5.5 Rotational Equilibrium and Newton's First Law in Rotational Form

Rotational equilibrium means a system keeps a constant angular velocity because the net torque on it is zero. This is the rotational version of Newton's first law, and a system can be in rotational equilibrium even when its center of mass is accelerating.

Why This Matters for the AP Physics 1 Exam

Torque and rotational dynamics make up about 10 to 15 percent of the exam, and this topic is where you learn the condition that keeps rotation steady. You will use the idea that net torque equals zero to set up balance problems, analyze force and free-body diagrams, and justify why a system's angular velocity stays constant.

This topic pairs naturally with functional reasoning, the kind tested in both multiple-choice and free-response. You might be asked to predict how a balance point shifts when a mass or distance changes, or to explain why an object spins faster when torques stop being balanced. Being able to set $\sum \tau = 0$ and solve for an unknown distance, force, or tension is a core skill here.

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Key Takeaways

• Rotational equilibrium means constant angular velocity, which requires net torque equal to zero: $\sum \tau_i = 0$.
• A system can be in rotational equilibrium without being in translational equilibrium, and vice versa. The two are independent.
• This is the rotational analog of Newton's first law: angular velocity stays constant only when net torque is zero.
• If torques are unbalanced, angular velocity must change (the rotational corollary to Newton's second law).
• Use force and free-body diagrams to identify every torque about your chosen axis before summing.
• Only the force component perpendicular to the position vector creates torque, and the lever arm is the perpendicular distance from the axis to the line of the force.

How to Use This on the AP Physics 1 Exam

Problem Solving

Most rotational equilibrium problems follow the same setup:

1. Pick a clear axis of rotation. Choosing the axis at the location of an unknown force (like a hinge or pivot) removes that force from your torque equation, since a force acting at the axis has zero lever arm.
2. Draw a force diagram showing where each force acts relative to that axis.
3. Write each torque as $\tau = rF_\perp = rF\sin\theta$, where $\theta$ is the angle between the force and the position vector.
4. Assign signs. Counterclockwise torques are usually positive and clockwise negative, but stay consistent.
5. Set $\sum \tau = 0$ and solve for the unknown.

Free Response

When a question asks you to explain rather than calculate, connect the math to the concept. State that zero net torque means constant angular velocity, and point to the specific torques that cancel. If a system is accelerating in a straight line but still not rotating faster, explain that translational and rotational equilibrium are separate conditions.

Common Trap

Watch for problems where forces are applied at angles. Only the perpendicular component contributes to torque, so a force pointing partly along the position vector does less than its full magnitude suggests. The exam only expects rotation about a single fixed axis, so you will not have to track rotation in more than one plane.

Conditions for Constant Angular Velocity

Rotational vs Translational Equilibrium

A system may show rotational equilibrium (constant angular velocity) without being in translational equilibrium, and the reverse is also true. The two kinds of motion are independent of each other.

• A spinning top keeps a constant angular velocity while its center of mass moves 🌀
• An object can have zero net force but nonzero net torque, or zero net torque but nonzero net force, depending on how forces are applied
• A wheel can rotate at constant angular velocity while being pulled so its center of mass accelerates forward, as long as the net torque about its center is zero

Free-body and force diagrams help you see the forces and torques acting on a system. These diagrams are essential for analyzing both types of equilibrium.

• Free-body diagrams show all external forces
• Force diagrams also show where each force acts relative to the axis of rotation
• Arrows should indicate both magnitude and direction of forces

Zero Net Torque

Rotational equilibrium is a configuration of torques such that the net torque on the system is zero. When all torques on an object balance perfectly, the object keeps its rotational state. This mirrors how balanced forces prevent acceleration in linear motion.

$\sum \tau_{i}=0$

• Balanced clockwise and counterclockwise torques produce rotational equilibrium ⚖️
• Net torque of zero means angular velocity stays constant
• Objects in rotational equilibrium do not speed up or slow down their rotation

Finding net torque means adding all individual torques about the axis of rotation, with signs.

• The magnitude of each torque is $\tau = rF_\perp = rF\sin\theta$, where $\theta$ is the angle between the force and the position vector from the axis to the point where the force acts
• Counterclockwise torques are typically taken as positive
• Clockwise torques are typically taken as negative
• The perpendicular distance from the axis to the line of the force is the lever arm

Rotational Analog of Newton's Laws

Newton's first law has a rotational version with clear symmetry to its linear form.

