Rotational Equilibrium and Newton’s First Law in Rotational Form: Study Guide (Physics C: Mechanics 5.5)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 5.5 with this title. This guide is the calculus-based one: it finds the torque of a non-uniform object by integration, proves why the torque point can be chosen freely, and derives results in symbols. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

Two kinds of equilibrium

A rigid system can do two things at once: its center of mass can move, and the system can turn about an axis. Each motion has its own condition.

• Translational equilibrium: the net external force is zero, ΣF = 0. The velocity of the center of mass is constant (Topic 2.4).
• Rotational equilibrium: the net external torque is zero, Στ = 0. The angular velocity is constant.

In this course every rotation happens in one plane, about one axis. So each torque is a signed number: we take counterclockwise as positive and clockwise as negative, and we say so at the start of every problem. Recall from Topic 5.3 that the size of a torque is τ = rF sin θ = r⊥F, where r⊥ is the perpendicular distance from the axis to the line of action of the force (the lever arm).

Written out, rotational equilibrium is simply

Στ = τ₁ + τ₂ + τ₃ + … = 0

Newton’s first law in rotational form

The rotational version of Newton’s first law says:

A system has a constant angular velocity only if the net torque exerted on it is zero.

“Constant” includes zero. A shelf bracket that never moves has ω = 0 at all times, and that is rotational equilibrium. A ceiling fan turning at a steady 12 rad/s is also in rotational equilibrium: the motor’s torque is exactly cancelled by the drag torque from the air. The fan does not need a net torque to keep turning, just as a puck on ice does not need a net force to keep sliding.

The other half of the statement is the corollary: if the torques on a rigid system are not balanced, its angular velocity must be changing, in size, in direction of turning, or both. Topic 5.6 turns this into an equation, Στ = Iα.

Extended force diagrams

A free-body diagram for translation can treat the object as a dot. For rotation that is not enough, because the point where a force acts changes its lever arm. So you draw an extended force diagram: the object’s outline, each force as an arrow starting at the point where it acts, and the axis or pivot marked.

Where does the weight act? Gravity pulls on every small piece of mass dm. Put the pivot at x = 0 on a horizontal object and take +x to the right. A piece at position x has weight g dm and lever arm x, and pulling down to the right of the pivot gives a clockwise (negative) torque. Adding all the pieces:

τ_grav = −∫ x g dm = −g ∫ x dm = −Mg x_cm

because ∫ x dm = M x_cm by the definition of the center of mass (Topic 2.1). So, for torque purposes, the whole weight acts at the center of mass. For a uniform object this is the geometric middle. For a non-uniform object you find x_cm by integration first, as in Worked example 1.

Choosing the point for torques

When an object is fixed on an axle, you take torques about the axle. But for a beam resting on supports, or a door on hinges, there is no single obvious axis. Which point should you use?

Suppose forces F₁, F₂, … act at positions r₁, r₂, … measured from a point P. The net torque about P is τ_P = Σ rᵢ × Fᵢ. Now move to a point Q at position d from P. The position of each force from Q is rᵢ − d, so

τ_Q = Σ (rᵢ − d) × Fᵢ = τ_P − d × ΣF

If the object is in translational equilibrium, ΣF = 0, and the last term vanishes: τ_Q = τ_P for every point Q. So when ΣF = 0 and Στ = 0 about one point, Στ = 0 about all points.

This gives you a free choice. Take torques about the point where an unknown force acts: that force then has zero lever arm and drops out of the equation. Use a second torque equation, or ΣF = 0, to check your answer.

Be careful with the reverse case. If ΣF ≠ 0, the net torque depends on the point you choose, and you must use the actual axis or the center of mass.

Worked example 1: a tapered beam on two supports

Question. Take +x to the right, origin at the left end, counterclockwise positive. A wooden beam is 2.0 m long and lies horizontally. It is thicker at the left end: its linear mass density is λ(x) = (6.0 kg/m) − (2.0 kg/m²)x for 0 ≤ x ≤ 2.0 m. It rests on support A at x = 0 and support B at x = 1.5 m. (a) Find the upward force from each support. (b) A can of paint is placed on the far right end, x = 2.0 m. What is the largest mass of can the beam can hold before it tips?

(a)

1. Mass: M = ∫₀^2.0 (6.0 − 2.0x) dx = 12 − 4.0 = 8.0 kg, so Mg = 78.4 N.
2. Center of mass: ∫₀^2.0 x(6.0 − 2.0x) dx = 12 − 16/3 = 6.67 kg·m, so x_cm = 6.67 ÷ 8.0 = 0.83 m (exactly 5/6 m).
3. Torques about A (N_A has zero lever arm and drops out): N_B(1.5) − (78.4)(0.833) = 0, so N_B = 65.3 ÷ 1.5 = 44 N (43.6 N).
4. Forces, vertical: N_A + N_B − 78.4 = 0, so N_A = 35 N (34.8 N).

