Press Next (or Play) to walk through each story one small step at a time. The numbers come last.
1. The old well
Read the story as text
An old well has a heavy wooden drum with a rope wound around it. A bucket hangs on the rope. When you let go of the crank, the bucket drops into the well, and the drum spins faster and faster.
But the bucket falls more slowly than if you just dropped it. The drum seems to hold it back, even with no friction at all.
How fast does the bucket speed up, and how does the drum's spin enter the story?
2. Dropped bucket against rope bucket
Quick check: a net torque of 2.0 N·m acts on a wheel with I = 0.50 kg·m². What is α?
More: Omar spins a bike wheel
More: Zoe pushes the merry-go-round
3. A heavy drum is hard to spin up
Quick check: a hoop and a solid disk have the same mass and radius. The same torque acts on each. Which spins up faster?
More: Kai stops the spinning wheel
4. Check yourself
Think of your answer first, then tap to see it.
a) The same torque acts on two wheels. Wheel B has twice the rotational inertia of wheel A. Which spins up faster?
Show answer
Wheel A, with twice the angular acceleration.
α = Στ ÷ I: double I, half α.
b) A bucket falls into a well, unwinding the rope from a drum. Is the rope tension bigger than, equal to, or smaller than the bucket's weight?
Show answer
Smaller. The bucket speeds up downward, so the net
force on it points down: weight is bigger than tension.
c) A wheel spins, and the net torque on it is zero. What happens to its spin?
Show answer
It keeps spinning at the same rate. Zero net torque
means α = 0, so ω stays constant (it does not have to stop).
Already know this?
Three quick questions. Get all three right first time and you can skip ahead.
A net torque of 6.0 N·m acts on a wheel with I = 2.0 kg·m². Its angular acceleration is:
The same torque acts on two wheels. Wheel B has the bigger rotational inertia. Which is true?
A bucket falls, unwinding a rope from a heavy drum. The rope tension is:
Net torque causes angular acceleration, and rotational inertia resists it: Στ = Iα.
Στ = Iα
In pictures
The rope pulls on the edge of the drum. That turning push (a torque) makes the drum spin faster and faster.
A heavy drum, with its mass far from the axle, is hard to spin up: the same torque gives a smaller angular acceleration.
Spinning up the drum uses some of the pull, so the bucket on the rope speeds up more slowly than a dropped bucket.
Rope and drum move together: the rope's speed is the drum's rim speed.
Below is the same idea in words, and then with numbers.
In straight-line motion, ΣF = ma. Rotation has the same law with the rotational partners: torque for force, rotational inertia for mass, angular acceleration for acceleration.
α = Στ / I (Στ = Iα)
A bigger net torque gives a bigger α. A bigger I gives a smaller α for the same torque. Zero net torque gives α = 0 (5.5).
Rope and pulley problems link the two laws. If a rope does not slip on a pulley of radius R, the rope's acceleration and the pulley's angular acceleration are tied together:
a = R α
Recipe: write ΣF = ma for each hanging block, Στ = Iα for the pulley, use a = Rα, then solve. With a massive pulley the rope tensions on the two sides are not equal: their difference is what gives the pulley its torque.
With numbers: the well (one hanging mass)
A 1.0 kg bucket hangs from a rope wound around a solid drum (disk) of mass 2.0 kg and radius 0.10 m. Take down as positive for the bucket and the drum's matching turning direction as positive.
Bucket: mg − T = ma, so 9.8 − T = 1.0a.Weight down, tension up, ΣF = ma.
Drum: the rope pulls at the rim with lever arm R: TR = Iα = I(a/R), so T = Ia/R² = (0.010/0.010)a = 1.0a.a = Rα because the rope does not slip.
Add the two equations: 9.8 = 2.0a, so a = 4.9 m/s².In general a = mg / (m + I/R²) = mg / (m + M/2) for a disk.
