5.6 Newton's Second Law in Rotational Form

Newton's second law in rotational form says a nonzero net torque causes angular acceleration. The angular acceleration is bigger when the net torque is bigger and smaller when the rotational inertia is bigger, all tied together by $\alpha_{sys} = \frac{\tau_{net}}{I_{sys}}$.

Why This Matters for the AP Physics 1 Exam

This topic is the rotational version of $F = ma$, and the AP Physics 1 exam tests it in both the multiple-choice section and free response. A common ask is functional dependence: if the net torque doubles, what happens to angular acceleration? That kind of reasoning shows up in the Qualitative/Quantitative Translation question, where you connect equations to predicted changes and explain phenomena using evidence.

You will also need to combine linear and rotational analysis. Problems like pulleys, falling masses on strings, and rolling objects expect you to apply both $\Sigma \vec{F} = m\vec{a}$ and $\alpha_{sys} = \frac{\tau_{net}}{I_{sys}}$, then link them with the constraint $a = r\alpha$.

Key Takeaways

• Angular velocity changes only when the net torque on the system is not zero.
• Angular acceleration is directly proportional to net torque and points in the same direction as the net torque.
• Angular acceleration is inversely proportional to rotational inertia, so more inertia means slower angular acceleration for the same torque.
• The core equation is $\alpha_{sys} = \frac{\tau_{net}}{I_{sys}}$, with $\alpha$ in rad/s², $\tau$ in N·m, and $I$ in kg·m².
• Fully describing a rotating system sometimes needs separate linear and rotational analyses that you then connect.
• This is the direct rotational analog of Newton's second law for linear motion.

Conditions for Angular Velocity Change

Nonzero net torque

The angular velocity of an object or system changes only when the net torque acting on it is not zero. This mirrors how linear velocity changes only when there is a net force.

• If the net torque is zero, angular velocity stays constant and there is no angular acceleration.
• The direction of angular acceleration matches the direction of the net torque.
• Angular velocity keeps changing as long as a net torque is applied.

How torque and angular acceleration connect

The angular acceleration of a rigid system is directly proportional to the net torque on it. This lets you predict rotational motion.

• Doubling the net torque doubles the angular acceleration.
• A counterclockwise net torque produces counterclockwise angular acceleration.
• Angular acceleration is inversely proportional to rotational inertia.
• Doubling the rotational inertia halves the angular acceleration for the same torque.

These relationships come together in:

$\alpha_{sys} = \frac{\Sigma \tau}{I_{sys}}$

• $\alpha_{sys}$ is the angular acceleration of the system (rad/s²)
• $\Sigma \tau$ is the net torque on the system (N·m)
• $I_{sys}$ is the rotational inertia about the axis of rotation (kg·m²)

A door opens with greater angular acceleration when you push farther from the hinge, since that increases the torque for the same force. 🚪

Independent linear and rotational analyses

To fully describe a rotating rigid system, you may need both a linear and a rotational analysis. The two approaches work together.

• Linear analysis looks at the translational motion of the center of mass using $\Sigma \vec{F} = m\vec{a}$.
• Rotational analysis looks at rotation about the axis using $\Sigma \tau = I\alpha$.

Sometimes you can run these analyses independently to simplify the problem, then connect them with a constraint like $a = r\alpha$. For example, a yo-yo needs both a center-of-mass analysis and a spin analysis.

How to Use This on the AP Physics 1 Exam

Problem Solving

1. Draw a force diagram and label where each force acts relative to the axis.
2. Find each torque with $\tau = rF_\perp = rF\sin\theta$, keeping track of signs (positive vs negative rotation).
3. Add torques to get the net torque, then divide by $I_{sys}$ to get angular acceleration.
4. If the system also moves linearly, write Newton's second law for each object and link the two analyses with $a = r\alpha$.

Free Response

For functional dependence questions, reason from the equation, not from a single calculation. Ask how $\alpha$ changes when only $\tau$ or only $I$ changes. For example, if you push a door twice as hard, $\tau$ doubles, so $\alpha$ doubles. If you instead move the same mass farther from the axis, $I$ grows and $\alpha$ drops for the same torque.

Common Trap

Combined linear and rotational problems are easy to set up wrong if you forget that the pulley's rotational inertia makes the two string tensions different. When $I$ is not zero, $T_1 \neq T_2$, and the net torque on the pulley is $(T_2 - T_1)r$.

Worked Examples

These examples apply the core ideas of this topic along with rotational inertia formulas from earlier topics. Use them to see how the equation works in different setups.

