Unit 5 · Topic 5.5

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

A rigid object is in rotational equilibrium when the net torque on it is zero, and then its angular velocity stays constant. Static equilibrium problems, like beams, seesaws and signs, need both the net force and the net torque to be zero.

Key terms

rotational equilibrium
translational equilibrium
static equilibrium
net torque
pivot point

What changed in the 2024 update

Before the Fall 2024 course update, this topic was part of one big Rotation unit (old Unit 5). That unit is now split in two: Unit 5 covers spinning motion, torque and rotational inertia, and Unit 6 covers rotational energy, angular momentum and rolling. Older videos and practice labeled "Unit 5: Rotation" still fit, but they may mix in Unit 6 ideas.

Newton's first law for rotation

If the net torque on a rigid object is zero, its angular velocity doesn't change. A wheel at rest stays at rest, and a wheel spinning at 5 rad/s keeps spinning at 5 rad/s. This is rotational equilibrium.

Zero net torque doesn't mean the object isn't spinning. It means its spin isn't changing. Flip it around: whenever the torques on an object fail to cancel, its spin is speeding up, slowing down or reversing.

Two separate conditions

Rotational equilibrium (zero net torque) and translational equilibrium (zero net force) are independent. An object can have one without the other.

Two equal and opposite forces applied at different points, like turning a steering wheel with both hands, give zero net force but a nonzero torque. The wheel's center stays put but it starts to turn. A ball falling straight down has a net force but no torque about its center.

For an object to stay completely at rest (static equilibrium), you need both: $\sum F_x = 0 \qquad \sum F_y = 0 \qquad \sum \tau = 0$

Solving statics problems

Follow the same steps every time:

Draw a force diagram with every force at its point of application. The object's weight acts at its center of mass.
Choose a pivot for your torque equation. For an object in equilibrium, the net torque is zero about every point, so you can pick any point.
Put the pivot where an unknown force acts, such as a hinge. That force has no lever arm, so it drops out of the torque equation.
Write the torque equation, with clockwise and counterclockwise torques given opposite signs, and solve for one unknown.
Use the force equations to find the rest.

When does it tip?

Picture a plank resting on two supports, with a person walking out past one of them. While the plank is balanced, both supports push up. As the person moves outward, the far support pushes up less and less.

The plank is just about to tip when the far support's force reaches zero. At that moment, the near support is the only upward force, and the plank acts like a seesaw balanced on it. To find how far the person can walk, take torques about the near support and set the far support's force to zero.

A support can only push, never pull. If your equations give a negative normal force, the object has already tipped, and the equilibrium you assumed isn't possible.

