Newton's Second Law in Rotational Form

Lesson ~11 min read 8 MCQs

In simple terms: In simple terms, this topic is about how unbalanced twists (torques) make things spin faster or slower, just like unbalanced forces make things speed up or slow down in a line.

Why this matters

Have you ever tried to open one of those heavy, old wooden doors, like at a library or a historic building in Boston? If you push right near the hinges, you have to heave with all your might. But if you push on the side farthest from the hinges, near the doorknob, it swings open easily. You're applying a force in both cases, but where you push and how it rotates are deeply connected. That same principle explains why a figure skater pulls their arms in to spin faster or how a wrench can loosen a stubborn bolt on a bike.

We've already mastered how forces cause objects to accelerate in a straight line. Now, we're going to build on that and establish the rules for what makes things spin. We'll see how Newton's famous second law has a rotational cousin that governs everything from spinning bike wheels to planets orbiting the sun.

Diagram

This diagram shows a physical setup and its corresponding free-body diagrams to explain Newton's Second Law for rotation. On the left, a block of mass 'm' hangs from a string that goes over a pulley of mass 'M'. On the right, two separate diagrams analyze the forces and torques: one for the block showing linear forces, and one for the pulley showing the torque from tension. Key equations for each part and the linking equation 'a = αR' are clearly labeled.

Concept map

flowchart TD
    A[Start: Analyze System] --> B{Object Type?};
    B -->|Linear Motion| C[Draw FBD & Set up F_net = ma];
    B -->|Rotational Motion| D[Draw FBD & Set up τ_net = Iα];
    C --> E{Is it a combined system?};
    D --> E;
    E -->|Yes| F[Identify Linking Equation e.g., a = αR];
    F --> G[Solve System of Equations for Unknowns];
    G --> H[End];
    E -->|No| I[Solve Single Equation];
    I --> H;

Core explanation

You've spent a lot of time with your good friend, Newton's Second Law: F_net = ma. It’s a powerful tool that tells us an object's acceleration is directly proportional to the net force on it and inversely proportional to its mass. More force means more acceleration. More mass means less acceleration. Simple, elegant, and true.

Well, get ready to meet its rotational twin. For every rule in linear motion, there's a parallel rule in rotational motion.

Linear Motion	Rotational Motion	Relationship
Force (`F`)	Torque (`τ`)	The rotational equivalent of force.
Mass (`m`)	Rotational Inertia (`I`)	The resistance to a change in motion.
Acceleration (`a`)	Angular Acceleration (`α`)	The rate of change of velocity.

This brings us to Newton's Second Law for Rotation:

α = Στ / I   or, more commonly written as,   Στ = Iα

Let's break this down.

Στ (Sigma Tau) is the net torque. Just like F_net, it's the unbalanced part that matters. If you and a friend push on a revolving door with equal force from opposite sides, the torques cancel out. The net torque is zero, and the door's rotation doesn't change (α = 0). It might be spinning at a constant rate, or not at all, but it won't accelerate. For the angular velocity to change, you need a non-zero net torque.
I is the rotational inertia. This is the object's inherent resistance to being spun up or slowed down. It's the "rotational mass." A heavy car tire (large I) is much harder to get spinning than a lightweight bicycle wheel (small I), even if you apply the same twisting force (torque).
α (alpha) is the angular acceleration. This is the rate at which the object's angular velocity (ω) changes. A positive α means it's spinning up (or slowing down its spin in the negative direction), and a negative α means it's slowing down (or speeding up its spin in the negative direction).

The equation Στ = Iα tells us that angular acceleration is directly proportional to the net torque and inversely proportional to the rotational inertia. More torque gets you more angular acceleration. More rotational inertia means you get less angular acceleration for the same torque.

Tying It All Together: Linear and Rotational Systems

This is where things get really interesting, and it's a huge part of the AP exam. Sometimes, you have a system where one part is moving in a line and another part is rotating. You can't just use one equation; you have to analyze them together.

The classic example is a mass hanging from a string wrapped around a pulley. Let's say it's a block of mass m hanging from a pulley of mass M and radius R.

Let's analyze this the right way. We have two objects, so we need two free-body diagrams and two applications of Newton's Second Law.

