular velocity is positive, so the marked line is rotating counterclockwise. The negative angular acceleration means that this positive rotation is slowing at that instant.
    ω(2.0)=0.60 rad/s,α(2.0)=−0.30 rad/s2\omega(2.0)=0.60\ \mathrm{rad/s},\quad\alpha(2.0)=-0.30\ \mathrm{rad/s^2}
Answer: At 2.0 s2.0\ \mathrm{s}, ω=0.60 rad/s\omega=0.60\ \mathrm{rad/s} counterclockwise and α=−0.30 rad/s2\alpha=-0.30\ \mathrm{rad/s^2}, where the negative acceleration is clockwise.
Check: The initial velocity from the differentiated expression is 1.20 rad/s1.20\ \mathrm{rad/s}, matching the linear-in-time position term. The acceleration is constant and negative, so velocity decreases from its initial positive value, consistent with the value at 2.0 s2.0\ \mathrm{s}. The derivative units are radians per second and radians per second squared.

Worked example

Use constant angular acceleration

A turntable starts at angular position θ0=0.30 rad\theta_0=0.30\ \mathrm{rad} with angular velocity ω0=2.0 rad/s\omega_0=2.0\ \mathrm{rad/s}. It has constant angular acceleration α=0.50 rad/s2\alpha=0.50\ \mathrm{rad/s^2} for 4.0 s4.0\ \mathrm{s}. A stationary observer uses the turntable's initial direction as the reference and counterclockwise as positive. Find its final angular velocity and angular displacement.
Turntable motion
Turntable motionReference rayPositive angular dirFixed-axis rotationTurntableInitial angular velocity: 2.0 rad/s

The stated angular velocity and acceleration are both positive.

  1. Identify knowns and unknowns
    The system is the turntable, viewed from a stationary room frame. The reference is its initial direction; counterclockwise is positive. The initial state is θ0=0.30 rad\theta_0=0.30\ \mathrm{rad} and ω0=2.0 rad/s\omega_0=2.0\ \mathrm{rad/s}; the final state is after 4.0 s4.0\ \mathrm{s}. Since acceleration is constant, use the constant-angular-acceleration relations.
  2. Find final angular velocity
    Use the velocity relation because the initial velocity, acceleration, and elapsed time are known. The positive result indicates counterclockwise rotation.
    ω=2.0+(0.50)(4.0)=4.0 rad/s\omega=2.0+(0.50)(4.0)=4.0\ \mathrm{rad/s}
  3. Find angular displacement and final position
    The angular displacement follows from the constant-acceleration position relation. Adding it to the initial position gives the final angular position measured from the chosen reference ray.
    Δθ=(2.0)(4.0)+12(0.50)(4.0)2=12 rad,θ=12.3 rad\Delta\theta=(2.0)(4.0)+\frac{1}{2}(0.50)(4.0)^2=12\ \mathrm{rad},\quad\theta=12.3\ \mathrm{rad}
Answer: The final angular velocity is 4.0 rad/s4.0\ \mathrm{rad/s} counterclockwise. The angular displacement is 12 rad12\ \mathrm{rad} counterclockwise, and the final angular position from the reference is 12.3 rad12.3\ \mathrm{rad}.
Check: Because positive acceleration acts in the same sense as positive initial velocity, angular speed increases from 2.02.0 to 4.0 rad/s4.0\ \mathrm{rad/s}. The average angular velocity is 3.0 rad/s3.0\ \mathrm{rad/s}, which over 4.0 s4.0\ \mathrm{s} gives 12 rad12\ \mathrm{rad}, agreeing with the displacement.

Worked example

Integrate a time-varying angular acceleration

A disk's angular acceleration is α(t)=(0.80 rad/s3)t\alpha(t)=(0.80\ \mathrm{rad/s^3})t, where time is measured in seconds from the start of observation. At t=0t=0, its angular velocity is −1.0 rad/s-1.0\ \mathrm{rad/s} and its angular position is 0.20 rad0.20\ \mathrm{rad}. A stationary observer uses the initial disk direction as reference and counterclockwise as positive. Find angular velocity and position at t=3.0 st=3.0\ \mathrm{s}, and determine its rotation direction then.
Disk rotation
Disk rotationReference rayPositive angular dirFixed-axis rotationDiskInitial angular velocity: -1.0 rad/s

The initial angular velocity is negative; the stated angular acceleration becomes positive for positive time.

