Interactive physics simulator
Angular Momentum
Explore the conservation of rotational momentum. Tuck a figure skater's arms to spin up instantly, sweep orbital paths to demonstrate Kepler's areal law, and collide particles with pivoted rods.
Angular Momentum Lab
Tweak settings in the control panel, choose the active graph representation, and click Simulate.
Live Telemetry
- Moment of Inertia (I)
- 0.00 kg·m²
- Angular Speed (ω)
- 0.00 rad/s
- Orbit Distance (r)
- 0.00 AU
- Linear Speed (v)
- 0.00 km/s
- Combined Inertia (I)
- 0.000 kg·m²
- Post-Impact Spin (ω)
- 0.00 rad/s
- Angular Momentum (L)
- 0.00 kg·m²/s
- System Status
- Ready
Introduction to Angular Momentum
Just as linear momentum describes an object's quantity of translational motion in a straight line, angular momentum describes an object's quantity of rotational motion about a specific pivot axis. It is the rotational equivalent of linear momentum, and plays a fundamental role in understanding planetary orbits, spinning gyroscopes, and acrobatics.
An object's angular momentum depends on its rotational speed, its total mass, and the distribution of that mass from the axis of rotation. The most remarkable feature of angular momentum is that, like energy and linear momentum, it is a conserved quantity. In the absence of external twisting forces (torques), the total angular momentum of any system remains constant.
Key Rotational Momentum Principles
1. Formulas and Formulations (L = Iω or L = r × p)
For a rigid rotating body, the angular momentum (L) is defined as the product of its moment of inertia (I) and its angular velocity (ω):
For a single point particle of mass m moving with linear velocity v at a position vector r relative to a pivot point, the angular momentum vector is the cross product of position and linear momentum:
Here, θ is the angle between the position vector r and the velocity vector v. The angular momentum reaches its maximum value when the velocity is perpendicular to the radius arm (θ = 90°).
2. Conservation of Angular Momentum
The rotational version of Newton's Second Law states that the net external torque is equal to the rate of change of angular momentum: Στ = dL/dt. When the net external torque is exactly zero, we find:
This explains how skaters spin up instantly when they pull their arms inward: they reduce their moment of inertia (I), forcing their angular speed (ω) to increase proportionally to keep L constant.
3. Kepler's Second Law (Elliptical Areal Conservation)
A planet orbiting a star experiences only a central gravitational force. Because this force points directly along the radial position vector, it generates zero torque. Thus, the planet's angular momentum is conserved at all points in its orbit. As the planet approaches the star at perihelion (distance r decreases), its speed v must increase. Conversely, at aphelion (distance r increases), it slows down. This conservation maintains a constant areal velocity, sweeping equal areas in equal time intervals.
Solved Numerical Examples
A figure skater spinning on ice with arms extended has a moment of inertia of 3.60 kg·m² and spins at an angular velocity of 2.50 rad/s. When they pull their arms in close, their moment of inertia decreases to 1.20 kg·m². (a) Calculate their new angular velocity. (b) Find the ratio of their final rotational kinetic energy to their initial rotational kinetic energy.
View Step-by-Step Solution
- Given: Initial inertia I₁ = 3.60 kg·m², initial angular speed ω₁ = 2.50 rad/s, final inertia I₂ = 1.20 kg·m².
- (a) Calculate New Angular Velocity (ω₂):
Since there is no external torque, angular momentum (L) is conserved: L₁ = L₂ ⇒ I₁ · ω₁ = I₂ · ω₂. - Solve for ω₂: ω₂ = (I₁ · ω₁) / I₂.
- Substitute values: ω₂ = (3.60 × 2.50) / 1.20 = 9.00 / 1.20 = 7.50 rad/s.
- The skater spins three times faster when they pull their arms in.
- (b) Compare Rotational Kinetic Energies (KE_rot):
Initial kinetic energy: KE₁ = ½ · I₁ · ω₁² = 0.5 × 3.60 × (2.50)² = 11.25 Joules. - Final kinetic energy: KE₂ = ½ · I₂ · ω₂² = 0.5 × 1.20 × (7.50)² = 33.75 Joules.
