Interactive physics simulator
Angular Acceleration
Explore how rotational speeds change over time. Analyze tangential linear acceleration ($a_t = r\alpha$), study constant acceleration kinematics curves with real-time graphs, and experiment with applied torques ($\tau = I\alpha$).
Angular Acceleration Lab
Configure parameters on the right and click Simulate to start loops or see real-time graphical plots.
Live Telemetry
- Angular Accel (α)
- 0.00 rad/s²
- Radius A (r_a)
- 0 px
- Radius B (r_b)
- 0 px
- Tangential Accel A
- 0 px/s²
- Tangential Accel B
- 0 px/s²
- Angular Accel (α)
- 0.00 rad/s²
- Angular Velocity (ω)
- 0.00 rad/s
- Displacement (θ)
- 0.00 rad
- Elapsed Time (t)
- 0.00 s
- Applied Torque (τ)
- 0.00 N·m
- Moment of Inertia (I)
- 0.0000 kg·m²
- Angular Accel (α)
- 0.00 rad/s²
- Angular Velocity (ω)
- 0.00 rad/s
Introduction to Angular Acceleration
When you start a ceiling fan, spin a bicycle wheel, or step on the accelerator of a car, the rotational speed of these systems changes. To measure how quickly the rate of rotation changes, we use **angular acceleration**. Just as linear acceleration measures the change in linear speed and direction per second, angular acceleration describes the rate of change of angular velocity over time.
Key Angular Acceleration Concepts
1. Definition of Angular Acceleration (α = Δω / Δt)
Average angular acceleration (denoted by the Greek letter alpha, α) is defined as the change in angular velocity (Δω) divided by the time interval (Δt) taken to make that change:
The standard SI unit for angular acceleration is radians per second squared (rad/s²). If the rate of spin is changing continuously, we define the instantaneous angular acceleration as the derivative of angular velocity with respect to time ($d\omega/dt$, or $d^2\theta/dt^2$).
2. Tangential Linear Acceleration (at = r · α)
When a wheel accelerates, particles on the wheel accelerate too. While all particles share the exact same angular acceleration (α), their linear acceleration components along the circumference—called **tangential acceleration** (at)—depend on how far they are from the center. Tangential acceleration is directly proportional to the radius of rotation:
This component of acceleration acts parallel (tangent) to the circular path and is responsible for changing the *speed* of the particle. (This is distinct from centripetal acceleration, which acts perpendicular/radial to change the *direction*).
3. Rotational Kinematics Equations
When a body rotates with a constant angular acceleration (α), we can write kinematics equations describing the position (θ) and velocity (ω) as a function of time (t), which are direct analogues to linear motion equations:
4. Dynamics and Torque (τ = I · α)
What causes angular acceleration? The rotational equivalent of force is **torque** (τ), and the rotational equivalent of mass is the **moment of inertia** (I). Newton\'s second law for rotation states:
This formula is fundamental to machine dynamics: for a given applied torque, a system with a larger moment of inertia (more mass distributed further from the axis) will accelerate much slower than one with a smaller moment of inertia.
Solved Numerical Examples
A heavy grinding wheel accelerates from an initial angular velocity of 2.0 rad/s to a final angular velocity of 8.0 rad/s over a time interval of 4.0 seconds. Determine: (a) its constant angular acceleration, and (b) the tangential linear acceleration of a particle on the rim at a radius of 0.30 meters.
View Step-by-Step Solution
- Given: Initial angular speed ω0 = 2.0 rad/s, final angular speed ω = 8.0 rad/s, elapsed time t = 4.0 s, radius r = 0.30 m.
- Use the first rotational kinematics equation to solve for angular acceleration (α):
ω = ω0 + α·t ⇒ α = (ω - ω0) / t. - Substitute values: α = (8.0 - 2.0) / 4.0 = 6.0 / 4.0 = 1.5 rad/s².
- Use the relationship between tangential acceleration and angular acceleration: at = r · α.
- Substitute values: at = 0.30 m × 1.5 rad/s² = 0.45 meters per second squared.
- Results: The grinding wheel has an angular acceleration of 1.5 rad/s², and particles on its rim accelerate tangentially at 0.45 m/s².
A solid cylinder has a mass of 4.0 kg and a radius of 0.20 meters. A mechanic applies a tangential force of 10.0 N to the outer edge of the cylinder using a pull-cord. Find: (a) the cylinder's moment of inertia, (b) the torque applied, and (c) the resulting angular acceleration. (Note: For a solid cylinder, I = 0.5 · M · R²).
View Step-by-Step Solution
- Given: Mass M = 4.0 kg, radius R = 0.20 m, tangential pull force F = 10.0 N.
- Calculate the cylinder's rotational inertia (moment of inertia): I = 0.5 · M · R².
I = 0.5 × 4.0 kg × (0.20 m)² = 2.0 × 0.040 = 0.080 kg·m². - Calculate the applied torque (τ): since the force is applied tangentially, the angle is 90°.
τ = R · F = 0.20 m × 10.0 N = 2.0 N·m. - Apply Newton's second law for rotation to find angular acceleration (α): τ = I · α ⇒ α = τ / I.
- Substitute values: α = 2.0 N·m / 0.080 kg·m² = 25.0 rad/s².
- Results: The moment of inertia is 0.080 kg·m², the torque is 2.0 N·m, and the resulting angular acceleration is 25.0 rad/s².
A wind turbine rotor spinning at 300 RPM decelerates to rest at a constant rate of α = -0.50 rad/s² due to emergency braking. Find: (a) the time taken to come to a complete stop, and (b) the total angular displacement in both radians and full revolutions during this braking period.
