AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/5/5-5)
Unit 5 · Topic 5.5
5.5 Rotational Equilibrium and Newton’s First Law in Rotational Form
An object's angular velocity stays constant only when the net torque on it is zero; that's rotational equilibrium, the rotational version of Newton's first law. Balanced forces and balanced torques are separate conditions, so a balanced beam or seesaw needs both the net force and the net torque to be zero.
Key terms
- rotational equilibrium
- net torque
- translational equilibrium
- static equilibrium
- free-body diagram
Newton's first law, rotational version
If the net torque on a rigid system is zero, its angular velocity doesn't change. It might be at rest, or it might spin steadily, like a well-oiled fan that keeps turning at the same rate while the motor's torque exactly cancels friction. Both are rotational equilibrium.
The flip side: unbalanced torques always mean a changing angular velocity. Seeing ω change tells you there's a net torque, and seeing ω stay constant tells you there isn't.
Two separate conditions
Translational equilibrium means the net force is zero, so the center of mass moves at constant velocity. Rotational equilibrium means the net torque is zero. One can happen without the other.
- Net force zero but net torque not: turn a steering wheel with one hand pushing up and the other pulling down. The forces cancel, but both turn the wheel the same way, so it starts spinning.
- Net torque zero but net force not: a book dropped flat without any spin. Gravity acts at its center of mass, so about that point there's no torque, but the book accelerates downward.
- Both zero (static equilibrium): a seesaw balanced and at rest, a shelf on brackets, a ladder against a wall.
Solving a static equilibrium problem
For an object in static equilibrium the net torque is zero about every point, so you're free to pick whichever axis makes the algebra easiest.
- Draw a force diagram with every force at the point where it acts. Put each object's weight at its center of mass.
- Write the force condition: the net force in each direction is zero (up forces = down forces, left = right).
- Choose an axis and write the torque condition: clockwise torques = counterclockwise torques.
- Put the axis where an unknown force acts. That force then has zero lever arm and drops out of the torque equation.
- Solve, then check that the answers make sense: support forces should push, and cables should pull.
On the verge of tipping
Questions often ask how far someone can walk along a plank before it tips. As the person moves past a support, the force from the other support shrinks. At the tipping point it reaches zero, because a support can only push, not pull. Set that force to zero and use the remaining support as your axis.
The same idea explains balance in everyday life. A stack of books stays put as long as its center of mass is above its base. Once the center of mass passes the edge, gravity's torque about that edge is unbalanced and it tips.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Balancing a seesaw
A uniform 20 kg seesaw is pivoted at its center. A 30 kg child sits 1.8 m from the pivot. Where should a 45 kg child sit to balance it, and what force does the pivot exert? Use g = 9.8 m/s².
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- Step 1: Take the pivot as the axis. The seesaw's weight acts at the pivot, so it gives no torque.
- Step 2: Counterclockwise torque = clockwise torque: (30)(9.8)(1.8) = (45)(9.8)x. The g cancels: x = (30)(1.8)/45 = 1.2 m on the other side.
- Step 3: Force balance: the pivot pushes up with the total weight, (30 + 45 + 20)(9.8) = 931 N.
Answer: 1.2 m from the pivot on the opposite side; the pivot pushes up with about 930 N
- Example 2Calculator allowed
A painter on a plank
A uniform 4.0 m plank of mass 20 kg rests on supports at its two ends. A 60 kg painter stands 1.0 m from the left end. Find the force from each support. Use g = 9.8 m/s².
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- Step 1: Take the axis at the left support so F_L drops out. Weights: plank 196 N acting at 2.0 m; painter 588 N acting at 1.0 m.
- Step 2: Torques about the left end: F_R(4.0) = (196)(2.0) + (588)(1.0) = 980 N·m, so F_R = 245 N.
- Step 3: Forces: F_L + F_R = 196 + 588 = 784 N, so F_L = 784 − 245 = 539 N.
- Step 4: Check: the painter is closer to the left end, so the left support should carry more. It does.
Answer: Left support ≈ 540 N; right support ≈ 245 N
- Example 3Calculator allowed
A hinged beam and a cable (exam level)
A uniform horizontal beam, 2.0 m long with mass 10 kg, is hinged to a wall at its left end. A cable from the right end runs up to the wall, making a 30° angle with the beam. A 20 kg sign hangs 1.5 m from the wall. Find the cable tension and the hinge force components. Use g = 9.8 m/s².
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- Step 1: Take the axis at the hinge, so the unknown hinge force drops out.
- Step 2: Clockwise torques from the weights: beam (98 N)(1.0 m) + sign (196 N)(1.5 m) = 98 + 294 = 392 N·m.
- Step 3: Counterclockwise torque from the cable: T(2.0) sin 30° = T(1.0 m). Setting them equal gives T = 392 N.
- Step 4: Horizontal: the cable pulls toward the wall with T cos 30° ≈ 339 N, so the hinge pushes away from the wall with about 339 N.
- Step 5: Vertical: the cable pulls up with T sin 30° = 196 N. Total weight is 98 + 196 = 294 N, so the hinge pushes up with 294 − 196 = 98 N.
Answer: T ≈ 390 N; hinge force ≈ 340 N horizontally away from the wall and 98 N up
Common mistakes
- Thinking an object with zero net torque can't be rotating. It can spin at a constant rate.
- Setting only the net torque to zero and forgetting the net force, or the other way round. Static equilibrium needs both.
- Choosing an axis that keeps two unknowns in the torque equation. Put the axis on an unknown force.
- Assuming a hinge force points along the beam. In general it has both horizontal and vertical components; find each from the force equations.
On the exam
- Free-response questions often ask you to start from a force diagram and write Στ = 0 symbolically before plugging in numbers. Show the axis you chose.
- Expect questions asking how a support force changes as a person walks along a beam. Describe the trend using torques about the other support.
Connected topics
Videos
Check yourself
4 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.
A 30 kg child sits 1.5 m from the pivot of a light seesaw. How far from the pivot, on the other side, must a 45 kg child sit to balance it?
A uniform meter stick rests on a narrow pivot at its 40 cm mark. A 0.20 kg mass hangs from the 10 cm mark. With this arrangement, the stick balances horizontally and stays at rest. Use g = 10 m/s².
Described experiment
What is the mass of the meter stick?
What is the magnitude of the force the pivot exerts on the stick?
The hanging mass is moved to the 0 cm mark and the stick is released from rest in a horizontal position. What happens immediately after release?
0 of 4 answered