10.5 Newton's Second Law for Rotation

The Equation That Was Waiting to Be Written

You have seen $F = ma$ predict translational acceleration. You apply a net force to an object, and the object accelerates --- proportionally to the force, inversely proportionally to the mass. One equation captures all of translational dynamics.

Now look at what you have built over the last four sections. You have torque --- the rotational effect of a force. You have moment of inertia --- the rotational resistance of a mass distribution. You have angular acceleration --- the rate at which rotational velocity changes.

Can we write a single equation that does for rotation what $F = ma$ does for translation?

Before you read on: A disk and a hoop have the same mass and the same radius. Both experience the same net torque. Which one has the larger angular acceleration?

Think carefully. What is different about these two objects, rotationally speaking?

If your instinct was "the disk," you are right --- and the reasoning matters. The disk has a smaller moment of inertia than the hoop (its mass is distributed closer to the axis on average), so the same torque produces a larger angular acceleration. The hoop, with all its mass at the rim, resists angular acceleration more stubbornly.

But notice that you just used a relationship --- more inertia means less acceleration for the same torque --- without having written it down formally yet. Let's fix that.

Exploring the Relationship

[Interactive: Torque-and-Spin Lab. A top-down view shows a disk mounted on a frictionless axle. Students can apply a tangential force at the rim using a slider (controlling the torque magnitude). A readout displays the moment of inertia $I$, the applied torque $\tau$, and the resulting angular acceleration $\alpha$.

Phase 1 --- Vary the torque: The object is fixed (constant $I$). Students adjust the applied torque from small to large and record the resulting angular acceleration in a data table. A live graph plots $\alpha$ vs. $\tau$.

Guided prompt: "What shape is the graph? If you double the torque, what happens to the angular acceleration?"

Phase 2 --- Vary the moment of inertia: The torque is fixed. Students switch between objects --- a solid disk, a hollow disk, a hoop, a solid sphere --- each with a different moment of inertia (displayed numerically). They record the angular acceleration for each. A live graph plots $\alpha$ vs. $I$.

Guided prompt: "What shape is this graph? If you double the moment of inertia, what happens to the angular acceleration?"

Phase 3 --- Check the product: For each trial, students compute $\tau$ and $I\alpha$. A new column appears in the data table.

Guided prompt: "Compare $\tau$ and $I\alpha$ across all your trials. What do you find?"]

If you did the exploration carefully, two patterns should have emerged:

1. With $I$ held fixed, angular acceleration is proportional to net torque. Double the torque, double the angular acceleration. The graph of $\alpha$ vs. $\tau$ is a straight line through the origin.

2. With $\tau$ held fixed, angular acceleration is inversely proportional to moment of inertia. Double the moment of inertia, halve the angular acceleration. The graph of $\alpha$ vs. $I$ is a hyperbola.

And in every single trial, the product $I\alpha$ equaled the applied torque.

The Rotational Second Law

These patterns combine into one equation:

$$\sum \tau = I\alpha$$

This is Newton's second law for rotation. The net torque on a rigid body equals its moment of inertia times its angular acceleration.

Read it again, slowly. Every word earns its place.

• $\sum \tau$: the net torque --- the sum of all torques about the chosen axis. Just as $F = ma$ requires the net force, not a single force, the rotational law requires the net torque.

• $I$: the moment of inertia about the axis of rotation. This is the rotational analog of mass --- it measures how much the object resists angular acceleration.

• $\alpha$: the angular acceleration. This is the rotational analog of $a$ --- it tells you how fast the angular velocity is changing.

The structure is identical to $F = ma$:

Translational	Rotational
Net force $\sum F$	Net torque $\sum \tau$
Mass $m$	Moment of inertia $I$
Acceleration $a$	Angular acceleration $\alpha$
$\sum F = ma$	$\sum \tau = I\alpha$

But here is something important: $\sum \tau = I\alpha$ is not a new postulate. It is not an independent law that Newton forgot to write down. It is a consequence of $F = ma$, applied to every particle in a rigid body.

Where Does It Come From?

The derivation is worth seeing, at least in outline, because it shows that rotational dynamics is not a separate branch of physics --- it is translational dynamics, cleverly reorganized.

Consider a rigid body rotating about a fixed axis. Pick one small piece of it --- a particle of mass $m_i$ at distance $r_i$ from the axis. This particle moves in a circle, so it has a tangential acceleration $a_{t,i} = r_i \alpha$, where $\alpha$ is the angular acceleration of the whole body (the same for every particle, because the body is rigid).

By Newton's second law applied to this particle in the tangential direction:

$$F_{t,i} = m_i a_{t,i} = m_i r_i \alpha$$

The torque this force produces about the axis is:

$$\tau_i = r_i F_{t,i} = m_i r_i^2 \alpha$$

Now sum over every particle in the body:

$$\sum_i \tau_i = \sum_i m_i r_i^2 \, \alpha = \left(\sum_i m_i r_i^2\right) \alpha$$

The left side is the net torque. The sum in parentheses on the right is the moment of inertia $I = \sum m_i r_i^2$. And $\alpha$ factors out because every particle shares the same angular acceleration (rigid body). So:

$$\sum \tau = I\alpha$$

That is the entire derivation. We applied $F = ma$ to each particle, multiplied by the distance from the axis, and summed. The internal forces between particles cancel in pairs (by Newton's third law), so only external torques survive.

