11.4 Angular Impulse and Angular Momentum Change

The Wrench, the Bolt, and the Passage of Time

A mechanic grips a wrench and pushes. The bolt does not move at first --- it is rusted in place. She holds steady, applying a constant torque of 40 N$\cdot$m for two full seconds. The bolt finally breaks free and begins to turn.

Before she pushed, the bolt had zero angular momentum. After two seconds of sustained torque, it has acquired some. The torque did not change the angular momentum instantaneously --- it accumulated over time, steadily building the bolt's rotational motion.

This is angular impulse: the total effect of a torque accumulated over a time interval. It is the rotational version of the impulse you studied in Section 8.2, where force accumulated over time produced a change in linear momentum. The structure is identical. The domain is new.

Before you read on: A spinning disk is braked by friction. The frictional torque is constant.

Does the angular momentum decrease linearly with time, or exponentially?

Commit to an answer before continuing.

Quick Recall: Angular Momentum and Linear Impulse

Two ideas from earlier chapters feed directly into this section.

Recall prompt: What is angular momentum for a rigid body? (Section 11.3.) What is the impulse-momentum theorem for translational motion? (Section 8.2.)

From Section 11.3: $\vec{L} = I\vec{\omega}$ for a rigid body rotating about a fixed axis. Angular momentum is the rotational analog of linear momentum $\vec{p} = m\vec{v}$.

From Section 8.2: the impulse-momentum theorem states $\vec{J} = \int \vec{F}\,dt = \Delta\vec{p}$. The integral of force over time equals the change in momentum. The area under the $F(t)$ curve is the impulse, and it tells you exactly how much the momentum changes.

If these feel fuzzy, revisit those sections now. Everything here is built on extending the impulse idea from forces to torques.

Exploration: Braking a Spinning Wheel

[Interactive: Angular Impulse Explorer. A spinning wheel is shown from the side, rotating at an initial angular velocity $\omega_0$ set by a slider (range: 5 to 30 rad/s). The wheel has a fixed moment of inertia $I = 0.4$ kg$\cdot$m$^2$.

A brake pad applies a constant frictional torque when engaged. A slider controls the braking torque magnitude (range: 1 to 10 N$\cdot$m). A "Start Brake" button engages the brake.

Three synchronized graphs update in real time as the brake is applied: - Top graph: $\omega(t)$ --- angular velocity versus time. Shows a straight line decreasing toward zero (for constant torque). - Middle graph: $L(t) = I\omega(t)$ --- angular momentum versus time. Decreases linearly, mirroring $\omega(t)$ since $I$ is constant. - Bottom graph: $\tau(t)$ versus time. A horizontal line at the braking torque value, with the area under the curve (angular impulse) shaded progressively.

A numerical readout displays: current $\omega$, current $L$, elapsed time, cumulative angular impulse (area under $\tau$-$t$ curve), and $\Delta L$ (change in angular momentum from start). The cumulative angular impulse and $|\Delta L|$ values are highlighted in matching colors to show they are equal at every instant.

Guided prompts: - "Start the wheel at 20 rad/s and apply a braking torque of 4 N$\cdot$m. Watch the three graphs as the wheel slows. Does $L$ decrease linearly or along a curve?" - "Read the shaded area under the $\tau(t)$ graph at the moment the wheel stops. Compare it to the total $\Delta L$. What do you notice?" - "Now reset and try a braking torque of 8 N$\cdot$m. The wheel stops in half the time. Is the total angular impulse different?" - "Try starting at different $\omega_0$ values with the same torque. How does the stopping time change? How does the total angular impulse change?"]

Here is what the interactive reveals. When a constant braking torque acts on the wheel:

1. The angular velocity decreases linearly with time. A constant torque produces a constant angular deceleration ($\alpha = \tau / I$), just as a constant force produces a constant linear deceleration.

2. The angular momentum $L = I\omega$ decreases linearly too, since $I$ is constant. No curves, no exponential decay --- just a steady, uniform decrease.