• First law (rotational form): A system keeps a constant angular velocity only if the net torque on it is zero. With no net torque, rotation does not change.
• Rotational corollary to the second law: If the net torque on a rigid system is not zero, the angular velocity must change. Unbalanced torque causes the rotation to speed up or slow down.
• For this topic, focus on the first-law idea: constant angular velocity requires zero net torque.

Applying these ideas requires a clear axis of rotation.

• The axis of rotation is the reference point for all torque calculations
• AP Physics 1 problems focus on fixed-axis rotation
• Rotational inertia depends on how mass is distributed relative to the axis

Practice Problem 1: Rotational Equilibrium

To balance the seesaw, find the position where the net torque about the pivot is zero.

Set up a coordinate system with the pivot at the origin and the positive direction to the right.

Step 1: Calculate the torque due to the 25 kg child. Position of child from pivot = 1.5 m - 2.5 m = -1.0 m (left of pivot)

Torque = mass × gravity × lever arm = 25 kg × 9.8 m/s² × (-1.0 m) = -245 N·m

Step 2: The seesaw's center of mass is at the pivot, so it creates no torque.

Step 3: Find where the 40 kg child should sit. Let x = distance from pivot (positive to the right) Torque due to second child = 40 kg × 9.8 m/s² × x = 392x N·m

Step 4: For equilibrium, the sum of torques equals zero. -245 N·m + 392x N·m = 0 392x = 245 x = 0.625 m

The 40 kg child should sit 0.625 m to the right of the pivot.

Practice Problem 2: Zero Net Torque

Step 1: Define the pivot point and coordinate system.

• Pivot at 50 cm mark (x = 0)
• Positions left of pivot are negative, right are positive
• Position of 200 g mass: x = -30 cm = -0.3 m

Step 2: Calculate the torque due to the 200 g mass.

• Torque = mass × gravity × lever arm
• Torque = 0.2 kg × 9.8 m/s² × (-0.3 m) = -0.588 N·m

Step 3: For zero net torque, the 150 g mass must create an equal and opposite torque.

• Let x = position of 150 g mass
• Torque = 0.15 kg × 9.8 m/s² × x = 1.47x N·m
• For equilibrium: -0.588 N·m + 1.47x N·m = 0
• Solving for x: x = 0.4 m = 40 cm from the pivot

The 150 g mass should be placed at the 90 cm mark of the meter stick.

Practice Problem 3: Applying Rotational Equilibrium

Step 1: Draw a free-body diagram. The forces on the beam are:

• The weight of the sign (80 N downward) acting at the midpoint of the beam (1.0 m from the hinge)
• The tension $T$ in the cable acting at the end of the beam (2.0 m from the hinge)
• The hinge force at the wall (which we can ignore by choosing the hinge as our pivot)

Step 2: Take torques about the hinge. Since the hinge force acts at the pivot, it produces zero torque.

• The sign's weight produces a clockwise torque: $\tau_{\text{sign}} = 80 \text{ N} \times 1.0 \text{ m} = 80 \text{ N·m}$
• The cable tension produces a counterclockwise torque. Only the vertical component of the tension ($T \sin 30°$) creates torque about the hinge, with a lever arm of 2.0 m: $\tau_{\text{cable}} = T \sin(30°) \times 2.0 \text{ m} = T(0.5)(2.0) = 1.0T \text{ N·m}$

Step 3: For rotational equilibrium, set $\sum \tau = 0$: $1.0T - 80 = 0$

$T = 80 \text{ N}$

The tension in the cable is 80 N. The sign stays at rest because the net torque about the hinge is zero, which is rotational equilibrium in action. Even though the hinge also exerts a force on the beam, choosing the hinge as the pivot point simplifies the problem since that force contributes no torque. This is a useful strategy whenever you can place the pivot at the location of an unknown force.

Common Misconceptions

• Rotational equilibrium does not mean the object is not moving. It means the angular velocity is constant, which includes spinning at a steady rate, not just being at rest.
• Zero net force and zero net torque are not the same condition. An object can satisfy one without the other, so always check both separately when a problem asks about full equilibrium.
• The full magnitude of a force does not always go into torque. Only the component perpendicular to the position vector counts, so angled forces contribute less than they first appear.
• A larger force does not automatically create a larger torque. A small force with a long lever arm can outweigh a big force applied close to the axis.
• Putting the pivot at an unknown force is a choice, not a rule. You can pick any axis, but choosing one at an unknown force makes its torque zero and simplifies the algebra.

Related AP Physics 1 Guides

• 5.1 Rotational Kinematics
• 5.2 Connecting Linear and Rotational Motion
• 5.3 Torque
• 5.4 Rotational Inertia
• 5.6 Newton's Second Law in Rotational Form