The key idea: $\sum \tau = I\alpha$ is $F = ma$ in disguise. The translational law, applied to every particle and reorganized using rotational variables, gives the rotational law. The structure of the equation is inherited from the structure of translational dynamics.

The Analogy and Its Limits

This is the culmination of the translational-rotational analogy that has been building since Chapter 9:

$$F \to \tau, \qquad m \to I, \qquad a \to \alpha$$

The structure is the same. The objects are different.

But there is one important difference that the analogy can obscure. Mass $m$ is a property of the object alone --- it does not depend on how you look at it or where you measure it. Moment of inertia $I$ depends on the axis. The same object has different moments of inertia about different axes. This means that when you write $\sum \tau = I\alpha$, you must be clear about which axis you are using --- and the torques, the moment of inertia, and the angular acceleration must all refer to the same axis.

This is not a minor bookkeeping detail. It is one of the most common sources of error in rotational dynamics problems, as you will see in the practice section below.

Applying the Law

Let's work through a concrete example to see $\sum \tau = I\alpha$ in action.

Example: A grinding wheel.

A solid disk of mass $M = 5.0 \text{ kg}$ and radius $R = 0.20 \text{ m}$ is initially at rest. A tangential force of $F = 8.0 \text{ N}$ is applied at the rim. Find the angular acceleration.

Step 1: Identify the net torque about the center of the disk.

$$\tau = FR = (8.0)(0.20) = 1.6 \text{ N}\cdot\text{m}$$

Step 2: Find the moment of inertia. For a solid disk rotating about its center:

$$I = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(5.0)(0.20)^2 = 0.10 \text{ kg}\cdot\text{m}^2$$

Step 3: Apply $\sum \tau = I\alpha$.

$$\alpha = \frac{\tau}{I} = \frac{1.6}{0.10} = 16 \text{ rad/s}^2$$

Now suppose the disk were replaced by a hoop of the same mass and radius. The moment of inertia would be $I = MR^2 = 0.20 \text{ kg}\cdot\text{m}^2$, and the angular acceleration would be $\alpha = 1.6 / 0.20 = 8.0 \text{ rad/s}^2$ --- exactly half. Same torque, twice the rotational inertia, half the angular acceleration. The law works exactly as advertised.

Practice

Layer 1: Concrete

A uniform solid cylinder (mass $M = 4.0 \text{ kg}$, radius $R = 0.15 \text{ m}$) is mounted on a frictionless axle through its center. A rope wrapped around the cylinder exerts a constant tangential force of $12 \text{ N}$.

(a) What is the moment of inertia of the cylinder about the axle?

(b) What is the net torque about the axle?

(c) What is the angular acceleration?

Check your answer

**(a)** For a solid cylinder about its central axis: $$I = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(4.0)(0.15)^2 = 0.045 \text{ kg}\cdot\text{m}^2$$ **(b)** The rope pulls tangentially at the rim, so: $$\tau = FR = (12)(0.15) = 1.8 \text{ N}\cdot\text{m}$$ **(c)** Applying $\sum \tau = I\alpha$: $$\alpha = \frac{\tau}{I} = \frac{1.8}{0.045} = 40 \text{ rad/s}^2$$ This is a large angular acceleration --- the cylinder spins up quickly because the rope exerts a substantial torque on an object with a relatively small moment of inertia.

Layer 2: Pattern

A flywheel (a uniform solid disk, $I = 0.50 \text{ kg}\cdot\text{m}^2$) starts from rest and has a constant net torque of $2.0 \text{ N}\cdot\text{m}$ applied to it.

(a) What is the angular acceleration?

(b) What is the angular velocity after $6.0 \text{ s}$?

(c) How many revolutions does the flywheel complete in those $6.0 \text{ s}$?

This problem requires you to combine the torque equation with rotational kinematics from Chapter 9. Use $\omega = \omega_0 + \alpha t$ and $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$.

Check your answer

**(a)** From $\sum \tau = I\alpha$: $$\alpha = \frac{\tau}{I} = \frac{2.0}{0.50} = 4.0 \text{ rad/s}^2$$ **(b)** Starting from rest ($\omega_0 = 0$): $$\omega = \omega_0 + \alpha t = 0 + (4.0)(6.0) = 24 \text{ rad/s}$$ **(c)** The angular displacement is: $$\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2 = 0 + \tfrac{1}{2}(4.0)(6.0)^2 = 72 \text{ rad}$$ Converting to revolutions: $$N = \frac{\theta}{2\pi} = \frac{72}{2\pi} \approx 11.5 \text{ revolutions}$$ Notice the workflow: the torque equation gives you $\alpha$, and then rotational kinematics takes over. This is exactly the same pattern as translational dynamics: $F = ma$ gives you $a$, and then $v = v_0 + at$ and $x = x_0 + v_0 t + \frac{1}{2}at^2$ do the rest.