  1. Define the initial and final states
    The system is the disk, and the observer is stationary in the room frame. The initial disk direction is the reference ray; counterclockwise is positive. The initial state is ω0=−1.0 rad/s\omega_0=-1.0\ \mathrm{rad/s} and θ0=0.20 rad\theta_0=0.20\ \mathrm{rad} at t=0t=0. Since acceleration varies with time, integrate it and apply these initial conditions.
  2. Integrate to find angular velocity
    Integrating acceleration gives velocity up to a constant. The condition at t=0t=0 sets that constant to the given initial angular velocity.
    ω(t)=−1.0+(0.40 rad/s3)t2\omega(t)=-1.0+(0.40\ \mathrm{rad/s^3})t^2
  3. Integrate to find angular position
    Integrating the velocity expression gives position up to a constant. Use the initial position to determine that constant, then evaluate both quantities at 3.0 s3.0\ \mathrm{s}.
    θ(t)=0.20 rad−(1.0 rad/s)t+(0.40 rad/s3)t33,ω(3.0)=2.6 rad/s,θ(3.0)=0.80 rad\theta(t)=0.20\ \mathrm{rad}-(1.0\ \mathrm{rad/s})t+(0.40\ \mathrm{rad/s^3})\frac{t^3}{3},\quad\omega(3.0)=2.6\ \mathrm{rad/s},\quad\theta(3.0)=0.80\ \mathrm{rad}
Answer: At 3.0 s3.0\ \mathrm{s}, the disk has angular velocity 2.6 rad/s2.6\ \mathrm{rad/s} counterclockwise and angular position 0.80 rad0.80\ \mathrm{rad} from the reference ray.
Check: At t=0t=0, the integrated expressions give the stated initial velocity and position. The velocity becomes zero when −1.0+0.40t2=0-1.0+0.40t^2=0, at t=2.5 st=\sqrt{2.5}\ \mathrm{s}; after that it is positive. Thus positive velocity at 3.0 s3.0\ \mathrm{s} is consistent with the changing acceleration and the initial clockwise rotation.

Common mistakes and how to avoid them

Treating angular position, angular displacement, and total angle turned as the same quantity.
Correction: Angular position is measured from a reference ray; displacement is final minus initial position. Total angle turned counts the motion along the rotation and can differ from net displacement.
Assuming a negative angular acceleration always means the object is slowing down.
Correction: Compare the signs of angular velocity and acceleration. Opposite signs mean angular speed is decreasing at that instant; matching signs mean it is increasing.
Using constant-angular-acceleration equations when acceleration varies with time.
Correction: For a stated variable acceleration, integrate it and use the initial conditions to determine the constants.
Dropping the sign of a clockwise quantity or switching the positive direction during a calculation.
Correction: Choose one positive angular direction at the start and use it consistently in position, velocity, and acceleration.

Lesson summary

  • Angular position is measured from a selected reference direction, and angular displacement is final position minus initial position.
  • Angular velocity is the time derivative of angular position; angular acceleration is the time derivative of angular velocity.
  • Differentiate a known position function, integrate a known acceleration function, or use constant-acceleration relations only when acceleration is constant.
  • Use radians, retain signed values, apply initial conditions, and check direction, units, and limiting behaviour.

Check your understanding

Question 1

Counterclockwise is positive. At an instant, a disk has ω=−3.0 rad/s\omega=-3.0\ \mathrm{rad/s} and α=+0.50 rad/s2\alpha=+0.50\ \mathrm{rad/s^2}. What is happening to its angular speed at that instant?
  1. It is rotating clockwise and slowing down.
  2. It is rotating clockwise and speeding up.
  3. It is rotating counterclockwise and slowing down.
  4. It is rotating counterclockwise at constant speed.
Show answer and explanation
It is rotating clockwise and slowing down.
Negative angular velocity means clockwise rotation. The acceleration has the opposite sign, so it reduces the magnitude of angular velocity at that instant.

Question 2

An object has constant angular velocity ω=1.5 rad/s\omega=1.5\ \mathrm{rad/s} and starts at θ0=0.20 rad\theta_0=0.20\ \mathrm{rad}. What is its angular position after 2.0 s2.0\ \mathrm{s}?
  1. 1.7 rad1.7\ \mathrm{rad}
  2. 3.0 rad3.0\ \mathrm{rad}
  3. 3.2 rad3.2\ \mathrm{rad}
  4. 0.20 rad0.20\ \mathrm{rad}
Show answer and explanation
3.2 rad3.2\ \mathrm{rad}
With constant angular velocity, the displacement is (1.5)(2.0)=3.0 rad(1.5)(2.0)=3.0\ \mathrm{rad}. Adding the initial position gives 3.2 rad3.2\ \mathrm{rad}.

Key terms

Angular position
The signed angle locating a rotating line relative to a chosen reference direction.
Angular displacement
The change in angular position between two states.
Angular velocity
The rate of change of angular position, including its direction sign.
Angular acceleration
The rate of change of angular velocity.
Initial condition
A known value of position or velocity at a stated starting time, used to determine constants in an integrated relationship.

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