9.2 Rotational Motion with Constant Angular Acceleration

A hard drive spins up

A hard drive sits motionless. You press the power button, and the platters begin to spin --- slowly at first, then faster and faster, until they reach 7200 revolutions per minute. The angular acceleration during spin-up is roughly constant. How long does it take? How many revolutions does the disk complete before it reaches full speed?

You already know how to answer these questions. Not because you have studied rotational formulas, but because you spent all of Chapter 2 mastering the mathematics of constant acceleration in a straight line. The structure of that mathematics does not care whether the thing accelerating is a car on a highway or a disk on a spindle. Constant acceleration is constant acceleration. The symbols change; the logic does not.

This section makes that claim precise. You will see that the kinematic equations for constant angular acceleration are not new formulas to memorize --- they are the same formulas you already know, wearing different clothes.

Prediction: wheels and cars

Before we develop anything formally, test your intuition on two problems side by side.

Prediction exercise:

Problem A (translational): A car starts from rest and accelerates at $2$ m/s$^2$. After 5 seconds, what is its velocity?

Problem B (rotational): A wheel starts from rest and accelerates at $2$ rad/s$^2$. After 5 seconds, what is its angular velocity?

Write down your answers to both before continuing.

[Interactive: Two input fields, side by side, labeled "Car velocity (m/s)" and "Wheel angular velocity (rad/s)." The student enters their predictions and clicks "Lock in." The answers are stored and displayed alongside the formal results below.]

If you answered $10$ m/s for the car and $10$ rad/s for the wheel, you are correct on both counts --- and you likely used the same reasoning for each. The numbers came out the same because the problems have the same structure. That is not a coincidence. It is the entire point of this section.

The guiding question

Which parts of one-dimensional constant-acceleration motion carry over to rotation unchanged? All of them? Some of them? And if the formulas really are identical, why are they identical?

The parallel derivation

In Section 2.2, you derived the translational kinematic equations by starting from a single assumption --- constant acceleration --- and integrating twice. Let's repeat that derivation, but now for rotation. Section 9.1 established the derivative chain for rotational quantities:

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

These are structurally identical to $a = dv/dt$ and $v = dx/dt$. The calculus does not know or care whether the quantities represent meters or radians. So if we assume $\alpha$ is constant, the integration proceeds in exactly the same way.

First integration. Start from

$$\frac{d\omega}{dt} = \alpha$$

where $\alpha$ is a constant. Integrate both sides from $t = 0$ to time $t$:

$$\int_0^t \frac{d\omega}{dt'}\,dt' = \int_0^t \alpha\,dt'$$

The left side is $\omega(t) - \omega_0$ by the fundamental theorem of calculus. The right side is $\alpha t$, since $\alpha$ is constant. Therefore:

$$\boxed{\omega = \omega_0 + \alpha t}$$

Compare with $v = v_0 + at$ from Section 2.2. The structure is identical: initial value plus rate of change times time.

Second integration. Now use $\omega = d\theta/dt$ with the expression we just found:

$$\frac{d\theta}{dt} = \omega_0 + \alpha t$$

Integrate from $0$ to $t$:

$$\int_0^t \frac{d\theta}{dt'}\,dt' = \int_0^t (\omega_0 + \alpha t')\,dt'$$

The left side is $\theta(t) - \theta_0$. On the right, $\int_0^t \omega_0\,dt' = \omega_0 t$ and $\int_0^t \alpha t'\,dt' = \frac{1}{2}\alpha t^2$. Therefore:

$$\boxed{\theta = \theta_0 + \omega_0 t + \tfrac{1}{2}\alpha t^2}$$

Compare with $x = x_0 + v_0 t + \frac{1}{2}at^2$. Again, identical structure.

Eliminating time. Solve the first equation for $t$:

$$t = \frac{\omega - \omega_0}{\alpha}$$

Substitute into the second equation and simplify (the algebra is the same manipulation you did in Section 2.2):

$$\boxed{\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)}$$

Compare with $v^2 = v_0^2 + 2a(x - x_0)$. Same structure, same derivation, same result.