3. The area under the $\tau(t)$ curve grows with time, and at every instant it exactly matches $|\Delta L|$. When the wheel stops, the total shaded area equals the entire angular momentum the wheel started with.

This is not a coincidence. It is a theorem.

Concept Reveal: The Angular Impulse-Momentum Theorem

Newton's second law for rotation says $\vec{\tau}_{\text{net}} = \frac{d\vec{L}}{dt}$. Rearranging and integrating over a time interval from $t_i$ to $t_f$:

$$\int_{t_i}^{t_f} \vec{\tau}{\text{net}}\,dt = \int$$}^{t_f} \frac{d\vec{L}}{dt}\,dt = \vec{L}(t_f) - \vec{L}(t_i) = \Delta\vec{L

The left side is the angular impulse --- the integral of net torque over time:

$$\vec{J}{\text{ang}} = \int\,dt$$}^{t_f} \vec{\tau}_{\text{net}

And the theorem says:

$$\boxed{\vec{J}_{\text{ang}} = \Delta\vec{L}}$$

The angular impulse equals the change in angular momentum. Graphically, the area under the $\tau(t)$ curve is the angular impulse, and it tells you exactly how much the angular momentum changes.

For a constant torque, the integral simplifies:

$$J_{\text{ang}} = \tau_{\text{net}} \cdot \Delta t = \Delta L$$

The Structural Parallel

Place this next to the translational theorem from Section 8.2:

Translational (Section 8.2)	Rotational (Section 11.4)
Force $\vec{F}$	Torque $\vec{\tau}$
Linear momentum $\vec{p} = m\vec{v}$	Angular momentum $\vec{L} = I\vec{\omega}$
Impulse $\vec{J} = \int \vec{F}\,dt$	Angular impulse $\vec{J}_{\text{ang}} = \int \vec{\tau}\,dt$
$\vec{J} = \Delta\vec{p}$	$\vec{J}_{\text{ang}} = \Delta\vec{L}$
Area under $F(t)$ graph $=$ impulse	Area under $\tau(t)$ graph $=$ angular impulse

Every structural element matches. The derivation is the same (integrate Newton's second law over time). The graphical interpretation is the same (area under the curve). The physical meaning is the same (accumulated influence over time produces a change in the quantity of motion).

This is not a surface-level analogy. It is the same mathematical architecture applied to a different physical setting. The conservation framework you learned in Chapter 8 does not need to be rebuilt --- it needs to be reused.

Returning to the Prediction

You were asked: when a constant frictional torque brakes a spinning disk, does the angular momentum decrease linearly or exponentially?

A constant torque means $\tau$ is constant. From $\tau = I\alpha$ (with $I$ constant), the angular deceleration $\alpha$ is constant. That means $\omega$ decreases at a steady rate:

$$\omega(t) = \omega_0 - \frac{\tau}{I}\,t$$

Since $L = I\omega$:

$$L(t) = I\omega_0 - \tau\,t$$

This is a linear decrease. The angular momentum drops at a constant rate of $\tau$ per second until the disk stops.

Exponential decay would require the torque to depend on the current angular velocity --- for example, viscous drag proportional to $\omega$, where $\tau = -b\omega$. With such a torque, the angular momentum would decay as $L(t) = L_0\, e^{-bt/I}$, approaching zero asymptotically but never quite reaching it. Constant friction does not behave this way. It delivers a fixed torque regardless of how fast the disk is spinning, producing a linear decrease and a definite stopping time.

The distinction matters physically. A disk braked by constant friction stops in finite time: $t_{\text{stop}} = I\omega_0 / \tau$. A disk subject to viscous drag never fully stops (in the idealized model). Recognizing which torque model applies is essential for predicting the motion.

