AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/5/5-6)
Unit 5 · Topic 5.6
5.6 Newton’s Second Law in Rotational Form
When the net torque on a rigid system isn't zero, its angular velocity changes at a rate α = τ_net/I. More torque means more angular acceleration and more rotational inertia means less. Pulleys with mass and similar problems need both the linear and the rotational form of Newton's second law, linked by the no-slip condition.
Key terms
- Newton's second law in rotational form
- angular acceleration
- net torque
- rotational inertia
- pulley with mass
The rotational second law
Newton's second law for straight-line motion is a = F_net/m. Its rotational partner is α = τ_net/I, where α is the angular acceleration, τ_net the net torque and I the rotational inertia about the same axis.
The angular acceleration is in the same sense as the net torque. A net counterclockwise torque speeds up a counterclockwise spin or slows down a clockwise one.
This makes two predictions you can test. Double the net torque and α doubles. Double I (by moving mass outward, say) at the same torque and α halves.
| Linear | Rotational |
|---|---|
| force F | torque τ |
| mass m | rotational inertia I |
| acceleration a | angular acceleration α |
| F_net = ma | τ_net = Iα |
Linking linear and rotational analyses
Many systems have one part that moves in a line and another that spins, like a block hanging from a rope wrapped around a heavy pulley. You analyze each part separately: F_net = ma for the block and τ_net = Iα for the pulley.
Then you connect them. If the rope doesn't slip, the block's acceleration equals the rim's tangential acceleration: a = Rα (from 5.2). That gives you enough equations to solve.
Here's the key physics. To give a heavy pulley angular acceleration, the rope tensions on its two sides must differ, since the difference creates the net torque. Only for a massless (ideal) pulley are the tensions on both sides equal.
A step-by-step method
- Draw a free-body diagram for each block and a force diagram for the pulley or wheel.
- Choose positive directions that match: if the block moving down makes the pulley turn clockwise, call down and clockwise both positive.
- Write F_net = ma for each block and τ_net = Iα for each rotating part.
- Add the no-slip link a = Rα and solve.
- Check limits: if the pulley's I went to zero, your answer should match the ideal pulley result from Unit 2.
Why a heavy pulley slows things down
For a block of mass m hanging from a rope wound on a pulley with rotational inertia I and radius R, the algebra gives a = mg/(m + I/R²). The pulley acts like an extra mass of I/R² that has to be accelerated, even though it doesn't fall. A bigger I means a smaller acceleration and a larger rope tension.
For a solid-disk pulley, I = ½MR², so I/R² = ½M: the pulley acts like half its own mass. For a hoop-like pulley it acts like all of its mass.
Seeing it on graphs
If a constant net torque acts, the ω–t graph is a straight line with slope α = τ_net/I. Changing the torque partway through changes the slope. If friction exerts a small opposing torque, the net torque is the applied torque minus the friction torque, so the measured α is a little less than predicted.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Applied torque plus friction
A wheel with I = 0.50 kg·m² starts from rest. A 4.0 N force is applied tangentially at its rim, 0.25 m from the axis. The axle exerts a friction torque of 0.20 N·m. Find the angular acceleration.
Show the solutionHide the solution
- Step 1: Applied torque: rF = (0.25)(4.0) = 1.0 N·m, in the direction of the spin.
- Step 2: Friction opposes the spin: τ_net = 1.0 − 0.20 = 0.80 N·m.
- Step 3: α = τ_net/I = 0.80 ÷ 0.50 = 1.6 rad/s².
Answer: 1.6 rad/s²
- Example 2Calculator allowed
A block on a heavy pulley (classic trap)
A 2.0 kg block hangs from a light rope wrapped around a pulley that is a uniform solid disk (I = ½MR²) of mass 4.0 kg and radius 0.10 m. The pulley turns on a frictionless axle and the rope doesn't slip. The block is released from rest. Find the block's acceleration, the rope tension and the pulley's angular acceleration. Use g = 9.8 m/s².
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- Step 1: Pulley: I = ½(4.0)(0.10)² = 0.020 kg·m².