The concept: same mathematics, different labels

Here is the complete correspondence, collected in one place:

Translational	Rotational
$v = v_0 + at$	$\omega = \omega_0 + \alpha t$
$x = x_0 + v_0 t + \frac{1}{2}at^2$	$\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$
$v^2 = v_0^2 + 2a(x - x_0)$	$\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$

Every $x$ became $\theta$. Every $v$ became $\omega$. Every $a$ became $\alpha$. Nothing else changed. The equations are not analogous --- they are structurally identical. The reason is straightforward: the derivation in both cases consists of integrating a constant twice. Calculus does not distinguish between linear displacement and angular displacement. The integral of a constant is a linear function. The integral of a linear function is a quadratic. This is a fact about mathematics, not about physics.

The physical content is in the assumption: that the angular acceleration $\alpha$ is constant. When that assumption holds --- a spinning hard drive during uniform spin-up, a flywheel subjected to a steady torque, a merry-go-round pushed at a constant rate --- these equations give exact predictions. When it does not hold, you must return to the integral relationships, just as you did for non-constant translational acceleration in Section 2.3.

Exploration: side by side

[Interactive: Translational-rotational parallel explorer. The screen is divided into two panels.

Left panel (Translational): A car on a straight track. Three sliders control $x_0$, $v_0$, and $a$. Below the track, synchronized graphs show $x(t)$, $v(t)$, and $a(t)$. The car animates along the track as time advances.

Right panel (Rotational): A wheel viewed from the side, with a radial marker line. Three sliders control $\theta_0$, $\omega_0$, and $\alpha$. Below the wheel, synchronized graphs show $\theta(t)$, $\omega(t)$, and $\alpha(t)$. The wheel rotates as time advances, with the marker line sweeping through angles.

Linking mode: A toggle labeled "Mirror values" locks the numerical values across panels: setting $a = 3$ m/s$^2$ on the left automatically sets $\alpha = 3$ rad/s$^2$ on the right, and vice versa. With mirroring on, students see that the graphs on both sides have exactly the same shape --- the same straight line for velocity/angular velocity, the same parabola for position/angular position.

Guided prompts appear below the panels: - "Set both sides to start from rest with acceleration 2 (m/s$^2$ and rad/s$^2$). Compare the graphs. Are they identical in shape?" - "Now set the initial velocity/angular velocity to $-4$. Watch for the moment of reversal. Does it happen at the same time on both sides?" - "Solve the translational problem: how far does the car travel in 3 seconds? Now look at the rotational side. How many radians does the wheel sweep in 3 seconds? Did the answer structure change?"]

The point of this exploration is visceral, not just intellectual. You should see that the graphs are the same shape, that the reversal happens at the same time, that the numerical answers match when the inputs match. The translational-rotational analogy is not a loose metaphor. It is exact.

A note on units: RPM, rad/s, and rev/s

Rotational problems in the real world rarely hand you angular velocity in rad/s. Hard drives are rated in RPM (revolutions per minute). Turntables spin at 33-1/3 RPM or 45 RPM. Engines are described in terms of RPM. Before you can use the kinematic equations, you must convert to radians per second, because the equations were derived using radian measure (and the derivative relationships $\omega = d\theta/dt$ and $\alpha = d\omega/dt$ require radians --- see Section 9.1).

The conversion chain is:

$$1 \text{ rev} = 2\pi \text{ rad}, \qquad 1 \text{ min} = 60 \text{ s}$$

So to convert RPM to rad/s:

$$\omega \;(\text{rad/s}) = \omega \;(\text{RPM}) \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}}$$

For example, 7200 RPM is:

$$7200 \times \frac{2\pi}{60} = 240\pi \approx 754 \text{ rad/s}$$

And to convert between rev/s and rad/s, simply multiply or divide by $2\pi$:

$$\omega \;(\text{rad/s}) = 2\pi \times \omega \;(\text{rev/s})$$

Forgetting this conversion is the single most common error in rotational kinematics problems. We will return to this in the Debug practice problem below.

Connection to Section 2.2

The parallel between this section and Section 2.2 is not accidental --- it is the entire pedagogical point. Section 2.2 derived the translational kinematic equations by integrating $a = dv/dt$ twice under the assumption that $a$ is constant. This section performed exactly the same derivation, integrating $\alpha = d\omega/dt$ twice under the assumption that $\alpha$ is constant. The mathematical steps were identical at every stage. Even the algebra for eliminating time was the same.

This means you do not need to memorize a second set of formulas. If you understand where $v = v_0 + at$ comes from, you already understand where $\omega = \omega_0 + \alpha t$ comes from. If you can derive $x = x_0 + v_0 t + \frac{1}{2}at^2$ from scratch, you can derive $\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$ from scratch. The translational-rotational analogy is a powerful learning tool precisely because it reduces what you need to learn.