Graphical Interpretation: Area Under the $\tau(t)$ Curve

This section uses exactly the same area-under-the-curve reasoning you have now seen twice before:

Section 7.2	Section 8.2	Section 11.4
$F(x)$ vs. position	$F(t)$ vs. time	$\tau(t)$ vs. time
Area $= \int F\,dx =$ work	Area $= \int F\,dt =$ impulse	Area $= \int \tau\,dt =$ angular impulse
Determines energy change	Determines momentum change	Determines angular momentum change

The pattern is always the same: when a quantity accumulates over some variable (position, time), the integral --- the area under the curve --- tells you the total accumulated effect. You read work from force-displacement graphs, impulse from force-time graphs, and now angular impulse from torque-time graphs.

For a constant torque, the $\tau(t)$ graph is a horizontal line, and the area is a rectangle: $J_{\text{ang}} = \tau \cdot \Delta t$.

For a torque that varies with time, you need the full integral or the geometric area of whatever shape the graph forms (triangles, trapezoids, or more complex shapes that require integration).

Worked Example: Braking a Grinding Wheel

Problem: A grinding wheel with $I = 0.60$ kg$\cdot$m$^2$ spins at $\omega_0 = 50$ rad/s. A brake applies a constant frictional torque of 15 N$\cdot$m. How long does it take the wheel to stop? What is the angular impulse delivered by the brake?

Solution:

The angular momentum before braking:

$$L_0 = I\omega_0 = 0.60 \times 50 = 30 \text{ kg} \cdot \text{m}^2/\text{s}$$

After the wheel stops, $L_f = 0$. The change in angular momentum:

$$\Delta L = L_f - L_0 = 0 - 30 = -30 \text{ kg} \cdot \text{m}^2/\text{s}$$

By the angular impulse-momentum theorem, $J_{\text{ang}} = \Delta L = -30$ kg$\cdot$m$^2$/s. The magnitude of the angular impulse is 30 kg$\cdot$m$^2$/s.

For a constant torque:

$$J_{\text{ang}} = -\tau \cdot \Delta t$$

$$-30 = -15 \cdot \Delta t$$

$$\Delta t = 2.0 \text{ s}$$

Sanity check: The angular deceleration is $\alpha = \tau / I = 15 / 0.60 = 25$ rad/s$^2$. Starting at 50 rad/s and decelerating at 25 rad/s$^2$, the wheel stops in $50/25 = 2.0$ s. Consistent.

Faded Example: Reading Angular Impulse from a Graph

A turntable with $I = 0.25$ kg$\cdot$m$^2$ is initially at rest. A motor applies a torque that varies with time:

• From $t = 0$ to $t = 2$ s: $\tau$ increases linearly from 0 to 6 N$\cdot$m.
• From $t = 2$ s to $t = 5$ s: $\tau$ remains constant at 6 N$\cdot$m.

Find the angular impulse delivered over the full 5-second interval and the final angular velocity.

Step 1: Angular impulse from 0 to 2 s.

The $\tau(t)$ graph is a triangle with base 2 s and height 6 N$\cdot$m.

$$J_1 = \frac{1}{2}(2)(6) = 6.0 \text{ N} \cdot \text{m} \cdot \text{s}$$

Step 2: Angular impulse from 2 to 5 s.

Your turn: The torque is constant at 6 N$\cdot$m for 3 seconds. What is the angular impulse during this interval?

Check your answer

$$J_2 = \tau \cdot \Delta t = 6 \times 3 = 18.0 \text{ N} \cdot \text{m} \cdot \text{s}$$ The $\tau(t)$ graph during this interval is a rectangle with width 3 s and height 6 N$\cdot$m.

Step 3: Total angular impulse and final angular velocity.

Your turn: Add the two angular impulses to find the total. Then use $J_{\text{ang}} = \Delta L = I\Delta\omega$ with $\omega_0 = 0$ to find the final angular velocity.