T = 1.0 × 4.9 = 4.9 N (less than mg = 9.8 N), and α = a/R = 4.9 / 0.10 = 49 rad/s².Check: τ = TR = 0.49 N·m, and Iα = 0.010 × 49 = 0.49 N·m. ✓
Atwood machine with a heavy pulley
The pulley is a solid disk (I = ½MR²). m₂ hangs on the right, m₁ on the left. Set m₁ = 0 for the "one bucket on a drum" case. Down is positive for m₂. Predict first, then play. The run ends when a block has moved 1.0 m.
Try: m₁ = 2, m₂ = 3, then raise the pulley mass M from 0 to 10 kg. a drops from 1.96 to 0.98 m/s², and T₁ and T₂ (readouts) move apart. With M = 0 the tensions are equal.
Worked examples
Torque on a wheel basic
A net torque of 3.0 N·m acts on a wheel with I = 0.60 kg·m². Find α.
Show solutionShow all steps as text
α = Στ / I = 3.0 / 0.60 = 5.0 rad/s².Same as a = F/m.
Stopping a grinding wheel medium
A grinding wheel (solid disk, 4.0 kg, radius 0.20 m) spins at 50 rad/s. A tool pressed on it gives a friction torque of 1.6 N·m. How long until it stops?
t = (0 − 50) / (−20) = 2.5 s.Rotational kinematics from 5.1.
Bucket in a well medium
A 5.0 kg bucket hangs from a rope wound on a solid cylinder drum (10 kg, radius 0.15 m). Find the bucket's acceleration and the rope tension.
Show solutionShow all steps as text
a = mg / (m + M/2) = (5.0)(9.8) / (5.0 + 5.0) = 4.9 m/s².From mg − T = ma and T = (M/2)a. The radius cancels.
T = m(g − a) = 5.0(9.8 − 4.9) = 24.5 N.Check: T = (M/2)a = 5.0 × 4.9 = 24.5 N. ✓
Atwood machine with a massive pulley AP
Blocks of 2.0 kg (left) and 3.0 kg (right) hang over a solid disk pulley of mass 2.0 kg and radius 0.10 m. Find a, both tensions, and compare with a massless pulley.
Show solutionShow all steps as text
Right block (down +): 3.0g − T₂ = 3.0a. Left block (up +): T₁ − 2.0g = 2.0a.Each block moves with the same size of a, in the directions shown.
Pulley: (T₂ − T₁)R = Iα = (½MR²)(a/R), so T₂ − T₁ = ½Ma = 1.0a.The two tensions pull opposite ways on the rim.
Add all three: 3.0g − 2.0g = (3.0 + 2.0 + 1.0)a, so a = 9.8 / 6.0 = 1.63 m/s².The internal tensions cancel when we add.
T₁ = 2.0(9.8 + 1.63) = 22.9 N; T₂ = 3.0(9.8 − 1.63) = 24.5 N.T₂ > T₁: the difference, 1.6 N, times R gives the torque that spins up the pulley.
Massless pulley: a = 9.8 / 5.0 = 1.96 m/s² and T₁ = T₂ = 23.5 N.A heavy pulley "uses" some of the net force to spin itself, so the blocks speed up less.
Practice
The same net torque acts on two wheels. Wheel B has twice the rotational inertia of wheel A. αB / αA is:
Show answer
α = τ/I: twice the I gives half the α.
A net torque of 12 N·m gives a flywheel an angular acceleration of 4.0 rad/s². Its rotational inertia is:
Show answer
I = τ/α = 12 / 4.0 = 3.0 kg·m².
In an Atwood machine with a massive pulley, the heavier block (m₂) moves down. Compare the tensions.
Show answer
The pulley speeds up its spin in the direction m₂ pulls, so the net torque (T₂ − T₁)R must be positive: T₂ > T₁. Equal tensions only for a massless (zero-I) pulley.
A block hangs from a string wrapped around a pulley with rotational inertia, and is released. The string tension is:
Show answer
The block accelerates downward, so mg − T = ma > 0: T < mg. The pulley needs a torque to spin up, so T > 0.