1
The Hanging Block (Linear Motion)
- Forces acting on it are gravity (mg) pulling down and tension (T) pulling up.
- Since the block is accelerating downwards, mg must be greater than T.
- Applying F_net = ma: mg - T = ma (We'll call the downward direction positive for the block).
2
The Pulley (Rotational Motion)
- The tension T from the rope pulls down on the edge of the pulley. This force creates a torque.
- The lever arm is the radius of the pulley, R.
- The torque is τ = T * R. This is the net torque causing the pulley to spin.
- Applying Στ = Iα: T * R = Iα (Here, I is the rotational inertia of the pulley, which for a solid disk is ½MR²).

The Connection: We have two equations but three unknowns (T, a, and α). We need one more relationship. The string connects the two objects! As the block falls a certain distance, the edge of the pulley must rotate a corresponding amount. The linear acceleration of the block, a, is tied to the angular acceleration of the pulley, α, by the "rolling condition":

a = αR

Now we have a system of three equations. We can solve for the acceleration of the system. This "hybrid" analysis—using both F=ma and τ=Iα on different parts of the same system—is a fundamental skill in this unit.

Worked examples

Example 1

Basic Angular Acceleration

Problem: A solid disk with a mass of 5.0 kg and a radius of 0.20 m is free to rotate about its center. A constant force of 10 N is applied tangent to its edge. What is the resulting angular acceleration of the disk?

Solution:

1
Identify the Goal
We need to find the angular acceleration, α. The key equation is Στ = Iα.
2
Calculate the Net Torque (Στ)
- Torque is force times the lever arm (τ = rF). The force is applied at the edge, so the lever arm is the radius.
- τ = (0.20 m)(10 N) = 2.0 N·m.
- Since this is the only torque, this is our net torque.
3
Calculate the Rotational Inertia (I)
- The problem states it's a solid disk. The formula for the rotational inertia of a solid disk rotating about its center is I = ½MR².
- I = ½ * (5.0 kg) * (0.20 m)² = ½ * (5.0 kg) * (0.04 m²) = 0.10 kg·m².
4
Solve for Angular Acceleration (α)
- Rearrange the main equation: α = Στ / I.
- α = (2.0 N·m) / (0.10 kg·m²) = 20 rad/s².

Why this matters: This shows the direct application of the formula. Notice how we needed to know the object's shape (a solid disk) to find its rotational inertia.

Example 2

The Pulley and Mass System

Problem: A 2.0 kg block (mass m) is connected by a string to a solid disk pulley with a mass of 4.0 kg (M) and a radius of 0.10 m (R). The block is released from rest. Find the linear acceleration of the block.

Solution:

This is a combined system. We need to analyze the block and the pulley separately and then link them.

1
Analyze the Block (Linear)
- Draw the free-body diagram. Gravity mg is down, Tension T is up. The block accelerates down, so we'll define down as the positive direction.
- F_net = ma becomes mg - T = ma.
- (2.0 kg)(9.8 m/s²) - T = (2.0 kg)a
- 19.6 - T = 2.0a (Equation 1)
2
Analyze the Pulley (Rotational)
- The tension T creates a torque. The lever arm is the radius R.
- Στ = Iα becomes T * R = Iα.
- We need I for the solid disk pulley: I = ½MR² = ½(4.0 kg)(0.10 m)² = 0.02 kg·m².
- Substituting into the torque equation: T * (0.10) = (0.02)α. (Equation 2)
3
Find the Linking Equation
- The string ensures the linear acceleration of the block is related to the angular acceleration of the pulley.
- a = αR => α = a / R = a / 0.10. (Equation 3)
4
Solve the System of Equations
- Now we substitute to eliminate variables. Let's plug Equation 3 into Equation 2.
- T * (0.10) = (0.02) * (a / 0.10)
- 0.10T = 0.2a => T = 2.0a
- This is a key insight! The tension is not mg (19.6 N). It's directly related to the acceleration.
- Now, substitute this expression for T into Equation 1.
- 19.6 - (2.0a) = 2.0a
- 19.6 = 4.0a
- a = 19.6 / 4.0 = 4.9 m/s².

Where students go wrong: The most common mistake is to assume T = mg. If that were true, the net force on the block would be zero, and it wouldn't accelerate! Because it is accelerating, T must be less than mg. Our calculation confirms this: T = 2.0a = 2.0 * 4.9 = 9.8 N, which is exactly half of mg.

Try it yourself

Problem 1: A playground merry-go-round, which can be modeled as a solid disk of mass 150 kg and radius 1.8 m, is initially at rest. Priya runs and pushes tangentially on the edge with a constant force of 80 N for 3.0 seconds. What is the final angular velocity of the merry-go-round?