Practice

Layer 1: Concrete --- standard calculations

Problem 1. A centrifuge starts from rest and accelerates at a constant angular acceleration of $50$ rad/s$^2$.

(a) What is its angular velocity after $8$ seconds?

(b) How many radians has it turned through in that time?

(c) How many complete revolutions is that?

Check your answer

Given: $\omega_0 = 0$, $\alpha = 50$ rad/s$^2$, $t = 8$ s. (a) $\omega = \omega_0 + \alpha t = 0 + 50 \times 8 = 400$ rad/s. (b) $\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + 0 + \frac{1}{2}(50)(64) = 1600$ rad. (c) Each revolution is $2\pi$ radians, so: $1600 / (2\pi) \approx 254.6$ revolutions.

Problem 2. A grinding wheel spinning at $900$ RPM is switched off and decelerates uniformly, coming to rest in $15$ seconds.

(a) What is the angular deceleration?

(b) How many revolutions does the wheel make while slowing down?

Check your answer

First, convert: $\omega_0 = 900 \times \frac{2\pi}{60} = 30\pi$ rad/s. The wheel comes to rest, so $\omega = 0$. (a) From $\omega = \omega_0 + \alpha t$: $$0 = 30\pi + \alpha(15) \implies \alpha = -2\pi \text{ rad/s}^2 \approx -6.28 \text{ rad/s}^2$$ (b) Use $\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$ with $\theta_0 = 0$: $$\theta = 30\pi(15) + \frac{1}{2}(-2\pi)(225) = 450\pi - 225\pi = 225\pi \text{ rad}$$ In revolutions: $225\pi / (2\pi) = 112.5$ revolutions. Alternatively, use the time-free equation: $\omega^2 = \omega_0^2 + 2\alpha\theta$ gives $0 = (30\pi)^2 + 2(-2\pi)\theta$, so $\theta = (30\pi)^2/(4\pi) = 900\pi^2/(4\pi) = 225\pi$ rad. Same answer.

Layer 2: Pattern --- unit conversions in context

Problem 3. A turntable accelerates from rest to $33\frac{1}{3}$ RPM in $2.0$ seconds with constant angular acceleration.

(a) What is the angular acceleration in rad/s$^2$?

(b) Through how many revolutions does it turn during spin-up?

(c) Repeat the calculation if the target speed is $45$ RPM with the same spin-up time.

Check your answer

(a) Convert: $33\frac{1}{3}$ RPM $= \frac{100}{3} \times \frac{2\pi}{60} = \frac{100\pi}{90} = \frac{10\pi}{9}$ rad/s $\approx 3.49$ rad/s. From $\omega = \omega_0 + \alpha t$: $\alpha = \frac{10\pi/9}{2.0} = \frac{10\pi}{18} = \frac{5\pi}{9} \approx 1.75$ rad/s$^2$. (b) $\theta = \frac{1}{2}\alpha t^2 = \frac{1}{2} \cdot \frac{5\pi}{9} \cdot 4 = \frac{10\pi}{9}$ rad. In revolutions: $\frac{10\pi/9}{2\pi} = \frac{10}{18} = \frac{5}{9} \approx 0.56$ revolutions. (c) $45$ RPM $= 45 \times \frac{2\pi}{60} = \frac{3\pi}{2}$ rad/s. $\alpha = \frac{3\pi/2}{2.0} = \frac{3\pi}{4} \approx 2.36$ rad/s$^2$. $\theta = \frac{1}{2} \cdot \frac{3\pi}{4} \cdot 4 = \frac{3\pi}{2}$ rad $= \frac{3}{4}$ revolutions. Notice the pattern: higher target speed with the same spin-up time requires larger angular acceleration and covers more angle.

Layer 3: Structure --- why the derivation works

Before you read on: Why does the derivation of the constant-angular-acceleration kinematic equations work in exactly the same way as the derivation of the constant-translational-acceleration equations in Chapter 2? Is this a deep physical fact, or a mathematical inevitability?

