Torque and Angular Momentum

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Torque and Angular Momentum

TL;DR

You'll learn how torque (the rotational push on an object) relates to angular momentum (the rotational equivalent of momentum), and why the second always equals the rate of change of the first. You'll see how this works for a single particle and a whole system, and why angular momentum is conserved when no external torque acts. You'll finish able to solve spinning-object problems by hand.

1. The Mental Model

Torque is what makes things start, stop, or change how they spin — it's a twisting force. Angular momentum is how much "spin" something has, and it takes torque to change it, just like it takes force to change ordinary momentum. If nothing twists a system from outside, its total spin stays locked in forever. Torque is to rotation what force is to motion — and angular momentum is what torque changes.

2. The Core Material

2.1 Torque and Angular Momentum for a Single Particle

For a particle with position vector $\vec{r}$ (measured from some chosen origin) and linear momentum $\vec{p} = m\vec{v}$, we define angular momentum about that origin as:

$$\vec{L} = \vec{r} \times \vec{p}$$

This is a cross product, so $\vec{L}$ is always perpendicular to the plane containing $\vec{r}$ and $\vec{v}$. Its magnitude is $L = rp\sin\theta$, where $\theta$ is the angle between $\vec{r}$ and $\vec{p}$. Notice $\vec{L}$ depends on where you put the origin — change the origin, and you can get a completely different $\vec{L}$, even for the same motion. This trips people up constantly: angular momentum isn't a property of the particle alone, it's a property of the particle relative to a chosen point.

Torque about the same origin is defined similarly:

$$\vec{\tau} = \vec{r} \times \vec{F}$$

Now here's the key relationship. Differentiate $\vec{L}$ with respect to time:

$$\frac{d\vec{L}}{dt} = \frac{d\vec{r}}{dt} \times \vec{p} + \vec{r} \times \frac{d\vec{p}}{dt}$$

The first term is $\vec{v} \times m\vec{v} = 0$ (a vector crossed with itself, or a parallel vector, is zero). The second term is $\vec{r} \times \vec{F} = \vec{\tau}$ (using Newton's second law, $d\vec{p}/dt = \vec{F}$). So:

$$\vec{\tau} = \frac{d\vec{L}}{dt}$$

This is the rotational version of $\vec{F} = d\vec{p}/dt$. It's not an assumption — it falls straight out of Newton's second law. That's worth sitting with: you don't need a new law of physics for rotation. You just need to take the cross product of the old one with $\vec{r}$.

2.2 Extending to a System of Particles

Now take $N$ particles. The total angular momentum of the system (about a fixed origin) is just the sum:

$$\vec{L}_{\text{total}} = \sum_{i=1}^{N} \vec{r}_i \times \vec{p}_i$$

Each particle feels two kinds of forces: external forces from outside the system (gravity, an applied push) and internal forces from the other particles in the system (particle 2 pulling on particle 1, for example). When you differentiate $\vec{L}_{\text{total}}$, you get a sum of torques — some external, some internal:

$$\frac{d\vec{L}_{\text{total}}}{dt} = \sum_i \vec{r}_i \times \vec{F}_i^{\text{ext}} + \sum_i \vec{r}_i \times \vec{F}_i^{\text{int}}$$

Here's the beautiful part. By Newton's third law, internal forces come in equal-and-opposite pairs along the line joining the two particles (this is the strong form of Newton's third law — it assumes the force is central, acting along the line connecting the particles, which is true for gravity and electrostatics). Consider particles $i$ and $j$: the torque from $i$ on $j$ plus the torque from $j$ on $i$ is:

$$\vec{r}_i \times \vec{F}_{ij} + \vec{r}_j \times \vec{F}_{ji} = (\vec{r}_i - \vec{r}_j) \times \vec{F}_{ij}$$

since $\vec{F}_{ji} = -\vec{F}_{ij}$. But $\vec{F}_{ij}$ points along the line joining the particles, which is parallel to $(\vec{r}_i - \vec{r}_j)$. A vector crossed with something parallel to itself is zero. So every internal torque pair cancels. All that survives is:

$$\vec{\tau}_{\text{ext}} = \frac{d\vec{L}_{\text{total}}}{dt}$$

This is the big result: only external torques can change the total angular momentum of a system. Internal forces — no matter how complicated, how many particles are flying around, how they collide or stick — never change the system's total $\vec{L}$.