Hints:

This is a two-step problem. First, use Newton's Second Law for rotation to find the angular acceleration (α).
Then, use your rotational kinematics equations from the previous topic to find the final angular velocity (ω_f) given that it starts from rest (ω_i = 0).

Problem 2: A 5.0 kg box sits on a frictionless horizontal table. It's connected by a string over a solid pulley (M=2.0 kg, R=0.15 m) to a 3.0 kg block hanging vertically. Find the tension in the horizontal part of the string.

Hints:

You have three objects/parts to analyze: the box on the table, the hanging block, and the pulley.
You will have two different tension forces! The tension in the horizontal string (T₁) is not the same as the tension in the vertical string (T₂) because the pulley has mass.
Set up three equations: one F=ma for each block, and one τ=Iα for the pulley. The net torque on the pulley will be (T₂ - T₁)R.

TL;DR

In simple terms, this topic is about how unbalanced twists (torques) make things spin faster or slower, just like unbalanced forces make things speed up or slow down in a line.

Syntax pattern

α = Στ / I   or, more commonly written as,   Στ = Iα

Key terms

AP Physics 1 Newton's Second Law for Rotation Torque Angular acceleration Rotational inertia Rotational dynamics Pulley problem Net torque AP Physics Unit 5 τ=Iα

You can now…

5.6.A: Describe the conditions under which a system's angular velocity changes.

Essential knowledge (exam-tested)

5.6.A.1: Angular velocity changes when the net torque exerted on the object or system is not equal to zero.
5.6.A.2: The rate at which the angular velocity of a rigid system changes is directly proportional to the net torque exerted on the rigid system and is in the same direction. The angular acceleration of the rigid system is inversely proportional to the rotational inertia of the rigid system. Relevant equation: α_sys = (Στ)/I_sys = τ_net/I_sys
5.6.A.3: To fully describe a rotating rigid system, linear and rotational analyses may need to be performed independently.

Concept map

flowchart TD
    A[Start: Analyze System] --> B{Object Type?};
    B -->|Linear Motion| C[Draw FBD & Set up F_net = ma];
    B -->|Rotational Motion| D[Draw FBD & Set up τ_net = Iα];
    C --> E{Is it a combined system?};
    D --> E;
    E -->|Yes| F[Identify Linking Equation e.g., a = αR];
    F --> G[Solve System of Equations for Unknowns];
    G --> H[End];
    E -->|No| I[Solve Single Equation];
    I --> H;

Read what Saavi narrates

(Sound of a gentle, thoughtful intro music fades)

Hello everyone, it's Saavi. Let's talk about spinning. Have you ever tried to open a really heavy, old door? If you push right near the hinges, it barely budges. But if you push far away from the hinges, it swings open easily. That's torque in action. We've mastered how forces make things move in a straight line with F equals m a. Now, we're going to see how twists, or torques, make things rotate.

The big idea is simple: just like an unbalanced force makes an object speed up or slow down, an unbalanced torque makes an object spin faster or slower. The rule that governs this is Newton's Second Law for Rotation: Net Torque equals I times alpha. Here, 'I' is rotational inertia, or how much an object resists spinning, and 'alpha' is its angular acceleration.

Let's look at a classic problem: a block hanging from a string wrapped around a pulley. Imagine a 2-kilogram block hanging from a 4-kilogram pulley. When you let it go, the block accelerates down, and the pulley starts to spin. We have to analyze both motions.

For the block, the net force is its weight minus the tension in the string. So, m g minus T equals m a.

For the pulley, the tension creates a torque that makes it spin. So, Torque, which is T times R, equals I times alpha.

The string connects them, so the block's acceleration 'a' equals the pulley's angular acceleration 'alpha' times its radius 'R'. We can use these three equations to solve for the acceleration.

Now, here is the number one mistake I see every year. Students assume the tension 'T' is just equal to the block's weight, 'm g'. But if that were true, the net force on the block would be zero, and it wouldn't move! Because it *is* accelerating, we know the weight must be winning the tug-of-war against tension. You have to solve for tension as part of the system.

By setting up both the linear and rotational equations and linking them together, you can solve for the true acceleration of the system. It's a powerful technique, and once you get the hang of it, you'll see these combined systems everywhere. Keep practicing, you've got this.

(Outro music fades in)

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Settings & Accessibility

Newton's Second Law in Rotational Form

Why this matters

Diagram

Concept map

Core explanation

Tying It All Together: Linear and Rotational Systems

Worked examples

Basic Angular Acceleration

The Pulley and Mass System

Try it yourself

Practice — 8 questions