9.3 Angular Quantities as Derivatives and Integrals

When Constant Acceleration Fails

A ceiling fan is switched off. It does not stop instantly --- it coasts, slowing down gradually. But the deceleration is not constant. Friction in the bearings increases as the lubricant heats up; air drag grows as the blades pass through their own turbulent wake. A reasonable model for the angular acceleration might be

$$\alpha(t) = -0.5t \quad (\text{in rad/s}^2)$$

At the moment you flip the switch ($t = 0$), the angular acceleration is zero --- the fan is still spinning freely. One second later, the angular deceleration is $-0.5$ rad/s$^2$. Two seconds later, $-1.0$ rad/s$^2$. The braking effect grows with time.

Try plugging this into the constant-acceleration equation from Section 9.2: $\omega = \omega_0 + \alpha t$. Which value of $\alpha$ do you use? The value at $t = 0$? That gives $\omega = \omega_0$ for all time --- the fan never slows down. The value at $t = 4$ s? That pretends the deceleration was $-2.0$ rad/s$^2$ all along, which it was not.

The constant-acceleration formulas break for the same reason they broke in Section 2.3: they were derived by pulling a constant out of an integral. When the quantity under the integral is not constant, the derivation does not apply. It is not a matter of picking a better value to plug in. The formula itself does not fit the situation.

But here is the key recognition: if you have already worked through Section 2.3, you know what to do. The approach that rescued translational kinematics --- replacing formulas with integrals --- works here with nothing more than a change of symbols.

Before you read on: If $\alpha(t)$ is not constant, can you still find $\omega(t)$ by integration? What initial information do you need besides $\alpha(t)$ itself?

Think back to Section 2.3. What did you need to know, besides $a(t)$, to determine $v(t)$? The rotational case is the same.

The Parallel You Already Know

Before we develop the rotational integral relationships, let's make sure the translational version is fresh. Answer from memory:

Recall prompt: In Section 2.3, you learned two general integral relationships for translational motion. Can you write them down? What does the dummy variable $t'$ mean? And what happens to these integrals when the acceleration is constant?

If those feel fuzzy, revisit Section 2.3 before continuing. This section follows the same logic, step by step, with rotational variables. If you understood the translational version, the rotational version will cost you almost no additional effort.

Exploration: Watching the Integral Chain Build

[Interactive: Angular Acceleration Integrator. The layout mirrors the Section 2.3 interactive. A text input field accepts a mathematical expression for $\alpha(t)$ (pre-loaded with $\alpha(t) = -0.5t$). Two additional fields allow the student to set initial conditions: $\omega_0$ and $\theta_0$. Below, three vertically stacked graphs update in real time: $\alpha(t)$ on top, $\omega(t)$ in the middle, and $\theta(t)$ on the bottom. The system numerically integrates $\alpha(t)$ to produce $\omega(t)$, then integrates $\omega(t)$ to produce $\theta(t)$. Shaded regions on each graph show the area being accumulated by the integral. A toggle labeled "Show translational version" splits the screen: the left side shows $\alpha \to \omega \to \theta$ and the right side shows $a \to v \to x$ with the same functional form and initial conditions. The graphs are visually identical except for the axis labels.]

Try the following guided explorations:

Step 1: Start with the default, $\alpha(t) = -0.5t$, and set $\omega_0 = 10$ rad/s, $\theta_0 = 0$.

Look at the angular velocity graph. With constant angular acceleration, $\omega(t)$ would be a straight line. Instead, it curves --- the fan slows gently at first, then more and more rapidly as $\alpha$ grows in magnitude. The angular velocity is not linear because the angular acceleration is not constant.

Step 2: Now toggle "Show translational version." Set $a(t) = -0.5t$, $v_0 = 10$ m/s, $x_0 = 0$ on the right side.

The two sets of graphs are identical in shape. The only differences are the axis labels: $\omega$ versus $v$, $\theta$ versus $x$. The calculus does not care which physical quantity it is operating on. An integral is an integral.

Step 3: Change $\alpha(t)$ to $3\sin(t)$. Set $\omega_0 = 0$.

