Virginia Research Institute
Virginia Research Institute
Virtual Laboratory  ·  Built by E2 Innovations
← Back to Virtual Labs|Angular Displacement and Angular Acceleration
Mechanics · Rotational Motion

Angular Displacement and Angular Acceleration

Send a ladybug around a circular track at a chosen angular velocity and watch the angle sweep out over time. Measure the angular displacement, see why a constant angular velocity means zero angular acceleration, and connect the angular quantities to the tangential speed and centripetal acceleration of the moving bug.

Theory — Angular Motion

Angular Displacement

When an object moves around a circle, its position is described by an angle θ measured from a starting line, in radians. The change in this angle is the angular displacement. One full revolution is 2π radians, so the angular displacement can be many radians if the object goes around several times.

Angular Displacement at Constant Angular Velocity θ = ω · t

θ = angular displacement (rad)
ω = angular velocity (rad/s)
t = time (s)
Number of revolutions = θ / (2π)

Angular Velocity and Angular Acceleration

The angular velocity ω is how fast the angle changes (radians per second). The angular acceleration α is how fast the angular velocity changes. If the object goes around at a steady rate — a constant angular velocity — then ω does not change, so the angular acceleration is zero.

Angular Acceleration α = Δω / Δt

If ω is constant, Δω = 0
Constant angular velocity → α = 0

Linking to Linear Quantities

A point on the rim at radius r has a tangential speed v = ωr directed along the circle. Even when the angular velocity is constant, the direction of this velocity keeps changing, so there is a centripetal acceleration ac = ω²r pointing toward the centre. This centripetal acceleration is a linear acceleration that changes the direction of motion; it is not the same as angular acceleration, which would change the rate of spin.

Tangential Speed and Centripetal Acceleration v = ω · r  (tangential speed, m/s)
a_c = ω² · r  (centripetal acceleration, m/s²)
Present even when α = 0 — direction changes, spin rate does not

More time

At constant ω, the angular displacement grows in direct proportion to time (θ = ωt).

Constant spin

If ω does not change, the angular acceleration α is zero — no matter how fast the spin.

Still accelerating

The bug has centripetal acceleration ω²r toward the centre even when α = 0.

QuantitySymbolEquation / meaning
Angular displacementθθ = ωt at constant ω (rad)
Angular velocityωrate of change of angle (rad/s)
Angular accelerationαα = Δω/Δt; zero if ω constant
Tangential speedvv = ωr (m/s)
Centripetal accelerationa_ca_c = ω²r (m/s²)
RevolutionsNN = θ/(2π)

Instructions — Running the Virtual Experiment

The Rotation tab spins the ladybug at a chosen angular velocity and reads the angle, time, and related quantities; the Graphs tab plots the angle and angular velocity against time. Record every reading in your lab notebook. Angles are shown in radians.

Experiment 1 — Angular Displacement (Rotation tab)
1
Open Simulation → Rotation. The track has a diameter of 3 m (radius 1.5 m). Set the angular velocity ω with the slider, or click a scenario button to load the angular velocity and run time for that scenario.
2
Press Play. The ladybug travels around the track and the readout shows the elapsed time, the accumulated angle θ in radians, the number of revolutions, and the angular velocity. When a target time is set, the run pauses automatically at that time so you can read the angular displacement.
3
Record the angular displacement θ for each scenario, then confirm it with the equation θ = ωt. Note that ω stays constant, so the angular acceleration is zero. The velocity vector (tangent) and acceleration vector (toward the centre) show the linear motion.
Experiment 2 — Angle and Angular Velocity vs Time (Graphs tab)
1
Open Graphs and run a scenario. The θ-vs-time graph is a straight line whose slope is the angular velocity ω, and the ω-vs-time graph is a flat horizontal line, confirming that the angular acceleration (the slope of ω vs t) is zero.
2
Read the slope of the θ-vs-time line and confirm it matches your set angular velocity. Tabulate θ, ω, and t for all three scenarios in your report.

Simulation — Ladybug on a Rotating Track

Angular Motion Virtual LabSet angular velocity, then press Play
velocity (tangent)
acceleration (centripetal)
Track diameter 3 m · angle in radians.

Controls

Scenarios (ω, run time)

Readout (radius 1.5 m)
Time t0.00 s
Angular displ. θ0.00 rad
Revolutions0.00
Angular velocity ω5.00 rad/s
Angular accel. α0.00 rad/s²
X position1.50 m
Y position0.00 m
Tangential v = ωr7.50 m/s
Centripetal a = ω²r37.50 m/s²
Set ω (or a scenario), then press Play.
θ vs time (slope = ω)
ω vs time (flat → α = 0)

Run a scenario, then watch the graphs

Graph readout
Slope of θ vs t— rad/s
= angular velocity ω— rad/s
Slope of ω vs t0 rad/s²