2.3 Conservation of Angular Momentum

The direct consequence: if the net external torque on a system is zero, $\vec{L}_{\text{total}}$ is constant.

$$\vec{\tau}_{\text{ext}} = 0 \implies \vec{L}_{\text{total}} = \text{constant}$$

This happens in two common situations. First, genuinely zero external forces (isolated system, deep space). Second — much more common in problems — forces act, but their torque about the chosen origin is zero, usually because the force is central (points straight at or away from the origin, so $\vec{r} \times \vec{F} = 0$). This is exactly why planets sweep out equal areas in equal times (Kepler's second law): gravity from the sun is central, so there's no torque about the sun, so $\vec{L}$ of the planet is conserved.

For rigid bodies spinning about a fixed axis, $L = I\omega$, where $I$ is the moment of inertia and $\omega$ is angular velocity. If $I$ changes (a skater pulling in their arms) while $L$ stays constant, $\omega$ must increase to compensate — that's why skaters spin faster when they pull their arms in. This is the rotational version of momentum conservation, and it's just as powerful: it lets you solve problems without ever working out the forces involved.

flowchart TD
    A["Newton's 2nd Law: F = dp/dt"] --> B["Cross with r: torque τ = dL/dt (single particle)"]
    B --> C["Sum over all particles in system"]
    C --> D{"Internal forces: Newton's 3rd Law pairs"}
    D -->|"Central forces cancel in pairs"| E["Internal torques sum to zero"]
    E --> F["τ_ext = dL_total/dt"]
    F --> G{"Is net external torque zero?"}
    G -->|Yes| H["L_total is conserved"]
    G -->|No| I["L_total changes over time"]

3. Worked Example

Setup: A merry-go-round (a uniform disk) has moment of inertia $I_1 = 300\ \text{kg·m}^2$ and spins at $\omega_1 = 2.0\ \text{rad/s}$. A child of mass $m = 40\ \text{kg}$ runs tangentially at $v = 3.0\ \text{m/s}$ and jumps onto the rim, at radius $R = 2.5\ \text{m}$ from the center, moving in the same direction the rim is already turning. Find the new angular velocity of the system.

Step 1 — Identify the system and check for external torque. Take the system to be disk + child. The forces at the moment of landing (friction with the ground supporting the disk's axle, gravity) act through or near the rotation axis, so they produce negligible torque about the center during this brief interaction. Angular momentum about the center is conserved through the collision.

Step 2 — Angular momentum before.
Disk: $L_{\text{disk}} = I_1 \omega_1 = 300 \times 2.0 = 600\ \text{kg·m}^2/\text{s}$

Child (treated as a particle): $L_{\text{child}} = m v R = 40 \times 3.0 \times 2.5 = 300\ \text{kg·m}^2/\text{s}$

(The child's angular momentum uses $L = mvR$ because she's running tangentially — her velocity is perpendicular to her radius vector at that instant, so $\sin\theta = 1$.)

Total before: $L_i = 600 + 300 = 900\ \text{kg·m}^2/\text{s}$

Step 3 — Moment of inertia after. Once she's standing on the rim, she moves with it, contributing $I_{\text{child}} = mR^2 = 40 \times (2.5)^2 = 250\ \text{kg·m}^2$.

New total moment of inertia: $I_2 = I_1 + I_{\text{child}} = 300 + 250 = 550\ \text{kg·m}^2$

Step 4 — Apply conservation.
$$L_i = L_f \implies 900 = I_2 \omega_2 \implies \omega_2 = \frac{900}{550} = 1.64\ \text{rad/s}$$

The merry-go-round slows down from 2.0 to 1.64 rad/s because the child's added mass increases the system's moment of inertia, even though her own angular momentum added to the total. Notice we never needed to compute the (messy, impulsive) friction force between her feet and the platform — conservation of $L$ sidesteps all of that.