View Step-by-Step Solution
- Given: Initial rotational speed = 300 RPM, final angular speed ω = 0 rad/s, angular acceleration α = -0.50 rad/s².
- Convert initial speed to radians per second:
ω0 = 300 × (2π / 60) = 10π rad/s ≈ 31.42 rad/s. - Use the kinematics equation ω = ω0 + α·t to find stop time (t):
0 = 31.42 + (-0.50)·t ⇒ t = 31.42 / 0.50 = 62.83 seconds. - Use the third rotational kinematics equation to find angular displacement (θ):
ω² = ω0² + 2α·θ ⇒ 0 = (31.42)² + 2·(-0.50)·θ.
θ = 986.96 / 1.0 = 986.96 radians. - Convert radians to revolutions: rev = θ / 2π = 986.96 / 6.283 ≈ 157.08 revolutions.
- Results: The turbine stops in approximately 62.83 seconds, sweeping through 987 radians (157.1 full revolutions) during deceleration.
Conceptual Practice
A ceiling fan is rotating at a constant rate of 60 RPM. Determine if its: (a) angular acceleration, (b) tangential acceleration, and (c) centripetal acceleration are zero or non-zero.
Show Explanation
Because the fan rotates at a constant speed, its angular velocity (ω) is constant. Therefore, its **angular acceleration (α = dω/dt) is exactly zero**. Consequently, its **tangential acceleration (at = rα) is also zero** since the speed along the circle does not change. However, because the particles are moving in a circular path, their velocity vectors are constantly changing direction. Thus, their **centripetal acceleration (ac = ω²r) is non-zero** and points toward the center axle.
Why is the moment of inertia often called "rotational mass"? Compare a solid metal cylinder and a hollow metal hoop of the exact same mass and outer radius. Which one is harder to spin up from rest?
Show Explanation
Moment of inertia (I) is the rotational analogue of mass; it measures a body's resistance to changes in its rotational state, just as mass resists changes in linear motion. A hollow hoop has all of its mass concentrated at the outer edge (large radius), whereas a solid cylinder has its mass distributed from the center to the edge. Since I = ∫r²dm, the **hollow hoop has a much larger moment of inertia (Ihoop = MR² > Icylinder = 0.5MR²)**. Consequently, the **hollow hoop is harder to spin up** (requires more torque to achieve the same angular acceleration).
A solid toy wheel rolls down an inclined ramp, accelerating. Explain what physical force provides the torque that causes the wheel's angular acceleration.
Show Explanation
The torque is provided by the **force of static friction** acting between the ramp surface and the bottom edge of the wheel. Gravity pulls the wheel downward, but without friction, the wheel would simply slide down without rotating. Static friction opposes sliding, acting upward along the ramp at the contact point. Since this friction force acts at a radius R from the wheel's center axis, it exerts a torque (τ = R · fs) that drives the wheel's angular acceleration, causing it to roll.
How do you determine the sign of angular acceleration? Can an object have a negative angular acceleration but still be speeding up?
Show Explanation
By convention, counter-clockwise (CCW) rotation is positive (+) and clockwise (CW) is negative (-). The sign of angular acceleration depends on the change in velocity. Yes, an object can have a negative angular acceleration and still speed up. If a wheel is rotating clockwise (negative velocity, -ω) and spins faster and faster, its velocity becomes more negative. The acceleration is therefore negative (-α), causing the speed to increase in the CW direction.
Frequently Asked Questions
What is angular acceleration?
Angular acceleration is the rate of change of angular velocity over time, describing how quickly an object speeds up or slows down its rotation.
What is the SI unit of angular acceleration?
The standard SI unit for angular acceleration is radians per second squared (rad/s² or rad/s²).
What is the formula for angular acceleration?
The average angular acceleration is α = Δω / Δt, where Δω is the change in angular velocity and Δt is the elapsed time.
How is linear acceleration related to angular acceleration?
The tangential linear acceleration (a<sub>t</sub>) of a point is related to angular acceleration by the formula a<sub>t</sub> = r · α, where r is the radius.
What is the difference between tangential and centripetal acceleration?
Tangential acceleration (a<sub>t</sub> = rα) changes the magnitude of linear speed and acts tangent to the path. Centripetal acceleration (a<sub>c</sub> = ω²r) changes the direction of motion and points toward the center.
What is Newton's second law for rotation?
Newton's second law for rotation states that the net torque applied to a body is equal to the product of its moment of inertia and its angular acceleration: τ = I · α.
What is moment of inertia?
Moment of inertia (I) is a measure of an object's resistance to rotational acceleration, depending on the object's mass and how that mass is distributed relative to the axis of rotation.
How do you convert degrees/s² to rad/s²?
Multiply the value in degrees per second squared by π / 180 (approximately 0.01745) to convert to radians per second squared.
Can a body have angular acceleration if its angular velocity is zero?
Yes, just like a ball thrown upward has a linear acceleration of gravity when it momentarily stops at the apex, a rotating body can have a non-zero angular acceleration at the instant its angular velocity is zero (e.g. when reversing direction).
What happens to angular acceleration if the applied torque is doubled?
According to τ = Iα, if the moment of inertia remains constant, doubling the torque will double the resulting angular acceleration.
How does mass distribution affect angular acceleration?
If mass is distributed further from the rotational axis, the moment of inertia increases. For a given torque, this larger inertia results in a smaller angular acceleration.
How is angular acceleration measured in vehicles and aircraft?
It is measured using angular accelerometers, gyroscopic sensors, or Inertial Measurement Units (IMUs) containing micro-machined silicon elements.