Team Questions

Question 1. A ladybug moves at a constant angular velocity of 5 rad/s for 12 s. What is its angular displacement? Use θ = ωt. (Type just the number in radians)
Question 2. What is the angular acceleration in Question 1, given that the angular velocity stays constant the whole time? (Type just the number in rad/s²)
Question 3. A ladybug moves at 6 rad/s for 22 s. Find the angular displacement. (Type just the number in radians)
Question 4. A ladybug moves at 7 rad/s for 6 s. Find the angular displacement. (Type just the number in radians)
Question 5. For the 60-rad displacement in Question 1, how many complete revolutions did the ladybug make? Use N = θ/(2π). (Type just the number, to one decimal place — e.g. 9.5)
Question 6. On a track of radius 1.5 m, what is the tangential (linear) speed of a ladybug moving at 5 rad/s? Use v = ωr. (Type just the number in m/s)
Question 7 — Challenge. The ladybug moves at constant angular velocity, so its angular acceleration is zero. Yet it still has a non-zero acceleration. What is this acceleration called, and which way does it point? (Name it in one or two words)

Example Lab Report

Sample report demonstrating the expected format and level of detail. Use as a guide for your own submission.

Angular Displacement and Angular Acceleration

Physics | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Purpose

To investigate the rotational motion of a ladybug on a circular track of diameter 3 m, to measure the angular displacement produced by a constant angular velocity over a set time, and to compare the measured angular displacement with the theoretical value θ = ωt. The experiment also examines why the angular acceleration is zero when the angular velocity is constant.

Theory

For motion at constant angular velocity, the angular displacement increases in direct proportion to time, θ = ωt, where θ is in radians. The angular acceleration is the rate of change of angular velocity, α = Δω/Δt; if the angular velocity is constant then α = 0. A point at radius r has tangential speed v = ωr and centripetal acceleration a_c = ω²r directed toward the centre.

θ = ω·t · α = Δω/Δt · N = θ/(2π)
v = ω·r · a_c = ω²·r (r = 1.5 m)

Calculations — Sample: ω = 5 rad/s, t = 12 s

Angular displacement: θ = ωt = 5 × 12 = 60 rad

Revolutions: N = 60 / (2π) = 9.55 rev

Angular acceleration: ω is constant, so α = 0 rad/s²

Tangential speed: v = ωr = 5 × 1.5 = 7.5 m/s; Centripetal accel: a_c = ω²r = 25 × 1.5 = 37.5 m/s²

Results Table (diameter 3 m, r = 1.5 m)

Scenarioω (rad/s)t (s)θ = ωt (rad)Revolutionsα (rad/s²)
1512609.550
262213221.010
376426.680

The angular displacement read from the simulation matched the theoretical θ = ωt for each scenario.

Discussion

The angular displacement measured from the simulation agreed with the theoretical values of 60, 132, and 42 rad, confirming that θ = ωt for motion at constant angular velocity. The θ-versus-time graph was a straight line whose slope equalled the set angular velocity, and the ω-versus-time graph was a flat horizontal line, showing that the angular velocity did not change. Because the angular velocity was constant, the angular acceleration was zero in every scenario (α = Δω/Δt = 0).

It is worth noting that a zero angular acceleration does not mean the ladybug had no acceleration at all: moving in a circle, it always had a centripetal acceleration a_c = ω²r directed toward the centre (37.5, 54, and 73.5 m/s² for the three scenarios). This is a linear acceleration that continually changes the direction of the velocity, distinct from angular acceleration, which would change the rate of spin. The simulation does not display angular acceleration directly, so only the angular displacement was compared between experiment and theory.

Conclusion

The experiment confirmed that angular displacement is proportional to time at constant angular velocity (θ = ωt), giving 60, 132, and 42 rad for the three scenarios, and that the angular acceleration is zero when the angular velocity is constant. The constant-slope θ-vs-time graph and flat ω-vs-time graph supported these conclusions.

Practice Questions

Show all work and include units in your answers. Use a radius of 1.5 m unless told otherwise.

Question 1
A wheel turns at a constant angular velocity of 8 rad/s for 5 s. Calculate the angular displacement and the number of revolutions.
Hint: θ = ωt; N = θ/(2π).
Question 2
A disc starts from rest and reaches 12 rad/s in 4 s. Calculate its angular acceleration. How does this differ from the ladybug scenarios in the lab?
Hint: α = Δω/Δt = (12 − 0)/4. In the lab ω was constant, so α was zero.
Question 3
A ladybug moves at 6 rad/s on a track of radius 1.5 m. Find its tangential speed and centripetal acceleration.
Hint: v = ωr; a_c = ω²r.
Question 4
If a ladybug completes 10 full revolutions at 5 rad/s, how long did it take?
Hint: θ = 10 × 2π rad; then t = θ/ω.
Question 5
Explain the difference between angular acceleration and centripetal acceleration. Can an object have one without the other?
Hint: angular acceleration changes the spin rate; centripetal acceleration changes direction. Constant-speed circular motion has the second but not the first.
Question 6 — Challenge
A ladybug's angular position is given by θ = 3t + 0.5t² (radians, with t in seconds). Find its angular velocity at t = 4 s and its angular acceleration. Is the angular acceleration zero here?
Hint: ω = dθ/dt = 3 + t; α = dω/dt = 1 rad/s². Here ω changes, so α is not zero.