Newton’s Second Law in Rotational Form: Study Guide (Physics C: Mechanics 5.6)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 5.6 with this title. This guide is the calculus-based one: it derives Στ = Iα, handles torques that change with time by integration, and uses the parallel axis theorem from Topic 5.4. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

When does angular velocity change?

Topic 5.5 gave the rule for a constant angular velocity: the net torque must be zero. Now take the other case.

The angular velocity of a rigid system changes only when the net external torque on it is not zero.

“Changes” means ω gets bigger, gets smaller, or reverses its sense of turning. This topic tells you how fast it changes.

Newton’s second law in rotational form

α = Στ / I (equivalently, Στ = Iα)

Three facts are packed into this equation:

1. Proportional to net torque. Double the net torque on the same system and α doubles.
2. Same sense. α points the same way as the net torque. If the net torque is counterclockwise, α is counterclockwise, even if the object is turning clockwise at that moment (then it is slowing down).
3. Inversely proportional to rotational inertia. The same torque gives a smaller α to a system with more rotational inertia about that axis.

Units: N·m ÷ kg·m² = (kg·m²/s²) ÷ (kg·m²) = s⁻² = rad/s². The radian is a ratio, so it can appear or disappear in units.

Using functional dependence. Suppose a string pulls with force F at the rim of a solid disk of mass M and radius R (I = ½MR²). Now use a disk with the same mass but twice the radius, pulled with the same force at its rim. The torque doubles (lever arm 2R) but I becomes ½M(2R)² = 4 × ½MR². So α changes by 2 ÷ 4 = ½. The larger disk turns more sluggishly even though the torque is bigger.

Where Στ = Iα comes from

Think of a rigid system turning about a fixed axis as many small pieces. Piece i has mass mᵢ at distance rᵢ from the axis. Because the system is rigid, every piece has the same angular acceleration α, and piece i has tangential acceleration a_t,i = rᵢα (Topic 5.2).

1. Newton’s second law for piece i, along its tangent: F_t,i = mᵢ a_t,i = mᵢ rᵢ α. Here F_t,i is the total tangential force on the piece, from outside and from other pieces.
2. Multiply by rᵢ to turn force into torque: τᵢ = rᵢ F_t,i = mᵢ rᵢ² α. (Radial force components have no lever arm, so they give no torque.)
3. Add over all pieces: Στᵢ = (Σ mᵢ rᵢ²) α = Iα.
4. Internal torques cancel. Two pieces push on each other with equal and opposite forces along the line joining them (Newton’s third law). Those two forces have the same lever arm and opposite senses, so their torques cancel. Only external torques survive.

So Στ_ext = Iα, with I = Σ mᵢ rᵢ² = ∫ r² dm. This is the rotational partner of ΣF = ma, built from it.

Linear and rotational analyses are separate

A rigid system obeys two equations at the same time:

a_cm = ΣF / M and α = Στ / I

They answer different questions and must be written separately. One can be zero while the other is not (Topic 5.5). An axle can push hard on a wheel without exerting any torque about the axle, because its force acts at the axis. When a problem asks for the force from an axle or pivot, you usually need both equations: Στ = Iα gives α, the geometry gives a_cm, and ΣF = Ma_cm then gives the pivot force.

Worked example 1: a falling signal arm

Question. Take +y upward and clockwise as positive for rotation (the sense the arm will turn). A uniform bar of mass M = 3.5 kg and length L = 1.6 m can turn freely about a horizontal axle through a point L/4 from its left end. It is held horizontal and released from rest. Find, just after release, (a) the angular acceleration, (b) the acceleration of its center of mass and (c) the force from the axle. Derive symbolic answers first.

(a) Rotation.

1. Rotational inertia about the axle (parallel axis theorem, Topic 5.4), with d = L/4: I = ML²/12 + M(L/4)² = 7ML²/48 = 7(3.5)(1.6)²/48 = 1.31 kg·m².
2. Torques about the axle: the axle force has zero lever arm; the weight acts L/4 to the right and turns the bar clockwise: Στ = MgL/4 = 3.5 × 9.8 × 0.40 = 13.7 N·m.
3. α = Στ/I = (MgL/4) ÷ (7ML²/48) = 12g/(7L) = 12(9.8) ÷ (7 × 1.6) = 10.5 rad/s², clockwise.

(b) Center of mass. At release ω = 0, so the center of mass has only tangential acceleration: a_cm = dα = (L/4)(12g/7L) = 3g/7 = 4.2 m/s², downward.

(c) Linear analysis, separately. ΣF_y = Ma_y: N − Mg = M(−3g/7), so N = 4Mg/7 = 4(34.3)/7 = 19.6 N, upward.