Watch the angular velocity oscillate. The angular acceleration pushes the fan one way, then the other. What does $\theta(t)$ look like? It grows on average but wobbles as it does so. Now predict: will $\theta(t)$ ever decrease? Look at the $\omega(t)$ graph --- does $\omega$ ever go negative? If it does, $\theta$ decreases during that interval. If it does not, $\theta$ only increases.

Before reading on: Does this interactive feel familiar? It should. The structure is not "like" the Section 2.3 interactive --- it is the Section 2.3 interactive with different labels. Why does that work?

The General Integral Relationships

The logic is identical to Section 2.3. We start from the definitions established in Section 9.1:

$$\omega(t) = \frac{d\theta}{dt}, \qquad \alpha(t) = \frac{d\omega}{dt}$$

The second equation says that $\alpha(t)$ is the rate of change of $\omega(t)$. To recover $\omega(t)$ from $\alpha(t)$, we integrate both sides from time $0$ to time $t$:

$$\int_0^t \alpha(t') \, dt' = \int_0^t \frac{d\omega}{dt'} \, dt' = \omega(t) - \omega(0)$$

Solving for $\omega(t)$:

$$\boxed{\omega(t) = \omega_0 + \int_0^t \alpha(t') \, dt'}$$

This says: the angular velocity at time $t$ equals the initial angular velocity plus the total accumulated angular acceleration between time $0$ and time $t$. The integral sums up all the little nudges that $\alpha$ delivers, moment by moment.

Similarly, since $\omega = d\theta/dt$, we integrate to find $\theta(t)$:

$$\boxed{\theta(t) = \theta_0 + \int_0^t \omega(t') \, dt'}$$

These two boxed equations are the general rotational kinematic relationships. They always work --- for constant $\alpha$, for time-varying $\alpha$, for oscillating $\alpha$, for $\alpha$ given as a graph rather than a formula. The constant-acceleration equations from Section 9.2 are the special case where the integrals collapse to simple algebra.

Compare with the translational versions from Section 2.3:

Translational (Section 2.3)	Rotational (this section)
$v(t) = v_0 + \int_0^t a(t') \, dt'$	$\omega(t) = \omega_0 + \int_0^t \alpha(t') \, dt'$
$x(t) = x_0 + \int_0^t v(t') \, dt'$	$\theta(t) = \theta_0 + \int_0^t \omega(t') \, dt'$

The structure is identical. The calculus does not know or care whether the quantity being integrated is a translational acceleration or an angular one. Differentiation and integration are operations on functions --- the physical interpretation is layered on top. This is why the translational-rotational analogy works so well: it rests on the universality of calculus, not on any coincidence of physics.

Connection: The Same Calculus, Twice

Section 2.3 built the chain $a(t) \to v(t) \to x(t)$ through integration. This section builds the chain $\alpha(t) \to \omega(t) \to \theta(t)$ through exactly the same integrals. The parallel is not approximate --- it is exact.

This means everything you learned in Section 2.3 transfers immediately:

If $\alpha(t)$ is a polynomial, the integrals produce higher-degree polynomials.
If $\alpha(t)$ is exponential, the integrals produce exponentials.
If the integral has no closed form, you can still evaluate it numerically or leave it in integral form.
The derivative chain runs in reverse: $\alpha = d\omega/dt$ and $\omega = d\theta/dt$.

You do not need to re-learn any of this. You need only recognize that the rotational versions are instances of the same underlying mathematics.

Worked Example: Slowing Fan

Let's return to the fan from the opening scenario and solve it completely.

Problem: A fan starts at $\omega_0 = 10$ rad/s with angular acceleration $\alpha(t) = -0.5t$ rad/s$^2$. Find $\omega(t)$ and $\theta(t)$ (assuming $\theta_0 = 0$). At what time does the fan stop?

Step 1: Find $\omega(t)$ by integrating $\alpha(t)$.

$$\omega(t) = \omega_0 + \int_0^t \alpha(t') \, dt' = 10 + \int_0^t (-0.5t') \, dt'$$

$$= 10 + \left[-0.25t'^2\right]_0^t = 10 - 0.25t^2$$

So $\omega(t) = 10 - 0.25t^2$ rad/s.

Step 2: Find when the fan stops.

