In pulley problems so far, the pulley was light, so one tension ran all the way around it. A real pulley has mass and has to be spun up, which takes a net torque. This topic joins $\Sigma\tau = I\alpha$ to $\Sigma F = ma$, using $a = R\alpha$.
Why it works
For the pulley to gain angular velocity, the torques on it can't cancel: $(T_2 - T_1)R = I\alpha$. So the tension on the side that is pulled harder must be larger. Only if $I = 0$ do the two tensions match. A cord that doesn't slip ties the motions together: the cord, the hanging masses and the pulley's rim all move with the same $a$, and $a = R\alpha$.
Key idea
Write one equation for each object: $\Sigma F = ma$ for each hanging mass, $\Sigma\tau = I\alpha$ for the pulley, and link them with $a = R\alpha$. For a uniform disk pulley ($I = \tfrac{1}{2}MR^2$), the pulley acts like an extra mass $\tfrac{1}{2}M$ that must be accelerated, and $R$ cancels.
Worked examples
Worked example: A bucket and a winch
A $2.0\ \text{kg}$ bucket hangs from a rope wound on a $4.0\ \text{kg}$ solid cylinder (radius $0.10\ \text{m}$) that turns freely. Find the bucket's acceleration and the rope's tension.
- Bucket (down positive): $mg - T = ma$.
- Cylinder: $TR = \tfrac{1}{2}MR^2 \cdot \dfrac{a}{R}$, so $T = \tfrac{1}{2}Ma = 2.0a$.
- Add: $mg = (m + \tfrac{1}{2}M)a$, so $a = \dfrac{(2.0)(9.8)}{2.0 + 2.0} = 4.9\ \text{m/s}^2$.
- $T = 2.0 \times 4.9 = 9.8\ \text{N}$, half the bucket's weight.
Answer: $a = 4.9\ \text{m/s}^2$, $T = 9.8\ \text{N}$
Worked example: An Atwood machine with a heavy pulley
Blocks of $2.0\ \text{kg}$ and $3.0\ \text{kg}$ hang over a $4.0\ \text{kg}$ disk pulley. Find $a$ and both tensions.
- Heavy block: $m_2g - T_2 = m_2a$. Light block: $T_1 - m_1g = m_1a$. Pulley: $T_2 - T_1 = \tfrac{1}{2}Ma$.
- Add all three: $(m_2 - m_1)g = (m_1 + m_2 + \tfrac{1}{2}M)a$, so $a = \dfrac{(1.0)(9.8)}{7.0} = 1.4\ \text{m/s}^2$.
- $T_1 = 2.0(9.8 + 1.4) = 22.4\ \text{N}$ and $T_2 = 3.0(9.8 - 1.4) = 25.2\ \text{N}$. Check: $T_2 - T_1 = 2.8 = \tfrac{1}{2}(4.0)(1.4)$.
Answer: $a = 1.4\ \text{m/s}^2$; $T_1 = 22.4\ \text{N}$, $T_2 = 25.2\ \text{N}$
Common mistake
Assuming the tension is the same on both sides of a pulley that has mass. Equal tensions would give zero net torque, and the pulley could never start turning. Use two different tensions and let the pulley's equation decide their difference. Check: with $M = 0$ your answers should turn into the light-pulley results, and with $M > 0$ the acceleration must come out smaller.