Two pulleys, a hoop and a solid disk, have the same mass and radius. The same hanging block is released from each. Which block has the larger acceleration?
Show answer
a = mg / (m + I/R²). The disk has the smaller I (½MR² vs MR²), so the block on the disk speeds up more.
A hanging mass m is wrapped on a solid disk pulley of mass M. If M is made very large compared with m, the block's acceleration:
Show answer
a = mg / (m + M/2). As M grows the bottom grows without limit, so a → 0. (With M = 0, a = g.)
Short answer. Show that for a block of mass m on a string wrapped around a pulley of rotational inertia I and radius R, a = mg / (m + I/R²).
Show answer
Block (down +): mg − T = ma. Pulley: TR = Iα, and no slipping means α = a/R, so T = Ia/R².
Substitute: mg − Ia/R² = ma, so mg = a(m + I/R²) and a = mg / (m + I/R²). The I/R² term acts like extra mass that must be accelerated.
Short answer. Explain why the angular acceleration of the pulley equals a/R, where a is the acceleration of the rope.
Show answer
If the rope does not slip, each bit of the pulley's rim moves with the rope. The rim's tangential acceleration is at = Rα (5.2), and it equals the rope's acceleration a. So α = a/R.
AP question types for this topic
Four short free response questions, one of each AP Physics 1 type: Mathematical Routines, Translation Between Representations, Experimental Design and Analysis, and Qualitative/Quantitative Translation. Try each one on paper before you open the worked answer.
1. Mathematical Routines (6 points)
Rafael hangs a block of mass m from a light string wrapped around a pulley, a uniform disk of mass M and radius R on a frictionless axle. He releases the block from rest.
Derive an expression for the block's acceleration a in terms of m, M and g.
m = 0.50 kg, M = 2.0 kg, R = 0.10 m. Calculate a, the pulley's angular acceleration α and the string tension.
Calculate the block's speed after it has fallen 1.0 m.
Show worked answer and scoring
(a) Block: mg − T = ma. Pulley: TR = Iα = (½MR²)(a/R), so T = ½Ma. Adding: mg = (m + ½M)a, so a = mg/(m + M/2).
(b) a = 0.50 × 9.8/(0.50 + 1.0) ≈ 3.3 m/s². α = a/R ≈ 33 rad/s². T = ½Ma = 1.0 × 3.27 ≈ 3.3 N (less than the block's weight, 4.9 N).
(c) v = √(2ah) = √(2 × 3.27 × 1.0) ≈ 2.6 m/s.
Scoring (6 points): (a) 3 points: 1 point for Newton's second law on the block, 1 point for τ = Iα on the pulley with I = ½MR², 1 point for combining with a = Rα. (b) 2 points: 1 point for a and α, 1 point for T. (c) 1 point for 2.6 m/s.
2. Translation Between Representations (6 points)
Yusuf applies a constant 3.0 N·m torque to a bicycle wheel (I = 0.60 kg·m²) that starts at rest, for 4.0 s. Then he lets go. The axle is frictionless.
ABCD
In words: Four graphs of omega against time over 8 seconds. A rises in a straight line for the first half, then stays flat. B rises in a straight line, then falls back to zero. C curves upward more and more steeply. D is flat.
Which graph (A to D) shows ω against t over 8.0 s? Justify.
Describe the graph of α against t, and how it relates to the slope of your ω graph.
Calculate ω at t = 4.0 s, and the angle turned in the first 4.0 s. Which feature of the ω graph gives the angle?
Show worked answer and scoring
(a)A. Constant net torque gives constant α = τ/I, so ω rises in a straight line. With no torque after 4.0 s, α = 0 and ω stays constant. B would need a backward torque, and C a growing torque.
(b) α = 3.0/0.60 = 5.0 rad/s² from 0 to 4.0 s, then drops to zero: a step down. α is the slope of the ω graph in each part.
(c) ω = 5.0 × 4.0 = 20 rad/s. θ = ½ × 5.0 × 4.0² = 40 rad, the area under the ω graph (a triangle ½ × 4.0 × 20).