Set $\omega(t) = 0$:

$$10 - 0.25t^2 = 0 \implies t^2 = 40 \implies t = \sqrt{40} \approx 6.32 \text{ s}$$

Step 3: Find $\theta(t)$ by integrating $\omega(t)$.

$$\theta(t) = \theta_0 + \int_0^t \omega(t') \, dt' = 0 + \int_0^t (10 - 0.25t'^2) \, dt'$$

$$= \left[10t' - \frac{0.25}{3}t'^3\right]_0^t = 10t - \frac{t^3}{12}$$

So $\theta(t) = 10t - \frac{t^3}{12}$ rad.

Step 4: Sanity check.

Differentiate $\theta(t)$: $\frac{d\theta}{dt} = 10 - \frac{3t^2}{12} = 10 - 0.25t^2 = \omega(t)$. Correct.

Differentiate $\omega(t)$: $\frac{d\omega}{dt} = -0.5t = \alpha(t)$. Correct.

Step 5: Total angle before stopping.

At $t = \sqrt{40}$: $\theta = 10\sqrt{40} - \frac{(\sqrt{40})^3}{12} = 10\sqrt{40} - \frac{40\sqrt{40}}{12} = \sqrt{40}\left(10 - \frac{40}{12}\right) = \sqrt{40} \cdot \frac{80}{12} \approx 6.32 \times 6.67 \approx 42.2$ rad.

That is about $42.2 / (2\pi) \approx 6.7$ revolutions. A fan coasting to a stop in a few seconds and making about 7 turns feels physically reasonable.

Faded Example: $\alpha(t) = 6e^{-2t}$

A motor starts from rest ($\omega_0 = 0$, $\theta_0 = 0$) with angular acceleration $\alpha(t) = 6e^{-2t}$ rad/s$^2$. Find $\omega(t)$ and $\theta(t)$.

Step 1: Find $\omega(t)$.

$$\omega(t) = \omega_0 + \int_0^t \alpha(t') \, dt' = 0 + \int_0^t 6e^{-2t'} \, dt'$$

Your turn: Evaluate $\int_0^t 6e^{-2t'} \, dt'$.

Check your answer

$$\omega(t) = 6 \cdot \left[-\frac{1}{2}e^{-2t'}\right]_0^t = -3e^{-2t} + 3 = 3(1 - e^{-2t})$$ So $\omega(t) = 3(1 - e^{-2t})$ rad/s. As $t \to \infty$, $\omega \to 3$ rad/s. The motor approaches a terminal angular speed --- the exponentially decaying acceleration adds less and less angular velocity over time.

Step 2: Find $\theta(t)$.

$$\theta(t) = 0 + \int_0^t \omega(t') \, dt' = \int_0^t 3(1 - e^{-2t'}) \, dt'$$

Your turn: Evaluate this integral.

Check your answer

$$\theta(t) = 3\int_0^t (1 - e^{-2t'}) \, dt' = 3\left[t' + \frac{1}{2}e^{-2t'}\right]_0^t = 3\left[\left(t + \frac{1}{2}e^{-2t}\right) - \left(0 + \frac{1}{2}\right)\right]$$ $$= 3t + \frac{3}{2}e^{-2t} - \frac{3}{2} = 3t + \frac{3}{2}(e^{-2t} - 1)$$ So $\theta(t) = 3t - \frac{3}{2}(1 - e^{-2t})$ rad.

Step 3: Sanity check.

Your turn: Differentiate $\theta(t)$ to verify you get $\omega(t)$, and differentiate $\omega(t)$ to verify you get $\alpha(t)$.