Check your answer

$$J_{\text{total}} = J_1 + J_2 = 6.0 + 18.0 = 24.0 \text{ N} \cdot \text{m} \cdot \text{s}$$ $$J_{\text{total}} = \Delta L = I\omega_f - I\omega_0 = I\omega_f$$ $$\omega_f = \frac{J_{\text{total}}}{I} = \frac{24.0}{0.25} = 96 \text{ rad/s}$$ The turntable reaches 96 rad/s (about 15 revolutions per second). The triangular ramp-up phase contributed one-quarter of the total angular impulse; the constant-torque phase contributed three-quarters. You can see both contributions directly as areas on the $\tau(t)$ graph.

Practice

Layer 1: Concrete

A potter's wheel with $I = 0.80$ kg$\cdot$m$^2$ is spinning at 12 rad/s. The potter applies a constant braking torque of 3.0 N$\cdot$m with her hand.

(a) What is the angular impulse needed to stop the wheel?

(b) How long does it take to stop?

Check your answer

**(a)** The initial angular momentum is $L_0 = I\omega_0 = 0.80 \times 12 = 9.6$ kg$\cdot$m$^2$/s. The final angular momentum is zero. $$J_{\text{ang}} = \Delta L = 0 - 9.6 = -9.6 \text{ kg} \cdot \text{m}^2/\text{s}$$ The angular impulse has magnitude 9.6 kg$\cdot$m$^2$/s, directed opposite to the wheel's rotation. **(b)** For a constant torque: $$|J_{\text{ang}}| = \tau \cdot \Delta t$$ $$\Delta t = \frac{|J_{\text{ang}}|}{\tau} = \frac{9.6}{3.0} = 3.2 \text{ s}$$ **Sanity check:** The angular deceleration is $\alpha = \tau / I = 3.0/0.80 = 3.75$ rad/s$^2$. Starting at 12 rad/s: $t = \omega_0 / \alpha = 12/3.75 = 3.2$ s. Consistent.

Layer 2: Pattern

Two different braking strategies are applied to the same flywheel ($I = 2.0$ kg$\cdot$m$^2$, $\omega_0 = 40$ rad/s).

• Strategy A: A large braking torque of 80 N$\cdot$m for a short time.
• Strategy B: A small braking torque of 10 N$\cdot$m for a longer time.

Both bring the flywheel to rest.

(a) What is the angular impulse in each case?

(b) How long does each strategy take?

(c) A third strategy applies a torque of 20 N$\cdot$m. Without calculating, predict the stopping time relative to Strategy B.

Check your answer

**(a)** The angular impulse is the same in both cases, because the change in angular momentum is the same: $$|J_{\text{ang}}| = |\Delta L| = I\omega_0 = 2.0 \times 40 = 80 \text{ kg} \cdot \text{m}^2/\text{s}$$ The total angular impulse does not depend on the braking strategy. It depends only on the initial angular momentum that must be removed. **(b)** Strategy A: $\Delta t = |J_{\text{ang}}| / \tau = 80/80 = 1.0$ s. Strategy B: $\Delta t = 80/10 = 8.0$ s. **(c)** Strategy C uses twice the torque of Strategy B, so it takes half the time: 4.0 s. This is the angular impulse-time tradeoff --- the rotational version of the impulse-time tradeoff from Section 8.2. For a fixed $\Delta L$, doubling the torque halves the time. The product $\tau \cdot \Delta t$ stays the same.

Layer 3: Structure

Why is the angular impulse-momentum theorem the rotational twin of the impulse-momentum theorem? Is this a coincidence, or does it follow from something deeper?

Check your answer

It is not a coincidence. It follows from the structural parallel between Newton's second law for translation and for rotation. The translational theorem starts from $\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$ and integrates over time: $$\int \vec{F}\,dt = \Delta\vec{p}$$ The rotational theorem starts from $\vec{\tau}_{\text{net}} = \frac{d\vec{L}}{dt}$ and integrates over time: $$\int \vec{\tau}\,dt = \Delta\vec{L}$$ The mathematical operation is identical: integrate both sides of Newton's second law over time. The physical quantities change ($\vec{F} \to \vec{\tau}$, $\vec{p} \to \vec{L}$), but the logical structure is the same. This is why the conservation architecture "repeats." It is not that someone discovered two separate theorems that happen to look alike. It is that the same derivation technique (integrating Newton's second law) applied to two different forms of that law (translational and rotational) produces two theorems with the same structure. Whenever you encounter a new domain where Newton's second law takes the form "rate of change of [some quantity] equals [some influence]," you can immediately write down the corresponding impulse theorem: the integral of [influence] over time equals the change in [quantity]. The pattern is general.