Layer 3: Structure

Why must we specify the axis when applying $\sum \tau = I\alpha$?

Consider a uniform rod of mass $M$ and length $L$. Its moment of inertia about the center is $\frac{1}{12}ML^2$. Its moment of inertia about one end is $\frac{1}{3}ML^2$.

If a force $F$ is applied perpendicular to the rod at its tip, and you compute the torque and the angular acceleration about each axis, do you get the same $\alpha$? Should you?

Check your answer

You should not expect the same $\alpha$ from both axis choices --- and here is why. **About the center:** $$\tau_{\text{center}} = F \cdot \frac{L}{2}, \qquad I_{\text{center}} = \frac{1}{12}ML^2$$ $$\alpha_{\text{center}} = \frac{F \cdot L/2}{ML^2/12} = \frac{6F}{ML}$$ **About the end (opposite from where the force is applied):** $$\tau_{\text{end}} = F \cdot L, \qquad I_{\text{end}} = \frac{1}{3}ML^2$$ $$\alpha_{\text{end}} = \frac{F \cdot L}{ML^2/3} = \frac{3F}{ML}$$ These are different. And they *should* be different, because the two calculations describe different physical situations. When you choose a different axis, the torques change, the moment of inertia changes, and generally the angular acceleration changes too. The equation $\sum \tau = I\alpha$ is valid about any axis, but all three quantities --- $\sum \tau$, $I$, and $\alpha$ --- must refer to the same axis. In practice, for a rigid body rotating about a fixed physical axis (like a hinge or an axle), you use that axis. The axis is not arbitrary in those problems --- it is determined by the physical setup. The freedom to choose arises mainly in statics problems (Section 10.7) or when analyzing the motion of the center of mass.

Layer 4: Debug

A student solves the following problem:

A thin hoop of mass $2.0 \text{ kg}$ and radius $0.30 \text{ m}$ is mounted on an axle through its center. A tangential force of $5.0 \text{ N}$ is applied at the rim. Find the angular acceleration.

The student writes:

$$I = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(2.0)(0.30)^2 = 0.090 \text{ kg}\cdot\text{m}^2$$

$$\alpha = \frac{\tau}{I} = \frac{(5.0)(0.30)}{0.090} = 16.7 \text{ rad/s}^2$$

The student's angular acceleration is wrong. Find the error and correct it.

Check your answer

The student used the moment of inertia formula for a **solid disk** ($I = \frac{1}{2}MR^2$) instead of the formula for a **hoop** ($I = MR^2$). A hoop has all its mass at the rim, so its moment of inertia is larger than a disk of the same mass and radius. The correct calculation: $$I = MR^2 = (2.0)(0.30)^2 = 0.18 \text{ kg}\cdot\text{m}^2$$ $$\alpha = \frac{\tau}{I} = \frac{(5.0)(0.30)}{0.18} = 8.3 \text{ rad/s}^2$$ The student's answer was exactly twice the correct value. This makes sense: the student used $I = \frac{1}{2}MR^2$ instead of $I = MR^2$, so their moment of inertia was half the true value, and their angular acceleration was double. This is a common error. The moment of inertia formula depends on the shape and mass distribution, not just on $M$ and $R$. A hoop and a disk of the same mass and radius have different moments of inertia because their mass is distributed differently. Always check that the formula you use matches the geometry of the object.

Reflection

Think back over what you have read and explored in this section.

In what sense is $\tau = I\alpha$ "the same" as $F = ma$, and in what sense is it different?

You might consider: the mathematical structure is identical --- a "cause" quantity equals an "inertia" quantity times a "response" quantity. But the physical ingredients are different. Force is a push or pull; torque depends on where and how that push is applied. Mass is a single number; moment of inertia depends on the axis. Acceleration describes motion of a point; angular acceleration describes the spinning of an extended body.

The analogy is powerful --- it lets you transfer everything you know about translational dynamics to rotational dynamics. But the differences are real, and ignoring them (especially the axis-dependence of $I$) leads to errors.

Looking Ahead

You now have the rotational counterpart of Newton's second law. With $\sum \tau = I\alpha$, you can predict angular acceleration from applied torques, just as $F = ma$ lets you predict translational acceleration from applied forces.

But most real objects do not just spin in place. A wheel rolls along the ground. A yo-yo drops while unwinding. A ball rolls down a ramp. These objects translate and rotate at the same time, and the two motions are linked --- the rate of spinning is tied to the rate of moving by a geometric constraint.

In the next section, you will learn to handle these coupled problems by writing both $F = ma$ and $\tau = I\alpha$ simultaneously, connected by the rolling or unwinding constraint. That is where rotational dynamics reaches its full power --- and where the analogy between translation and rotation stops being a metaphor and becomes a working tool.