Scoring (6 points): (a) 2 points: 1 point for A, 1 point for constant α then zero torque. (b) 2 points: 1 point for the step from 5.0 to 0, 1 point for α = slope. (c) 2 points: 1 point for 20 rad/s and 40 rad, 1 point for area under the graph.
3. Experimental Design and Analysis (7 points)
Ingrid wants the rotational inertia of a wheel on a low friction axle. She wraps a string around the wheel's spindle (radius 0.050 m), hangs a mass m, and times how long it takes to fall 1.0 m from rest.
m (kg)
0.10
0.20
0.30
0.40
t (s)
4.07
2.89
2.37
2.06
In words: Masses of 0.10, 0.20, 0.30 and 0.40 kilograms fell 1.0 metre in 4.07, 2.89, 2.37 and 2.06 seconds.
Explain how she can find the angular acceleration α and the torque τ on the wheel from her measurements.
What should she graph to get a straight line whose slope is the wheel's rotational inertia? Why is the tension not just mg?
Use the data to find I.
Show worked answer and scoring
(a) From rest, h = ½at², so a = 2h/t². The string unwinds from the spindle, so α = a/r. The block obeys mg − T = ma, so T = m(g − a), and the torque is τ = Tr = m(g − a)r.
(b) τ = Iα, so graph τ against α: a straight line through the origin with slope = I. The tension is less than mg because the block accelerates downward; T = m(g − a).
Scoring (7 points): (a) 3 points: 1 point for a = 2h/t², 1 point for α = a/r, 1 point for τ = m(g − a)r. (b) 2 points: 1 point for τ against α with slope I, 1 point for why T < mg. (c) 2 points: 1 point for the processed values, 1 point for I ≈ 0.020 kg·m².
Zara has a hoop and a solid disk, each 1.5 kg with radius 0.20 m, on frictionless axles. She claims: “If I give both the same torque, the disk speeds up twice as fast.”
Explain in words why Zara is right.
Derive the ratio αdisk/αhoop for equal torques.
Each gets a 0.60 N·m torque from rest for 2.0 s. Calculate both final angular speeds, and connect the numbers to (a).
Show worked answer and scoring
(a) The hoop has all its mass at the rim, so it has more rotational inertia and resists changes in spin more. The same torque therefore gives the disk, with its mass closer in, a larger angular acceleration.
(c) Ihoop = 1.5 × 0.20² = 0.060 kg·m², α = 10 rad/s², ω = 20 rad/s. Idisk = 0.030 kg·m², α = 20 rad/s², ω = 40 rad/s. The factor 2 is exactly the “less inertia, more acceleration” of (a), put in numbers.
Scoring (6 points): (a) 2 points: 1 point for the hoop having more inertia, 1 point for same torque giving the disk more α. (b) 2 points: 1 point for α = τ/I, 1 point for the ratio 2. (c) 2 points: 1 point for 20 and 40 rad/s, 1 point for the link to (a).
Common mistakes
The mistake: using the same tension on both sides of a massive pulley.
Why it is wrong: if T₁ = T₂ the pulley has zero net torque and could not spin up.
How to spot it: the pulley has a mass or an I. Then use T₁ and T₂.
The mistake: setting the tension equal to mg for a hanging block that is accelerating.
Why it is wrong: T = mg only when a = 0. A falling block has T < mg.
How to spot it: write ΣF = ma for the block first.
The mistake: mixing up a and α, e.g. writing Στ = Ia.
Why it is wrong: α is in rad/s², a in m/s². They are linked by a = Rα.
How to spot it: check units: τ (N·m) = I (kg·m²) × α (1/s²).
The mistake: inconsistent sign choices between the blocks and the pulley.
Why it is wrong: if "m₂ down" is positive, then "m₁ up" and the pulley's matching turning direction must also be positive, or the equations fight each other.
How to spot it: your answer gives a negative a when the heavier side clearly falls.