4. Production Pitfalls & Best Practices

4.1 Real-World Best Practices

• Always state your origin before computing $\vec{L}$ or $\vec{\tau}$. Since both depend on the choice of reference point, write "angular momentum about point O" explicitly in every solution.
• Check whether Newton's third law pairs are central before assuming internal torques cancel. For most mechanics problems (gravity, contact forces, springs along a rod) this holds, but magnetic forces between moving charges can violate it.
• Decompose $\vec{L}$ into "orbital" and "spin" parts for extended bodies: $\vec{L}_{\text{total}} = \vec{L}_{\text{cm}} + \vec{L}_{\text{about cm}}$, where the first term treats the whole mass as concentrated at the center of mass, and the second is the rotation of the body about its own center. This split makes rolling and rigid-body problems tractable.
• Use conservation of $L$ whenever collisions or shape changes happen with no external torque — it avoids ever modeling the internal, impulsive forces.
• Match the axis for $I$ and $\omega$. When using $L = I\omega$, both must be defined about the exact same rotation axis, or the formula is simply wrong.
• Watch units: angular momentum is $\text{kg·m}^2/\text{s}$, torque is $\text{N·m}$ (same dimensions as energy but never called joules).

4.2 Common Bugs & Anti-Patterns

Mistake 1: Forgetting $\vec{L}$ depends on origin choice.

# WRONG: computing L about center of mass, then comparing to
# a torque calculated about a different pivot point
L_cm = r_from_cm.cross(p)
tau_pivot = r_from_pivot.cross(F)
# these don't relate to each other — mismatched origins!

# FIXED: pick ONE origin and use it for everything
origin = pivot_point
L = r_from_pivot.cross(p)
tau = r_from_pivot.cross(F)
# now tau = dL/dt is valid

Mistake 2: Assuming conservation of $L$ when there IS external torque.

# WRONG: a falling cat or a pendulum losing energy to air drag —
# assuming L is conserved because "nothing touches it directly"
L_final = L_initial  # ignores torque from gravity about a non-CM pivot!

# FIXED: check where gravity acts relative to your chosen origin.
# If origin is NOT directly below/above the center of mass, gravity
# produces real torque and L changes.
tau_gravity = r_cm.cross(m * g_vector)
dL_dt = tau_gravity  # must integrate, not assume constant

Mistake 3: Using scalar $L = I\omega$ for non-fixed or tilted axes.

# WRONG: applying L = I*omega to a wobbling gyroscope
L = I * omega  # scalar form hides the fact that L and omega
               # aren't even parallel for a tilted spinning body

# FIXED: use the full tensor relation for tilted/asymmetric bodies
# L_vector = I_tensor @ omega_vector  (requires moment of inertia tensor)

Mistake 4: Double-counting internal torques in a system.

# WRONG: summing torque from particle A on B AND B on A separately
# without recognizing they cancel, then adding both to "external" torque
tau_total = tau_ext + tau_AB + tau_BA  # should cancel but code adds them

# FIXED: only sum torques from forces originating outside the system
tau_total = tau_ext  # internal pairs are proven to cancel analytically

4.3 When to Use What (Decision Matrix

5. Now Try It

Problem: A uniform rod of length $\ell = 1.2\ \text{m}$ and mass $M = 2.0\ \text{kg}$ hangs from a frictionless pivot at one end, initially at rest. A small ball of mass $m = 0.5\ \text{kg}$ moving horizontally at $v = 6.0\ \text{m/s}$ strikes and sticks to the rod's free end (the end opposite the pivot). The rod's moment of inertia about the pivot is $I_{\text{rod}} = \frac{1}{3}M\ell^2$.

Find the angular velocity of the rod+ball system immediately after the collision.

A top-band answer must:
1. State clearly why angular momentum (not linear momentum, and not energy) is conserved about the pivot during the collision.
2. Compute $L_i$ of the ball about the pivot before impact ($L = mv\ell$, since velocity is perpendicular to the rod).
3. Compute the combined moment of inertia after impact: $I_{\text{total}} = I_{\text{rod}} + m\ell^2$.
4. Solve $\omega_f = L_i / I_{\text{total}}$ and give a numeric answer with correct units (rad/s).
5. Briefly note that kinetic energy is NOT conserved here (it's an inelastic collision), explaining why you couldn't use energy conservation instead.

Success looks like: you arrive at $\omega_f \approx 3.16\ \text{rad/s}$, with every step justified by the physics, not just the arithmetic.

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