Example 1: Rotational Inertia Effect

Use Newton's second law in rotational form.

Given:

• Same torque applied to both disks
• Disk A accelerates twice as fast as Disk B ($\alpha_A = 2\alpha_B$)
• Both disks have the same radius

From $\alpha = \frac{\tau}{I}$, the same torque with twice the angular acceleration means Disk A has half the rotational inertia of Disk B:

$I_A = \frac{1}{2}I_B$

Because both disks have the same radius and are the same type of object (same geometric factor), rotational inertia is proportional to mass. So:

$M_A = \frac{1}{2}M_B$

Equivalently, $M_B = 2M_A$. Disk B has twice the mass, which gives it greater rotational inertia and slower angular acceleration for the same torque.

Example 2: Door Torque

Use the rotational form of Newton's second law after finding the torque.

Given:

• Door mass: 20 kg
• Door width: 1.0 m
• Applied force: 15 N perpendicular to the door at its edge
• Rotational inertia about the hinge: $I = \frac{20}{3} \text{ kg}\cdot\text{m}^2$

Step 1: Calculate the torque. The force is perpendicular at 1.0 m from the hinge, so: $\tau = rF = 1.0 \text{ m} \times 15 \text{ N} = 15 \text{ N}\cdot\text{m}$

Step 2: Calculate the angular acceleration using $\alpha = \frac{\tau}{I}$: $\alpha = \frac{15 \text{ N}\cdot\text{m}}{\frac{20}{3} \text{ kg}\cdot\text{m}^2} = \frac{15 \times 3}{20} = \frac{45}{20} = 2.25 \text{ rad/s}^2$

The door experiences an angular acceleration of 2.25 rad/s².

Example 3: Pulley System with Rotational Inertia

This needs both linear and rotational analysis since the pulley's rotational inertia affects the blocks' motion.

Given:

• m₁ = 2.0 kg
• m₂ = 3.0 kg
• Pulley radius = 0.10 m
• Pulley rotational inertia = 0.002 kg·m²

Step 1: Identify the tensions. Let T₁ be the tension on the 2.0 kg side and T₂ be the tension on the 3.0 kg side.

Step 2: Apply Newton's second law to each block. Let the 3.0 kg block accelerate downward and the 2.0 kg block accelerate upward with magnitude a. For the 2.0 kg block: $T_1 - m_1g = m_1a$

For the 3.0 kg block: $m_2g - T_2 = m_2a$

Step 3: Relate linear and angular acceleration: $a = r\alpha$, so $\alpha = a/r$.

Step 4: Apply the rotational form of Newton's second law to the pulley: $(T_2 - T_1)r = I\alpha = I(a/r)$

Step 5: Substitute $T_1 = m_1(g + a)$ and $T_2 = m_2(g - a)$:

$[m_2(g - a) - m_1(g + a)]r = I(a/r)$

Step 6: Solve: $[3.0(9.8 - a) - 2.0(9.8 + a)](0.10) = 0.002(a/0.10)$

$[(29.4 - 3.0a) - (19.6 + 2.0a)](0.10) = 0.02a$

$(9.8 - 5.0a)(0.10) = 0.02a$

$0.98 - 0.50a = 0.02a$

$0.98 = 0.52a$ $a = 1.88 \text{ m/s}^2$

Then: $\alpha = \frac{a}{r} = \frac{1.88}{0.10} = 18.8 \text{ rad/s}^2$

The angular acceleration of the pulley is approximately 18.8 rad/s².

Common Misconceptions

• Thinking force alone causes angular acceleration. Only the torque matters, which depends on the perpendicular force component and its distance from the axis.
• Believing zero net torque means zero angular velocity. Zero net torque means constant angular velocity, which can still be a nonzero spin.
• Assuming a heavier object always has more rotational inertia. Inertia depends on how mass is distributed, so where the mass sits relative to the axis matters as much as how much there is.
• Treating the two string tensions over a real pulley as equal. When the pulley has rotational inertia, the tensions differ, and that difference produces the net torque.
• Forgetting to connect linear and rotational motion. In rolling or pulley problems, you still need the constraint $a = r\alpha$ to tie the two analyses together.

Related AP Physics 1 Guides

• 5.1 Rotational Kinematics
• 5.2 Connecting Linear and Rotational Motion
• 5.3 Torque
• 5.4 Rotational Inertia
• 5.5 Rotational Equilibrium and Newton's First Law in Rotational Form