Check your answer

$$\frac{d\theta}{dt} = 3 - \frac{3}{2} \cdot (-2)e^{-2t} \cdot (-1) = 3 - 3e^{-2t} = 3(1 - e^{-2t}) = \omega(t) \quad \checkmark$$ Wait --- let's be more careful. From $\theta(t) = 3t - \frac{3}{2}(1 - e^{-2t})$: $$\frac{d\theta}{dt} = 3 - \frac{3}{2}(0 - (-2)e^{-2t}) = 3 - 3e^{-2t} = 3(1 - e^{-2t}) = \omega(t) \quad \checkmark$$ $$\frac{d\omega}{dt} = 3 \cdot 2e^{-2t} = 6e^{-2t} = \alpha(t) \quad \checkmark$$

Graphical Interpretation

The relationships between $\alpha$, $\omega$, and $\theta$ have a clean graphical reading, identical in structure to the translational case from Section 2.3:

The slope of $\theta(t)$ at any instant is $\omega(t)$. If $\theta(t)$ is getting steeper, the object is spinning faster. If the curve levels off, the angular velocity is approaching zero.
The slope of $\omega(t)$ at any instant is $\alpha(t)$. If $\omega(t)$ is increasing, the angular acceleration is positive. If $\omega(t)$ is decreasing, $\alpha < 0$.
The area under $\alpha(t)$ from $0$ to $t$ equals $\omega(t) - \omega_0$. The accumulated angular acceleration gives the change in angular velocity.
The area under $\omega(t)$ from $0$ to $t$ equals $\theta(t) - \theta_0$. The accumulated angular velocity gives the angular displacement.

These are the same four statements you learned for $a$, $v$, and $x$ in Sections 2.1 through 2.3. If you drew three stacked graphs --- $\alpha(t)$, $\omega(t)$, $\theta(t)$ --- you could read the relationships off visually: slopes going down, areas going up.

Practice

Layer 1: Concrete

A flywheel starts from rest ($\omega_0 = 0$, $\theta_0 = 0$) with angular acceleration $\alpha(t) = 12t - 3t^2$ (in rad/s$^2$).

(a) Find $\omega(t)$.

(b) Find $\theta(t)$.

(c) At what time does the angular acceleration return to zero? What is the angular velocity at that moment?

Check your answer

**(a)** $\omega(t) = \int_0^t (12t' - 3t'^2) \, dt' = 6t^2 - t^3$ rad/s. **(b)** $\theta(t) = \int_0^t (6t'^2 - t'^3) \, dt' = 2t^3 - \frac{t^4}{4}$ rad. **(c)** Set $\alpha(t) = 0$: $12t - 3t^2 = 3t(4 - t) = 0$, so $t = 0$ or $t = 4$ s. At $t = 4$ s: $\omega(4) = 6(16) - 64 = 96 - 64 = 32$ rad/s. **Sanity check:** $d\omega/dt = 12t - 3t^2 = \alpha(t)$ and $d\theta/dt = 6t^2 - t^3 = \omega(t)$. Consistent.

Layer 2: Pattern

Below are three angular velocity graphs for different spinning objects. For each, sketch (or describe) the corresponding $\alpha(t)$ and $\theta(t)$ graphs.

(a) $\omega(t)$ is a straight line with positive slope, starting from a positive value.

(b) $\omega(t)$ starts at a large positive value and curves downward, leveling off at a smaller positive value (like a decaying exponential approaching a horizontal asymptote).

(c) $\omega(t)$ oscillates sinusoidally about zero.

Check your answer

**(a)** $\omega(t)$ is linear with positive slope, so $\alpha(t)$ is a positive constant (horizontal line). Since $\omega > 0$ and increasing, $\theta(t)$ is a parabola opening upward --- the object spins faster and faster, accumulating angle at a growing rate. **(b)** $\omega(t)$ starts high and curves downward toward an asymptote. The slope of $\omega(t)$ is negative and decreasing in magnitude, so $\alpha(t)$ is a negative function that approaches zero --- something like $\alpha = -ke^{-t/\tau}$. Since $\omega(t) > 0$ throughout, $\theta(t)$ always increases, but its slope decreases over time. $\theta(t)$ climbs steeply at first, then approaches a constant slope (since $\omega$ approaches a constant). **(c)** $\omega(t) = A\sin(\beta t)$. Then $\alpha(t) = A\beta\cos(\beta t)$ --- a cosine, shifted ahead by a quarter period. And $\theta(t) = \theta_0 - \frac{A}{\beta}\cos(\beta t) + \frac{A}{\beta}$ --- it oscillates, meaning the object rocks back and forth without completing full revolutions (if the amplitude is small enough). **The key skill:** slopes of one graph give the values of the graph above it; areas under one graph give the changes in the graph below it.