- Ratio: KE₂ / KE₁ = 33.75 / 11.25 = 3.00.
- Note: Rotational kinetic energy increased by a factor of 3. This extra energy comes from the internal muscular work performed by the skater pulling their arms inward against centrifugal effects.
- Results: (a) The new angular velocity is 7.50 rad/s. (b) The ratio of final to initial kinetic energy is 3.00.
A satellite of mass 250 kg orbits a planet in an eccentric elliptical orbit. At its closest approach (perihelion), its distance from the center of the planet is 1.20 × 10⁷ meters and its orbital speed is 8.00 km/s. Calculate the speed of the satellite at aphelion (its farthest point) where its distance from the planet center is 3.60 × 10⁷ meters.
View Step-by-Step Solution
- Given: Satellite mass m = 250 kg, perihelion distance rp = 1.20 × 10⁷ m, perihelion speed vp = 8.00 km/s = 8000 m/s, aphelion distance ra = 3.60 × 10⁷ m.
- Recall that gravity acts along the line connecting centers, producing zero external torque on the satellite. Thus, its angular momentum (L) is conserved: Lp = La.
- At aphelion and perihelion, the velocity vector is perpendicular to the position vector (θ = 90°). The angular momentum equation simplifies to: L = m · v · r.
- Write the conservation equation: m · vp · rp = m · va · ra.
- Cancel mass (m) from both sides: vp · rp = va · ra.
- Solve for aphelion speed va: va = (vp · rp) / ra.
- Substitute values: va = (8.00 km/s × 1.20 × 10⁷ m) / (3.60 × 10⁷ m).
- Solve: va = 9.60 × 10⁷ / 3.60 × 10⁷ = 2.67 km/s.
- Result: The orbital speed at the farthest point is 2.67 km/s (or 2670 m/s).
A stationary uniform bar of mass 4.0 kg and length 1.8 meters is pivoted at its center, allowing it to rotate freely in a horizontal plane. The moment of inertia of the bar is I_rod = 1.08 kg·m². A projectile of mass 0.50 kg is launched horizontally at a speed of 15.0 m/s perpendicular to the bar. It strikes and sticks to the bar at a distance of 0.60 meters from the pivot. Find the angular velocity of the bar-projectile system immediately after the collision.
View Step-by-Step Solution
- Given: Bar mass M = 4.0 kg, length L = 1.8 m, Irod = 1.08 kg·m², projectile mass m = 0.50 kg, speed v = 15.0 m/s, impact distance r = 0.60 m.
- We treat the pivot as the center of coordinate torque axis. Since there are no external torques acting on the rod-projectile system during collision, angular momentum is conserved: Linitial = Lfinal.
- (1) Calculate Initial Angular Momentum (L_initial):
Before impact, only the flying projectile has angular momentum relative to the pivot: Linitial = m · v · r. - Substitute values: Linitial = 0.50 kg × 15.0 m/s × 0.60 m = 4.50 kg·m²/s.
- (2) Calculate Final Moment of Inertia (I_final):
After impact, the projectile sticks to the rod, contributing a point-mass term: Ifinal = Irod + m · r². - Calculate: Ifinal = 1.08 kg·m² + [0.50 kg × (0.60 m)²] = 1.08 + (0.50 × 0.36) = 1.08 + 0.18 = 1.26 kg·m².
- (3) Solve for Post-Collision Angular Speed (ω_final):
Lfinal = Ifinal · ωfinal ⇒ ωfinal = Linitial / Ifinal. - Calculate: ωfinal = 4.50 / 1.26 ≈ 3.57 rad/s.
- Result: The angular velocity of the system after collision is approximately 3.57 rad/s.
Conceptual Practice
What is angular momentum, and how does it differ from linear momentum? State their formulas and SI units.