Layer 4: Transfer

A helicopter's main rotor blade has a moment of inertia of $I = 120$ kg$\cdot$m$^2$. The engine must spin the blade from rest to an operating angular velocity of 40 rad/s.

(a) What angular impulse must the engine deliver?

(b) If the engine produces a constant torque of 960 N$\cdot$m on the rotor, how long does spin-up take?

(c) In reality, aerodynamic drag on the blade produces a resistive torque that increases with $\omega$. Would the actual spin-up time be shorter or longer than your answer in (b)? Explain using the angular impulse-momentum theorem.

Check your answer

**(a)** The blade starts from rest ($\omega_0 = 0$) and reaches $\omega_f = 40$ rad/s. $$J_{\text{ang}} = \Delta L = I\omega_f - I\omega_0 = 120 \times 40 = 4800 \text{ kg} \cdot \text{m}^2/\text{s}$$ **(b)** With a constant engine torque and no drag: $$\Delta t = \frac{J_{\text{ang}}}{\tau} = \frac{4800}{960} = 5.0 \text{ s}$$ **(c)** The spin-up time would be **longer**. The angular impulse-momentum theorem says $\int \tau_{\text{net}}\,dt = \Delta L$. The required $\Delta L$ is the same (the blade must still reach 40 rad/s). But aerodynamic drag reduces the net torque: $\tau_{\text{net}} = \tau_{\text{engine}} - \tau_{\text{drag}}$. With a smaller net torque at each instant, the integral accumulates more slowly. It takes more time for the area under the $\tau_{\text{net}}(t)$ curve to reach 4800 kg$\cdot$m$^2$/s. Put differently: the engine still delivers 960 N$\cdot$m, but some of that torque goes to fighting drag rather than building angular momentum. The effective torque for changing $L$ is less than 960 N$\cdot$m, so the spin-up takes longer than 5 seconds.

Reflection

Which parts of this section felt like "new learning" and which felt like "translating something you already knew"?

If you studied Section 8.2 carefully, much of this section probably felt familiar. The impulse-momentum theorem became the angular impulse-momentum theorem. Force became torque. Momentum became angular momentum. The graphs changed labels, but the reasoning was the same.

That feeling of recognition is the point. The conservation architecture of physics is not a collection of unrelated theorems. It is a single logical structure --- integrate Newton's second law over time to get an impulse theorem --- applied repeatedly in different domains. Once you see the pattern, each new domain becomes a translation exercise rather than a fresh learning challenge.

Pay attention to what was new. The physical intuition is different: torques act on rotating objects, angular momentum involves moments of inertia, and braking a spinning wheel is not quite the same experience as catching a ball. The mathematics may be parallel, but the physics has its own texture. Learning to feel that texture while recognizing the underlying structure is what mastery looks like.

Looking Ahead

You now know how torques accumulated over time change angular momentum, completing the angular impulse-momentum theorem. This is the tool for analyzing how rotational motion is built up or destroyed by sustained torques.

But what happens when there are no external torques at all? In Section 8.3, you saw that when the net external force is zero, linear momentum is conserved. The same logic applies here: when the net external torque is zero, angular momentum is conserved. In the next section, you will see this principle produce some of the most dramatic effects in all of physics --- from a figure skater who doubles her spin rate by pulling in her arms, to a collapsing star that accelerates to hundreds of rotations per second. Conservation of angular momentum is where the architecture you have been building pays its biggest dividend.