Layer 3: Structure

Why is the calculus structure the same for translation and rotation? A student says: "It's just a coincidence that the equations look the same." Is this right?

Check your answer

It is not a coincidence. The calculus structure is the same because the **definitions** are constructed in parallel: - Velocity is the time derivative of position: $v = dx/dt$. - Angular velocity is the time derivative of angular position: $\omega = d\theta/dt$. - Acceleration is the time derivative of velocity: $a = dv/dt$. - Angular acceleration is the time derivative of angular velocity: $\alpha = d\omega/dt$. Once you define quantities through the same derivative chain, the integral relationships follow automatically. Integration is the inverse of differentiation --- that is a fact about calculus, not about physics. So the integral $\omega(t) = \omega_0 + \int_0^t \alpha(t') \, dt'$ is not a separate physical law; it is the mathematical consequence of having defined $\alpha$ as $d\omega/dt$. The parallel is exact whenever you have a quantity, its first time derivative, and its second time derivative. The same structure appears in any context with that pattern --- electrical circuits (charge, current, rate of change of current), population dynamics (population, growth rate, rate of change of growth rate), and many others. Calculus provides the universal scaffolding; physics provides the particular quantities that fit into it.

Layer 4: Transfer

A motor's angular speed decays exponentially after being switched off: $\omega(t) = \omega_0 e^{-t/\tau}$, where $\omega_0 = 120$ rad/s and $\tau = 5$ s.

(a) Find the angular acceleration $\alpha(t)$.

(b) Find the total angle traversed from $t = 0$ to $t = T$.

(c) Find the total angle traversed as $T \to \infty$. Does the motor spin forever?

Check your answer

**(a)** Differentiate: $\alpha(t) = \frac{d\omega}{dt} = -\frac{\omega_0}{\tau}e^{-t/\tau} = -24e^{-t/5}$ rad/s$^2$. The angular acceleration is always negative (the motor is always decelerating), and it decays to zero as the motor slows down. **(b)** Integrate the angular velocity: $$\theta(T) - \theta_0 = \int_0^T \omega_0 e^{-t/\tau} \, dt = \omega_0 \left[-\tau e^{-t/\tau}\right]_0^T = \omega_0 \tau (1 - e^{-T/\tau})$$ $$= 120 \cdot 5 \cdot (1 - e^{-T/5}) = 600(1 - e^{-T/5}) \text{ rad}$$ **(c)** As $T \to \infty$: $\theta_{\text{total}} = \omega_0 \tau = 600$ rad $\approx 95.5$ revolutions. The motor does spin forever in this idealized model --- $\omega$ never quite reaches zero. But the total angle is finite. After a few time constants ($\sim 5\tau = 25$ s), the motor is barely turning, and the remaining angle is negligible. This is analogous to the convergent-integral examples from Section 2.3: an ever-decreasing rate can still produce a finite total.

Reflection

Think about what you have worked through in this section, then consider these questions:

How does the translational-rotational analogy reduce your learning burden? You did not learn new calculus techniques in this section. You did not encounter new types of integrals. You applied the same integral chain from Section 2.3 to different physical quantities. The effort of understanding this section was almost entirely the effort of recognizing a pattern you already knew.

That recognition is itself a skill worth cultivating. In the chapters ahead --- when torque, angular momentum, and rotational energy appear --- the analogy will continue to carry weight. Each time, the question will be the same: Is this really new, or is this a familiar structure wearing different symbols?

Looking Ahead

You now have the complete calculus toolkit for rotational kinematics: definitions from Section 9.1, constant-acceleration formulas from Section 9.2, and general integral relationships from this section. These describe how an object rotates --- the angles, speeds, and accelerations.

But so far, everything has been purely angular. A point on a spinning wheel does not just have an angular velocity --- it has a tangential speed, measured in meters per second, that depends on how far it sits from the axis. In Section 9.4, we connect the angular world to the translational one through geometry: $s = r\theta$, $v_t = r\omega$, and $a_c = r\omega^2$. These relationships bridge the two descriptions and let you answer questions like: how fast is the tip of a helicopter blade actually moving through the air?