Show Explanation
Linear momentum ($p = mv$, unit: kg·m/s) measures translational mass in motion. **Angular momentum** ($L = Iomega$, unit: kg·m²/s or N·m·s) measures rotational inertia in motion. While linear momentum depends solely on mass and linear speed, angular momentum depends on mass, rotational speed, and how far the mass is distributed from the axis of rotation ($L = r imes p$).
Explain why rotational kinetic energy increases when a spinning skater tucks their arms, even though angular momentum is conserved.
Show Explanation
Angular momentum is conserved ($L = Iomega = ext{const}$). When the skater pulls their arms in, their moment of inertia ($I$) decreases, and angular speed ($omega$) increases proportionally. Rotational kinetic energy is defined as $KE = rac{1}{2} Iomega^2 = rac{L^2}{2I}$. Because $I$ is in the denominator, decreasing $I$ at constant $L$ increases the kinetic energy. The extra energy is supplied by the skater doing positive mechanical work as their muscles pull their arms inward against the outward apparent centrifugal force.
How does the conservation of angular momentum explain Kepler's Second Law (equal orbital areas swept in equal times)?
Show Explanation
A planet orbiting a star experiences gravitational pull acting directly along the radial position vector connecting their centers. This means the force is parallel to the radius vector, producing zero torque ($ ec{ au} = ec{r} imes ec{F} = 0$). With zero external torque, the planet's angular momentum ($L = m v r sinphi$) is conserved. The rate at which orbital area is swept (areal velocity) is $dA/dt = rac{L}{2m}$. Since $L$ and $m$ are constant, $dA/dt$ is constant, yielding equal areas swept in equal times.
Describe the Right-Hand Rule for angular momentum. What does the direction of the vector represent?
Show Explanation
The angular momentum vector of a particle is defined as the cross product of position and linear momentum: $ ec{L} = ec{r} imes ec{p}$. To find its direction using the Right-Hand Rule, point the fingers of your right hand in the direction of the position vector $ ec{r}$ (from pivot to particle), then curl them toward the momentum vector $ ec{p}$ (velocity direction). Your extended thumb points in the direction of $ ec{L}$. For counter-clockwise rotation, the vector points upward/out-of-screen; for clockwise, it points downward/into-of-screen. The vector defines the axis of rotation.
Write the rotational analogue of Newton's Second Law. How does it justify conservation of angular momentum?
Show Explanation
Newton's Second Law in translational motion is $F_{net} = rac{dp}{dt}$ (force is the rate of change of linear momentum). The rotational analogue is $ au_{net} = rac{dL}{dt}$ (net torque is the rate of change of angular momentum). If the net external torque acting on a system is zero ($ au_{net} = 0$), then $ rac{dL}{dt} = 0$, meaning the total angular momentum of the system remains constant over time. This is the Principle of Conservation of Angular Momentum.
Frequently Asked Questions
What is angular momentum?
Angular momentum is the rotational equivalent of linear momentum, representing the quantity of rotation of a body. It depends on the object's rotational speed, mass, and mass distribution relative to the axis.
What is the formula for angular momentum?
For a rigid body, the formula is L = I·ω, where I is the moment of inertia and ω is the angular velocity. For a point mass, it is L = r × p = m·v·r·sin(θ).
What is the SI unit of angular momentum?
The SI unit of angular momentum is the kilogram-meter squared per second (kg·m²/s), which is equivalent to Newton-meter-seconds (N·m·s).
When is angular momentum conserved?
Angular momentum is conserved in any closed system where there is no net external torque acting on the system: Στ = 0.
Is angular momentum a vector?
Yes, angular momentum is a vector quantity. Its direction is perpendicular to the plane of rotation and is determined using the Right-Hand Rule.
Why do skaters pull their arms in to spin faster?
By pulling their arms in, skaters decrease their moment of inertia (I). Because angular momentum (L = I·ω) must remain constant, their angular velocity (ω) must increase, making them spin faster.
How does torque affect angular momentum?
According to the rotational version of Newton's Second Law, net torque is equal to the rate of change of angular momentum (Στ = dL/dt). Applying torque changes the rotation speed or axis direction.