Check. Torques about B: −N_A(1.5) + 78.4(1.5 − 0.833) = −52.3 + 52.3 = 0. If you had put the weight at the middle (x = 1.0 m), you would get N_B = 52 N, which is wrong because the beam is not uniform.

(b) As the can gets heavier, the beam presses less on A. It is about to tip about B when N_A = 0. Take torques about B. The beam’s weight acts 1.5 − 0.833 = 0.667 m to the left of B (counterclockwise); the can acts 0.50 m to the right (clockwise):

(8.0)(9.8)(0.667) − m(9.8)(0.50) = 0, so m = (8.0 × 0.667) ÷ 0.50 = 11 kg (10.7 kg).

Notice that g cancels. Any can lighter than this leaves N_A > 0 and the beam stays put.

Worked example 2: the forces on a door’s hinges

Question. Take +x away from the wall (to the right), +y upward, counterclockwise positive as seen facing the door. A uniform door of mass M and width w hangs from two hinges on its left edge, a vertical distance h apart. The door is at rest. (a) Derive the horizontal force each hinge exerts on the door. (b) Evaluate it for M = 24 kg, w = 0.90 m and h = 1.6 m. (c) What can and cannot be found about the vertical hinge forces?

(a)

1. Unknowns: a horizontal and a vertical force at each hinge (four in all). Equilibrium gives three equations: ΣF_x = 0, ΣF_y = 0, Στ = 0.
2. Torques about the bottom hinge. Both bottom-hinge forces drop out. The vertical top-hinge force acts along the hinge line, so its lever arm is zero too. The weight acts w/2 to the right: torque −Mg(w/2). The top hinge’s horizontal force H_top acts at height h; pointing towards the wall (−x) it gives a counterclockwise torque +H_top h.
3. Στ = 0: H_top h − Mg(w/2) = 0, so H_top = Mgw / (2h), towards the wall.
4. ΣF_x = 0: the bottom hinge pushes with the same size, away from the wall.

(b) H = (24)(9.8)(0.90) ÷ (2 × 1.6) = 211.7 ÷ 3.2 = 66 N.

(c) ΣF_y = 0 gives only V_top + V_bottom = Mg = 235 N. The three equations cannot split this sum between the hinges; that depends on how tightly each hinge is fitted. Such a problem is called statically indeterminate. If you are told the hinges share the weight equally, each carries 118 N upward, and the top hinge’s total force on the door is √(66² + 118²) ≈ 135 N.

Interpret. H ∝ 1/h: double the hinge spacing and the horizontal forces halve, to 33 N. That is why tall, heavy doors have hinges near the top and bottom. Taking torques about the top hinge instead gives the same H: a useful check that follows from the “any point” result above.

Equilibrium of one kind only

The two conditions are independent. Check each one separately.

• Translational but not rotational equilibrium. A driver pushes one side of a steering wheel up and the other side down, each with 20 N, at 0.19 m from the center. The forces cancel, so ΣF = 0. But both torques are in the same sense: Στ = 2 × 20 × 0.19 = 7.6 N·m. Such a pair of equal, opposite, offset forces is a couple. The wheel’s angular velocity changes even though its center does not accelerate.
• Rotational but not translational equilibrium. A spanner is tossed so that it tumbles end over end. Ignore air resistance. The only force is gravity, which acts at the center of mass, so its torque about the center of mass is zero. The spanner keeps turning at a constant angular velocity about its center of mass, while its center of mass accelerates downward at g. In 0.50 s its vertical velocity changes by 4.9 m/s, but its spin does not change.
• Both. The steady ceiling fan, or the beam in Worked example 1.
• Neither. A door swinging shut while a person pushes it.

Whenever the torques do not balance, expect a changing ω. That is the corollary, and the start of Topic 5.6.

Common misconceptions

• “A spinning object is not in equilibrium.” Constant ω, even a large one, means Στ = 0.
• “ΣF = 0, so nothing can start to rotate.” A couple has ΣF = 0 and a nonzero torque.
• “The weight of a beam acts at its middle.” Only for a uniform beam. Otherwise find x_cm by integration (Worked example 1).
• Using the force’s full distance instead of the lever arm. Use r⊥, or r sin θ.
• Thinking the answer depends on the torque point. When ΣF = 0 it does not. Choose the point that removes unknowns.
• Forgetting the pivot force. It has no torque about the pivot, but it still belongs in ΣF = 0.
• Expecting statics to find every force. Three equations cannot find four unknowns (Worked example 2).

Where this leads

Earlier: Topic 5.4, Rotational Inertia. Next, Topic 5.6, Newton’s Second Law in Rotational Form, tells you how quickly ω changes when the torques do not balance: α = Στ/I. Unit 6 then adds energy and angular momentum to rotating systems. Try the practice questions now, then use the revision notes and the checklist. You can also return to the course roadmap.