Rotation · F=ma/USAPhO · 6 min
A massive pulley needs unequal tensions to angularly accelerate.
01 The trap
Wrong path
Students set both string tensions equal in every pulley problem.
Why it feels right
Intro problems usually use massless, frictionless pulleys, where the tension is the same on both sides.
02 Correct model
If the pulley has rotational inertia, the tension difference supplies torque: (T2 - T1)R = Iα.
03 Mini-example
Prompt
An Atwood machine uses a pulley with moment of inertia I. Are the two tensions equal?
Common wrong answer
Yes, because it is one string.
Correct reasoning
Not generally. Unequal tensions create the torque that spins the pulley.
Diagnostic cue
If a pulley has a listed I or mass, it probably needs its own rotational equation.
04 Guided practice
Warm-up isolates the principle. Bridge changes the context. Contest-style requires a complete setup on less familiar geometry.
An ideal massless, frictionless pulley redirects a light string. Are the tensions on both sides equal?
A massless pulley cannot require net torque to angularly accelerate.
Yes, in the ideal massless/frictionless model the string tension is the same throughout the string.
Trap: Overcorrecting and saying tensions are never equal.
A pulley has moment of inertia I and radius R. The string does not slip. Why can T_left and T_right differ?
The pulley needs angular acceleration.
The torque equation is (T_right - T_left)R = Iα. If I and α are nonzero, a tension difference is required.
Trap: Using one-string-equals-one-tension even when the pulley has rotational inertia.
Two masses m_2 > m_1 are connected by a light string over a pulley of radius R and moment of inertia I. The string does not slip. (a) Derive the acceleration magnitude. (b) Explain in words why the pulley contributes an I/R² term.
Write Newton's second law for each mass and the torque equation (T_2 - T_1)R = Iα.
For the descending m_2 side: m_2g - T_2 = m_2a. For the rising m_1 side: T_1 - m_1g = m_1a. The pulley gives (T_2 - T_1)R = I(a/R), so T_2 - T_1 = Ia/R². Adding the two mass equations after substituting the tension difference gives (m_2 - m_1)g = (m_1 + m_2 + I/R²)a, so a = (m_2 - m_1)g/(m_1 + m_2 + I/R²). The term I/R² is rotational inertia translated into the same linear acceleration coordinate as the two masses.
Trap: Adding the pulley's mass directly without converting rotational inertia to an equivalent linear term.
In an Atwood machine with a massive pulley, the right-hand mass descends and the pulley rotates clockwise. Compare the tension on the descending side, T_right, with the tension on the rising side, T_left.
A nonzero clockwise torque requires a tension difference.
T_right > T_left because (T_right - T_left)R = I|α| supplies the clockwise angular acceleration.
Trap: Using equal tension because the same string touches both sides of the pulley.
A light string is wrapped around a fixed drum of radius R and moment of inertia I. A mass m hangs from the free end and is released from rest. The string does not slip and axle friction is negligible. (a) Derive the mass's acceleration and string tension. (b) Find the drum's angular speed after the mass descends a distance h. (c) Check the limits I → 0 and I → ∞.
Write mg - T = ma, TR = Iα, and a = αR.
Combining the equations gives a = mg/(m + I/R²) and T = mgI/(mR² + I). Constant acceleration gives ω = √(2ah)/R. As I → 0, a → g and T → 0; as I → ∞, a → 0 and T → mg.
Trap: Setting T = mg even though the hanging mass accelerates.
05 Mock checks