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In most early pulley problems the pulley is "light and frictionless", so the tension is the same on both sides. Real pulleys have mass and must be spun up, which takes a net torque, so the tensions are different. JEE questions use this to test whether you can join translational and rotational equations. This post gives a three-equation method, two worked examples with energy checks, and the mistakes to avoid.
Why does a pulley's mass change the answer?
A massless pulley needs no net torque to speed up, so T1 = T2. A pulley with moment of inertia I obeys τ = Iα, which NCERT's rotational motion chapter calls "Newton's second law for rotational motion about a fixed axis". If the string does not slip, the net torque comes from the difference in tensions: (T1 − T2)R = Iα. So the tensions must differ whenever the pulley speeds up or slows down.
The JEE (Main) 2026 syllabus lists "rigid body rotation and equations of rotational motion" and moments of inertia in Unit 5. Check your exam year's syllabus.
What is the method?
Write three kinds of equation:
Each block: Newton's second law along its motion.
The pulley: net torque = Iα about its axle.
The string: if it does not slip, a = αR.
Count unknowns (accelerations and tensions) and check you have as many equations. For constraint relations with several strings, see our post on pulley constraint relations.
Worked example 1: an Atwood machine with a heavy pulley
Blocks of 3.0 kg and 1.0 kg hang over a uniform disc pulley of mass 4.0 kg. Take g = 10 m/s². NCERT's Table 6.1 gives a disc's moment of inertia as MR²/2, so I/R² = 2.0 kg.
3.0 kg block: 3.0g − T1 = 3.0a
1.0 kg block: T2 − 1.0g = 1.0a
Pulley: (T1 − T2)R = (MR²/2)(a/R), so T1 − T2 = 2.0a
Adding all three: 2.0g = 6.0a, so a = 20/6 ≈ 3.33 m/s². Then T1 = 3.0(10 − 3.33) = 20 N and T2 = 1.0(10 + 3.33) ≈ 13.3 N.
Check the pulley: T1 − T2 ≈ 6.67 N, and 2.0a ≈ 6.67 N. With a massless pulley you would get a = 5.0 m/s² and one tension of 15 N, so the pulley's mass slows everything down and splits the tension. Note that R cancelled: only I/R² matters.
The general result is a = (m1 − m2)g/(m1 + m2 + I/R²). The pulley adds an "effective mass" I/R² to the denominator: half its mass for a disc, its full mass for a ring.
Worked example 2: a block unwinding a cylinder
A 2.0 kg block hangs from a light string wound round a solid cylinder of mass 4.0 kg and radius 0.20 m, free to turn on a fixed axle.
Block: 2.0g − T = 2.0a
Cylinder: TR = (½ × 4.0 × R²)(a/R), so T = 2.0a
Adding: 20 = 4.0a, so a = 5.0 m/s², T = 10 N and α = a/R = 25 rad/s².
Energy check after the block falls 1.0 m. v² = 2 × 5.0 × 1.0 = 10. Block KE = ½ × 2.0 × 10 = 10 J. Cylinder KE = ½ × (½ × 4.0 × 0.04) × (10/0.04) = 10 J. Total 20 J, which equals mgh = 2.0 × 10 × 1.0. Correct.
Compare NCERT's Example 6.12, where a steady 25 N pull unwinds a flywheel. There the force on the cord is given. With a hanging block the tension is not the block's weight: here it is 10 N, not 20 N.
Why do students struggle with this?
Close, Gomez and Heron (2013) studied students applying Newton's second law to rigid bodies. They found a widespread tendency to claim that forces which cause rotation have a smaller effect on translational motion, and developed a tutorial that improved students' answers. The habit to build is to write the translational and rotational equations separately and trust both.
Which mistakes cost marks?
Using one tension on both sides of a massive pulley.
Setting the tension equal to the hanging weight.
Using MR² for a disc pulley, or MR²/2 for a ring.
Wrong torque direction: decide which way the pulley turns and use that sign throughout.
Forgetting a = αR, which needs the string not to slip.
A practice routine
Re-derive a = (m1 − m2)g/(m1 + m2 + I/R²) without notes.
Solve worked example 1 with a ring pulley of the same mass. Predict before solving: a = 20/8 = 2.5 m/s².
Add friction torque at the axle to one problem and see which equation changes.
Finish every problem with an energy check like the one above.
When your answer and the key disagree and you cannot find the step, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Write the three equations for an Atwood machine with a heavy pulley today, and check your tensions with an energy balance.