Checks. If the axle were at the center (d = 0), α = 0 and N = Mg: rotational equilibrium. If it were at the end (d = L/2), the same method gives α = 3g/(2L) and N = Mg/4. Our case lies between. Using I = ML²/3 (the end-axis value) would give 4.6 rad/s², which is wrong because the axle is not at the end.

These answers hold only just after release. As the bar swings down, the weight’s lever arm shrinks, so α changes; the energy methods of Unit 6 handle that motion.

Torques that change with time

When the net torque depends on time, the law holds at each instant:

α(t) = Στ(t) / I

Then the calculus of Topics 1.2 and 5.1 does the rest:

ω(t) = ω₀ + (1/I) ∫₀ᵗ Στ dt and θ(t) = θ₀ + ∫₀ᵗ ω dt

So the change in angular velocity equals the signed area under the net-torque–time graph, divided by I. Two consequences:

• ω keeps increasing while Στ is positive, even if Στ is getting smaller. The greatest ω occurs when Στ passes through zero: a moment of rotational equilibrium.
• When Στ is negative, ω decreases, even if a motor is still pushing forward.

Worked example 2: a pottery wheel with a fading motor

Question. Take counterclockwise (seen from above) as positive. A pottery wheel and its clay have I = 0.25 kg·m² about the axle. Starting from rest at t = 0, its motor exerts a torque τ_m(t) = (2.4 N·m) − (0.40 N·m/s)t for 0 ≤ t ≤ 6.0 s, then switches off. A constant friction torque of 0.40 N·m opposes the rotation throughout. Find α(t), ω(t) and θ(t) for the first 6.0 s, the greatest angular velocity, the angular velocity and angle at 6.0 s, and when the wheel stops.

1. Net torque: Στ = 2.4 − 0.40t − 0.40 = 2.0 − 0.40t (N·m, t in s).
2. α = Στ/I = (2.0 − 0.40t) ÷ 0.25 = 8.0 − 1.6t (rad/s²).
3. ω = 0 + ∫₀ᵗ (8.0 − 1.6t) dt = 8.0t − 0.80t² (rad/s).
4. θ = 0 + ∫₀ᵗ (8.0t − 0.80t²) dt = 4.0t² − 0.267t³ (rad).
5. Greatest ω when Στ = 0: t = 5.0 s, where ω = 40 − 20 = 20 rad/s. The motor still pushes (0.40 N·m), but no harder than friction.
6. At 6.0 s: ω = 48 − 28.8 = 19.2 rad/s; θ = 144 − 57.6 = 86.4 rad.
7. After 6.0 s, only friction acts: α = −0.40 ÷ 0.25 = −1.6 rad/s², constant. Time to stop: 19.2 ÷ 1.6 = 12 s, so the wheel stops at t = 18 s, after a further 19.2² ÷ (2 × 1.6) = 115 rad.

Check. The area under the net-torque graph from 0 to 6.0 s is ∫₀⁶ (2.0 − 0.40t) dt = 12 − 7.2 = 4.8 N·m·s, and 4.8 ÷ 0.25 = 19.2 rad/s. Treating the starting net torque of 2.0 N·m as constant for 6.0 s would give 48 rad/s, far too large.

Measuring rotational inertia in the lab

Wind a string round a spindle of radius r on the rotating part. Hang a mass m on the string and release it. A rotary motion sensor gives α. Two corrections make the measurement honest:

1. Tension is not mg. The hanging mass accelerates downward at a = rα, so m g − T = m rα and T = m(g − rα). The applied torque is τ = Tr.
2. Friction torque. Στ = Tr − τ_f = Iα, so Tr = Iα + τ_f.

Plot Tr (vertical) against α (horizontal) for several hanging masses. The slope is I and the vertical intercept is the friction torque τ_f. Plotting α against Tr instead gives slope 1/I. Change only m between runs, and repeat each run.

Common misconceptions

• “α points the way the object turns.” It points along the net torque. A wheel turning clockwise while braking has a counterclockwise α.
• “Biggest torque means biggest ω.” The biggest net torque gives the biggest α. In Worked example 2, ω was greatest when the net torque was zero.
• “The axle force has no effect, so I can ignore it.” It has no torque about the axle, but it appears in ΣF = Ma_cm (Worked example 1).
• Using the wrong I. I depends on the axis. Use the parallel axis theorem when the axle is not at the center of mass.
• Setting T = mg for a hanging mass that accelerates. T = m(g − a).
• Using constant-α equations with a changing torque. If Στ depends on t, integrate.
• Writing τ = Iα with internal torques included. Only external torques count.

Where this leads

Earlier: Topic 5.5, Rotational Equilibrium and Newton’s First Law in Rotational Form, the Στ = 0 case. Next, Topic 6.1, Rotational Kinetic Energy, starts Unit 6, where energy and angular momentum let you follow rotations whose torque changes with angle, such as the swinging arm in Worked example 1. Try the practice questions now, then use the revision notes and the checklist. You can also return to the course roadmap.