Check your answer

It is a mathematical inevitability. The derivation in both cases follows the same abstract pattern: 1. Start with a quantity whose time derivative is constant. 2. Integrate once to get a linearly growing quantity. 3. Integrate again to get a quadratically growing quantity. 4. Eliminate time algebraically to get a third relation. The calculus does not care what the symbols represent. Whether you write $a = dv/dt$ or $\alpha = d\omega/dt$, the integration of a constant produces a linear function, and the integration of a linear function produces a quadratic. The *structure* of the result is determined entirely by the mathematical operation, not by the physical meaning of the variables. This is why the translational-rotational analogy is so powerful: it is not a loose resemblance. It is the same theorem applied to different variables. Any system of quantities connected by $\text{(something)} = d(\text{something else})/dt$ will produce the same kinematic equations under the assumption that the top-level quantity is constant.

Layer 4: Debug --- the unit conversion trap

A student is solving the hard drive problem from the opening of this section. The hard drive accelerates from rest at $\alpha = 150$ rad/s$^2$ and needs to reach $7200$ RPM. The student writes:

$$\omega = \omega_0 + \alpha t$$ $$7200 = 0 + 150 \cdot t$$ $$t = 48 \text{ s}$$

The student concludes it takes $48$ seconds to spin up. A real hard drive reaches full speed in about $5$ seconds. Something is clearly wrong.

Where is the error, and what is the correct answer?

Check your answer

The student used $\omega = 7200$ directly, but $7200$ is in RPM, not rad/s. The kinematic equations require consistent units, and since $\alpha$ is in rad/s$^2$, the angular velocity must be in rad/s. Converting properly: $$\omega = 7200 \text{ RPM} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = 240\pi \approx 754 \text{ rad/s}$$ Now: $$754 = 0 + 150 \cdot t$$ $$t \approx 5.0 \text{ s}$$ The student's answer was off by a factor of about $9.6$ --- which is $\frac{7200}{754} \approx 9.55$. The error is not small or subtle. Failing to convert RPM to rad/s produces answers that are wildly wrong, and the discrepancy is large enough that a reality check ("does a hard drive really take 48 seconds to spin up?") should catch it immediately. The lesson: whenever a problem states angular velocity in RPM or rev/s, convert to rad/s before substituting into any equation. This is not optional.

Graphs of constant angular acceleration

The graphical interpretation of these equations is the same as the translational case, and reading rotational graphs is a skill worth practicing.

When $\alpha$ is constant:

• The $\alpha$-vs-$t$ graph is a horizontal line at height $\alpha$.
• The $\omega$-vs-$t$ graph is a straight line with slope $\alpha$ and intercept $\omega_0$. The area under this line between $t = 0$ and $t$ equals the angular displacement $\theta - \theta_0$.
• The $\theta$-vs-$t$ graph is a parabola. If $\alpha > 0$ and $\omega_0 \geq 0$, the parabola opens upward. If $\alpha > 0$ but $\omega_0 < 0$, the wheel initially turns in the negative direction, slows, reverses, and then accelerates in the positive direction --- just like a ball thrown upward under gravity.

Every graphical interpretation you learned for translational kinematics in Sections 2.2 and 2.4 carries over directly. The slope of $\theta(t)$ is $\omega(t)$. The slope of $\omega(t)$ is $\alpha$. The area under $\omega(t)$ is $\Delta\theta$. The area under $\alpha(t)$ is $\Delta\omega$. Same rules, different axis labels.

Reflection

You have now seen the constant-acceleration kinematic equations twice --- once for straight-line motion in Chapter 2, and once for rotation in this section. The derivations were identical. The formulas are identical in structure. The graphical interpretations are identical.

Did you need to memorize new formulas, or just translate old ones?

Think about what this means for your study strategy. Instead of maintaining two separate formula sheets --- one for translation, one for rotation --- you can carry a single set of relationships and swap symbols as needed. The analogy is not a shortcut or a trick. It is the mathematically correct observation that constant-rate-of-change problems all have the same solution, regardless of what is changing.

Looking Ahead

The constant-acceleration equations you derived here are exact and powerful, but they apply only when $\alpha$ is truly constant. Many real rotating systems do not have this luxury: a fan slowing under air resistance experiences an angular deceleration that depends on how fast it is spinning, and a motor whose torque varies with speed has a non-constant $\alpha(t)$. In the next section, you will handle these more general cases by returning to the integral relationships $\omega(t) = \omega_0 + \int_0^t \alpha(t')\,dt'$ and $\theta(t) = \theta_0 + \int_0^t \omega(t')\,dt'$ --- the same calculus approach you used for non-constant translational acceleration in Section 2.3. Once again, the structure carries over. The only difference is the symbols.