- Step 2: Block (down positive): mg − T = ma, so 19.6 − T = 2.0a.
- Step 3: Pulley: the only torque about the axle is from the rope: TR = Iα. With α = a/R this gives T = Ia/R² = (0.020/0.010)a = 2.0a.
- Step 4: Substitute: 19.6 − 2.0a = 2.0a, so a = 4.9 m/s². Then T = 2.0(4.9) = 9.8 N, and α = a/R = 49 rad/s².
- Step 5: The trap is assuming T = mg = 19.6 N. If it were, the block wouldn't accelerate at all. The tension is less than the weight because the block speeds up downward.
Answer: a = 4.9 m/s² downward; T = 9.8 N; α = 49 rad/s²
- Example 3Calculator allowed
Atwood machine with a massive pulley (exam level)
Blocks of 3.0 kg and 1.0 kg hang on either side of a pulley of radius 0.10 m and rotational inertia 0.010 kg·m². The rope doesn't slip and the axle is frictionless. Find the acceleration and the tension on each side. Use g = 9.8 m/s².
Show the solutionHide the solution
- Step 1: The 3.0 kg block moves down. Take its direction of motion as positive for each block.
- Step 2: 3.0 kg: (3.0)(9.8) − T₁ = 3.0a. 1.0 kg: T₂ − (1.0)(9.8) = 1.0a. Pulley: (T₁ − T₂)R = Iα = Ia/R, so T₁ − T₂ = (I/R²)a = 1.0a.
- Step 3: Add all three: 29.4 − 9.8 = (3.0 + 1.0 + 1.0)a, so a = 19.6/5.0 = 3.92 m/s².
- Step 4: T₁ = 3.0(9.8 − 3.92) ≈ 17.6 N. T₂ = 1.0(9.8 + 3.92) ≈ 13.7 N.
- Step 5: Check: (T₁ − T₂)R = (3.92)(0.10) = 0.392 N·m, and α = 0.392/0.010 = 39.2 rad/s² = a/R.
Answer: a ≈ 3.9 m/s²; T₁ ≈ 17.6 N on the heavy side, T₂ ≈ 13.7 N on the light side
Common mistakes
- Assuming the tension equals the hanging block's weight. That's only true if the block isn't accelerating.
- Using equal tensions on both sides of a pulley that has mass. The difference in tensions is what makes it spin up.
- Using I about one axis and torques about another. Both must be about the same axis.
- Choosing signs that don't match. If down is positive for the block, the pulley's positive sense must be the one that block motion produces.
On the exam
- Free-response questions often ask you to derive an expression for acceleration in terms of m, M, R and g. Start from Newton's laws for each part, and state the no-slip condition.
- A common qualitative question: how does replacing a light pulley with a heavy one (or a disk with a hoop) change the acceleration and tension? Use the effective extra mass I/R² to justify your answer.
Connected topics
Videos
Check yourself
5 questions on 5.6 Newton’s Second Law in Rotational Form. Pick an answer to see if you got it, and why.
A grindstone with rotational inertia 3.0 kg·m² starts from rest. A constant net torque of 12 N·m acts on it. How long does it take to reach an angular speed of 20 rad/s?
A uniform disk of mass 4.0 kg and radius 0.50 m turns on a frictionless axle through its center. Its rotational inertia is ½MR². A string wrapped around the rim is pulled with a constant 10 N force. What is the disk's angular acceleration?
A 3.0 kg block and a 1.0 kg block hang from opposite ends of a light string that passes over a pulley. The pulley is a uniform disk of mass 2.0 kg and radius 0.10 m, with rotational inertia ½MR² = 0.010 kg·m². It turns on a frictionless axle, and the string doesn't slip on it. The blocks are released from rest. Use g = 10 m/s².
Described scenario
What is the magnitude of each block's acceleration?
Which statement about the tensions is correct?
The pulley is replaced by another uniform disk with the same mass but twice the radius. How does the blocks' acceleration change?
0 of 5 answered