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A string wrapped round a cylinder turns a straight-line problem into a rotation problem. JEE questions use it in three main forms: a string pulled by a force, a block hanging from a wound pulley, and a cylinder falling as its string unwinds, which is the yo-yo. One method handles all three. This post sets it out, works each case with numbers, and lists the slips that cost marks.
Where is this in the syllabus?
The Syllabus for JEE (Main) 2026 lists moment of inertia, torque and "rigid body rotation and equations of rotational motion" in Unit 5. The NCERT Class 11 chapter on rotational motion covers τ = Iα in Section 6.11 and lists moments of inertia: MR²/2 for a solid cylinder or disc about its axis and MR² for a hollow cylinder. Check the syllabus for your own exam year.
What is the method?
OpenStax University Physics states Newton's second law for rotation: the net torque about a fixed axis equals Iα. With a string, use three equations:
Force equation for each body moving in a straight line: net force = ma.
Torque equation for the cylinder: TR = Iα, where T is the tension at the rim, radius R.
Constraint: if the string does not slip, the string's acceleration equals the rim's tangential acceleration, so a = αR.
Choose directions so that a, α and the motion agree, then solve.
Case 1: pulling the string on a fixed cylinder
NCERT Exercise 6.13: a rope is wound round a hollow cylinder of mass 3 kg and radius 40 cm, and pulled with 30 N. Find α and the rope's acceleration.
I = MR² = 3 × (0.40)² = 0.48 kg m².
Torque = FR = 30 × 0.40 = 12 N m.
α = 12 ÷ 0.48 = 25 rad/s².
Rope acceleration a = αR = 25 × 0.40 = 10 m/s².
NCERT's Example 6.12 does the same with a 20 kg flywheel of radius 20 cm and a 25 N pull, getting α = 12.5 rad/s², and shows that the work done by the pull equals the wheel's gain in kinetic energy.
Case 2: a block hanging from a wound disc
A block of mass m = 1 kg hangs from a light string wound on a uniform disc of mass M = 2 kg and radius R = 0.10 m, free to turn on a fixed, frictionless axle. Take g = 9.8 m/s².
Block: mg - T = ma.
Disc: TR = (½MR²)α, and with a = αR this gives T = ½Ma.
Add them: mg = (m + ½M)a, so a = mg ÷ (m + ½M) = 9.8 ÷ 2 = 4.9 m/s².
Tension: T = ½ × 2 × 4.9 = 4.9 N. Angular acceleration α = 4.9 ÷ 0.10 = 49 rad/s².
The tension is 4.9 N, not mg = 9.8 N. If T were mg, the block would not accelerate.
Case 3: a cylinder falling as its string unwinds
Now hold the free end of the string still and let a solid cylinder of mass m and radius R fall, unwinding as it goes. The cylinder both falls and spins, so it gets both equations.
Falling: mg - T = ma.
Spinning about its centre: TR = (½mR²)α, and a = αR gives T = ½ma.
Combine: mg = (3/2)ma, so a = 2g/3 ≈ 6.5 m/s², and T = mg/3.
For a thin ring, I = mR², the same steps give a = g/2 and T = mg/2. In general, a = g ÷ (1 + I/mR²): the more the mass sits at the rim, the slower the fall.
Energy check. For the solid cylinder falling h = 0.60 m from rest, mgh = ½mv² + ½Iω² = (3/4)mv², so v = √(4gh/3) = √(4 × 9.8 × 0.60 ÷ 3) = √7.84 = 2.8 m/s. Using v² = 2ah with a = 2g/3 gives the same 2.8 m/s.
Why does a real yo-yo fall so slowly?
A toy yo-yo winds its string on a thin axle of radius r, not on its outer rim. The torque arm is now r, so a = g ÷ (1 + I/mr²). With I ≈ ½mR², this is a = g ÷ (1 + R²/2r²).
Take sample dimensions R = 3.0 cm and r = 0.50 cm. Then R²/2r² = 9 ÷ 0.5 = 18, so a = g/19 ≈ 0.52 m/s², and T = mg - ma = (18/19)mg ≈ 0.95mg. The string carries almost the full weight, and the yo-yo drops gently while spinning fast.
Which mistakes cost marks?
Assuming T = mg for a hanging or falling body.
Using the wrong radius. For a yo-yo, the torque arm and the constraint use the axle radius r.
Using MR² for a solid cylinder. It is MR²/2 about its axis; MR² is for a hollow one.
Mixing signs. Pick one positive direction for a and the matching sense for α.
Forgetting the constraint a = αR, which is what ties the equations together.
A practice routine for this topic
Draw separate diagrams for each body before writing any equation.
Solve Case 2 with symbols, then check the limits: if M = 0, a should be g; if M is huge, a should be near 0.
Redo Case 3 for a solid sphere (I = 2mR²/5, from NCERT's table) and compare its fall with the cylinder's.
Check one answer by energy each session, as in the energy check above.
Rolling and unwinding are easier to grasp when you watch them. The Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read. For the kinematics behind α and ω, see our guide to rotational kinematics equations, and for strings over pulleys, pulley constraint relations.
Tonight, work out a and T for a ring falling as its string unwinds, and check that T is half its weight.
