Pulleys and Constraints in Rotation | JEE Notes

A no-slip condition connects accelerations; it does not make tensions equal. A pulley with rotational inertia requires a torque to accelerate.

For a taut, inextensible string passing over a fixed pulley, both hanging ends have equal acceleration magnitudes. If the string does not slip on a pulley of radius $R$ , its tangential acceleration is $a = R α$ . A massive pulley needs $(T_{2} - T_{1}) R = I α$ , so its tensions usually differ. “Massless string” means constant tension along a free segment; contact with the pulley can transmit tangential forces that make the two segments' tensions different.

Try the simulator: Massive Atwood machine

The missing energy is in the pulley

Masses $m_{1} = 3 k g$ and $m_{2} = 5 k g$ hang from a light inextensible string over a uniform solid-disk pulley of mass $4 k g$ and radius $0.20 m$ . The axle is fixed and frictionless, the string remains taut and does not slip, and $g = 10 m s^{- 2}$ . Find the acceleration, both tensions and pulley angular acceleration.

Hint: use three equations for the two masses and the pulley

Take downward positive for the heavier mass, upward positive for the lighter mass, and pulley rotation towards the heavier side as positive.

Full solution

m_{2} g - T_{2} = m_{2} a, T_{1} - m_{1} g = m_{1} a, (T_{2} - T_{1}) R = I \frac{a}{R} .

Add the first two equations and eliminate the tension difference:

a = \frac{(m_{2} - m_{1}) g}{m_{1} + m_{2} + I / R^{2}} .

Here $I = \frac{12}{(} 4) (0.20)^{2} = 0.080 k g m^{2}$ , so $I / R^{2} = 2 k g$ . Therefore

a = 2.0 m s^{- 2}, T_{1} = 3 (10 + 2) = 36 N, T_{2} = 5 (10 - 2) = 40 N,

α = \frac{2}{0.20} = 10 r a d s^{- 2} .

After either mass travels $1 m$ from rest, $v^{2} = 2 a s = 4 m^{2} s^{- 2}$ . The gravitational-energy decrease is $(5 - 3) 10 (1) = 20 J$ ; translation receives $\frac{12}{(} 8) (4) = 16 J$ , and pulley rotation receives $\frac{12}{(} 0.080) (10^{2}) = 4 J$ .

Check: $(40 - 36) (0.20) = 0.80 N m = I α$ . As $I \to 0$ , the tensions become equal and the usual massless-pulley acceleration is recovered.

The quantity $I / R^{2}$ behaves as an extra inertial mass in the acceleration denominator. It is not extra gravitational mass: it stores kinetic energy without contributing to the hanging weights' potential-energy difference. The no-slip assumption also requires sufficient contact traction; an axle being frictionless says nothing about string–rim traction.

Advanced extension: two strings, two winding radii

Suppose two separate strings are wound in opposing senses around coaxial drums of radii $r_{1}$ and $r_{2}$ , rigidly joined with total axle inertia $I$ . Let positive rotation lower $m_{2}$ and raise $m_{1}$ . Their accelerations are now $a_{2} = r_{2} α$ and $a_{1} = r_{1} α$ , which are generally unequal. Assuming taut strings and no slip,

m_{2} g - T_{2} = m_{2} r_{2} α, T_{1} - m_{1} g = m_{1} r_{1} α, T_{2} r_{2} - T_{1} r_{1} = I α .

Elimination gives

α = \frac{g (m_{2} r_{2} - m_{1} r_{1})}{I + m_{1} r_{1}^{2} + m_{2} r_{2}^{2}} .

The sign is governed by the competing weight moments, not by which hanging mass is heavier. A lighter mass on a sufficiently larger drum can descend. If the numerator is zero, a system initially at rest can remain in equilibrium. Setting $r_{1} = r_{2} = R$ recovers the ordinary Atwood result after multiplying by $R$ . Notice that these are two wound strings; blindly assigning unequal accelerations to the two ends of the single inextensible string in Example 16 would violate its geometry.

A yo-yo unwinds from a fixed string

A yo-yo of mass $M = 0.30 k g$ has outer radius $R = 0.040 m$ and inertia $I_{C} = \frac{12}{M} R^{2}$ . A light string is wound around an axle of radius $r = 0.020 m$ . Its upper end is fixed; the yo-yo is released and descends while the string unwinds without slipping. Neglect losses and string thickness. Find its acceleration and string tension, using $g = 10 m s^{- 2}$ .

Hint: use the centre of mass for torque

Take downward translation and the unwinding sense of rotation as positive. The correct constraint uses the winding radius $r$ , not outer radius $R$ .

Full solution

The straight string segment is stationary. Its no-slip contact condition gives $v_{C} = r ω$ and $a = r α$ . Newton's law and torque about the centre are

M g - T = M a, T r = I_{C} α = I_{C} \frac{a}{r} .

Hence

a = \frac{g}{1 + I_{C} / (M r^{2})}, T = \frac{M g I_{C}}{I_{C} + M r^{2}} .

Here $I_{C} = 0.000240 k g m^{2}$ , and $I_{C} / (M r^{2}) = 2$ . Thus

a = \frac{10}{3} m s^{- 2}, T = 2.0 N, α = \frac{a}{r} = 166.7 r a d s^{- 2} .

For descent through height $h$ , energy provides an independent check:

M g h = \frac{12}{M} v_{C}^{2} + {\frac{12}{I}}_{C} {(\frac{v_{C}}{r})}^{2}, v_{C}^{2} = \frac{2 g h}{3} = 2 a h .

Tension removes translational power $T v_{C}$ but supplies rotational power $T r ω$ . These cancel because $v_{C} = r ω$ : the instantaneous string-contact point has zero velocity, so ideal tension does zero net work on the whole yo-yo.

Check: $0 < a < g$ and $0 < T < M g$ . Decreasing the winding radius while holding $I_{C}$ fixed reduces the descent acceleration. Using $R$ instead of $r$ would predict the wrong motion.

A reliable order for a coupled problem

Draw separate free-body diagrams; define positive directions; write translation for each moving centre; write torque about the centre or a stationary axle; derive the string or rolling constraint; solve the equations together. Then check units, contact feasibility and an energy balance. These checks expose sign errors much faster than restarting the algebra.

Massive pulleys and an unwinding yo-yo

In 60 seconds

The missing energy is in the pulley

Advanced extension: two strings, two winding radii

A yo-yo unwinds from a fixed string

A reliable order for a coupled problem