Limits on what's tested

All the forces and rotation stay in one plane. You won't need to analyze rotation about two different axes at once.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Balancing a seesaw
A 30 kg child sits 1.5 m to the left of a seesaw's pivot. Where must a 45 kg child sit to balance it? What force does the pivot exert? Ignore the board's mass.
Show the solutionHide the solution
1. Step 1: Torques about the pivot must cancel: $m_1 g\,r_1 = m_2 g\,r_2$ . The g cancels: $(30)(1.5) = (45)r_2$ , so $r_2 = 1.0$ m to the right.
2. Step 2: Forces must also balance. The pivot pushes up with the total weight: (30 + 45)(9.8) = 735 N.
Answer: 1.0 m to the right of the pivot; the pivot pushes up with about 735 N
Example 2Calculator allowed
Beam held by a cable (exam-level)
A uniform 4.0 m beam of mass 20 kg sticks out horizontally from a wall, attached by a hinge. A cable runs from the beam's far end up to the wall, making a 30° angle with the beam. A 50 kg sign hangs 3.0 m from the wall. Find the cable tension and the hinge force.
Show the solutionHide the solution
1. Step 1: Take torques about the hinge, so the unknown hinge force drops out. The beam's weight acts at its center, 2.0 m out; the sign hangs at 3.0 m. Only the cable tension's vertical part, $T\sin 30^\circ$ , has a lever arm about the hinge, and that lever arm is 4.0 m.
2. Step 2: Counterclockwise = clockwise: $T\sin 30^\circ\,(4.0) = (20)(9.8)(2.0) + (50)(9.8)(3.0) = 392 + 1470 = 1862$ N·m, so $T = \dfrac{1862}{2.0} = 931$ N.
3. Step 3: Horizontal forces: the hinge pushes out from the wall to balance $T\cos 30^\circ = 806$ N.
4. Step 4: Vertical forces: $H_y + T\sin 30^\circ = (70)(9.8)$ , so $H_y = 686 - 465.5 \approx 220$ N upward.
5. Step 5: Hinge force: $\sqrt{806^2 + 220^2} \approx 836$ N, about 15° above the horizontal, pointing away from the wall.
Answer: T ≈ 930 N; hinge force ≈ 840 N (806 N out from the wall and 220 N up)
Example 3
Spinning at a steady rate (classic trap)
A ceiling fan spins at a constant 20 rad/s. Its motor exerts a torque of 0.30 N·m. What is the frictional and air-drag torque on the blades?
Show the solutionHide the solution
1. Step 1: The trap is thinking a spinning fan must have a net torque on it. A constant ω means zero angular acceleration.
2. Step 2: So the fan is in rotational equilibrium, and the net torque is zero.
3. Step 3: The resisting torques must add to 0.30 N·m, in the direction opposite the motor's torque.
Answer: 0.30 N·m, opposite the motor's torque

Common mistakes

Thinking zero net torque means the object isn't rotating. It means the angular velocity is constant, which can be nonzero.
Checking only the torques or only the forces. Static equilibrium needs both.
Picking a pivot that leaves two unknowns in the torque equation. Put the pivot where an unknown force acts.
Forgetting the beam's own weight, or putting it at the end instead of the center of mass.

On the exam

Free-response statics questions usually ask for a force diagram first, then a torque equation. Label the pivot you chose and keep the signs consistent.
Symbolic answers are common. Write the torque balance in letters, such as T sin θ · L = Mg(L/2) + mgx, before solving.

Connected topics

Videos

Topic 5.5 - Newton's Laws of Rotation
Lessons With LondotWatch on YouTube (opens in a new tab)
Rotational Equilibrium Introduction (and Static Equilibrium too!!)
Flipping PhysicsWatch on YouTube (opens in a new tab)
AP Physics 1 - Unit 5 Lesson 5 - Rotational Statics Explained
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Statics: Crash Course Physics #13
CrashCourseWatch on YouTube (opens in a new tab)
Physics 15 Torque Example 1 (1 of 7) Mass on Rod and Cable
Michel van BiezenWatch on YouTube (opens in a new tab)

Check yourself

5 questions on 5.5 Rotational Equilibrium and Newton’s First Law in Rotational Form. Pick an answer to see if you got it, and why.

Question 1 of 5Calculator allowed

A 30 kg child sits 1.5 m to the left of a seesaw's pivot. The seesaw board is uniform and balanced on its center. Where must a 45 kg child sit to balance it?

Question 2 of 5Calculator allowed

A uniform 4.0 m beam of mass 20 kg rests on supports at its two ends. A 30 kg box sits on the beam 1.0 m from the left end. Use g = 10 m/s². What force does the right support exert on the beam?

A uniform horizontal strut of mass 10 kg and length 2.0 m is attached to a wall by a hinge at its left end. A cable runs from the strut's right end up to a point on the wall, making a 30° angle with the strut. A 20 kg sign hangs from the right end of the strut. Everything is at rest. Use g = 10 m/s² and sin 30° = 0.50.

Described situation

Question 3 of 5Calculator allowed

What is the tension in the cable?

Question 4 of 5Calculator allowed

What is the horizontal component of the force the hinge exerts on the strut?

Question 5 of 5Calculator allowed

What is the vertical component of the force the hinge exerts on the strut?

0 of 5 answered