At a village well in Rajasthan, a bucket hangs from a rope wound on a heavy wooden drum with a handle. One day the handle slips from a woman's hands. The bucket drops into the well and the drum spins with a clatter. Yet the bucket does not fall like a stone dropped down the well. It falls noticeably more slowly, and the rope stays tight all the way.
The link between a string and a wheel
In the chapter on Accelerated Objects, a pulley was always light. It only turned the string round a corner, and the tension was the same on both sides. A real pulley, a well drum or the flywheel of a machine has mass. It must be made to turn, and turning it needs a torque. Now the string does two jobs at once: it pulls the block, and it turns the wheel.
Suppose the string does not slip on the rim. Then the string and the rim move together. The rim moves at \(v = \omega R\), so the string moves at \(\omega R\) too. Differentiate with respect to time: the acceleration of the string is \(a = \alpha R\). This is the constraint that ties the wheel to the block. It is the rotational cousin of the string constraints of the Atwood machine.
One procedure for every problem
A problem with a turning body and sliding bodies can always be solved in the same five steps. The same steps will solve the rolling problems of later chapters.
- Draw a free-body diagram of each body separately.
- For each turning body, write \(\tau_{net} = I\alpha\) about its fixed axis. Choose the positive sense once.
- For each body that moves in a line, write \(F_{net} = ma\) along its motion.
- Link each \(a\) to each \(\alpha\) by the constraint, \(a = \alpha R\).
- Solve, then check the signs and the limiting cases.
A string that does not slip is wound on a wheel of radius \(0.2\,\text{m}\). The wheel has an angular acceleration of \(15\,\text{rad/s}^2\). What is the acceleration of the block hanging from the string?
A block of mass \(1\,\text{kg}\) hangs from a light string wound on a wheel on a fixed axle. The wheel is a uniform disc of mass \(2\,\text{kg}\) and radius \(0.2\,\text{m}\). The block is released from rest and the string does not slip. Take \(g = 10\,\text{m/s}^2\). Find the acceleration of the block, the tension and the angular acceleration of the wheel.
On the block act the tension \(T\) upward and \(mg\) downward; it accelerates down at \(a\). On the wheel, only the tension at the rim has a moment arm about the axle. \(I = \tfrac12 MR^2 = 0.04\,\text{kg m}^2\).
Block: \(mg - T = ma\). Wheel: \(TR = I\alpha\). Link: \(a = \alpha R\), so \(T = Ia/R^2\).
Add the block's equation to \(T = (I/R^2)a\): \(mg = (m + I/R^2)a\). Here \(I/R^2 = M/2 = 1\,\text{kg}\), so \(10 = 2a\) and \(a = 5\,\text{m/s}^2\). Then \(T = 1 \times (10 - 5) = 5\,\text{N}\) and \(\alpha = a/R = 25\,\text{rad/s}^2\).
The block falls at \(5\,\text{m/s}^2\), half of \(g\). The tension is \(5\,\text{N}\), half the weight. Check the wheel: \(TR = 5 \times 0.2 = 1\,\text{N m}\), and \(I\alpha = 0.04 \times 25 = 1\,\text{N m}\).
Read Fig 2.1 body by body. The block feels less upward pull than its weight, so it accelerates down. The wheel feels the same tension pulling down at its rim, and that is the only force with a moment arm. The axle must hold up both the wheel's weight and that tension, \(20 + 5 = 25\,\text{N}\). The general result is \(a = mg/(m + I/R^2)\). The wheel acts like an extra mass \(I/R^2\) that must be accelerated, but that gravity does not pull.
A \(3\,\text{kg}\) block hangs from a string wound on a wheel whose mass, \(2\,\text{kg}\), is all in its thin rim. The axle is fixed and the string does not slip. Take \(g = 10\,\text{m/s}^2\). Find the acceleration of the block, in m/s².
A block hangs from a string wound on a heavy wheel and is falling with increasing speed. How does the tension in the string compare with the weight of the block?
The Atwood machine with a real pulley
Hang \(m_1\) and \(m_2\), with \(m_1 > m_2\), from a string over a pulley of moment of inertia \(I\) and radius \(R\). The string does not slip. Now the tension need not be the same on both sides. Call the tension on the side of \(m_1\) \(T_1\), and on the side of \(m_2\) \(T_2\). The heavier block falls, the lighter one rises, and the pulley turns towards \(m_1\).
Write one equation per body. For \(m_1\): \(m_1 g - T_1 = m_1 a\). For \(m_2\): \(T_2 - m_2 g = m_2 a\). For the pulley: \((T_1 - T_2)R = I\alpha = Ia/R\). Divide the last equation by \(R\) and add all three. The tensions cancel, and \[ a = \dfrac{(m_1 - m_2)\,g}{m_1 + m_2 + I/R^2}. \] With \(I = 0\) this is the light-pulley result of the chapter on Accelerated Objects.
Play the machines in the figure above with \(M = 2\,\text{kg}\). The disc has \(I/R^2 = M/2 = 1\,\text{kg}\), so \(a = 10/(2 + 1 + 1) = 2.5\,\text{m/s}^2\), against \(3.33\,\text{m/s}^2\) for the light pulley. The tensions are \(T_1 = 2 \times 7.5 = 15\,\text{N}\) and \(T_2 = 1 \times 12.5 = 12.5\,\text{N}\). On the light pulley both are \(13.3\,\text{N}\). Push the slider to \(6\,\text{kg}\): the machine slows to \(1.67\,\text{m/s}^2\) and the tensions spread to \(16.7\) and \(11.7\,\text{N}\).
Why the tensions must differ
If the string pulled equally on both sides of the pulley, the net torque on it would be zero. A pulley with zero net torque cannot gain angular velocity. So a pulley with mass turns faster and faster only because \(T_1 > T_2\), and the difference is exactly what it needs: \((T_1 - T_2)R = I\alpha\). Check the numbers: \(T_1 - T_2 = 2.5\,\text{N}\), and \((I/R^2)a = 1 \times 2.5 = 2.5\,\text{N}\).
"One string, one tension" is true only when nothing with inertia sits between the two parts of the string. A light, smooth pulley passes the tension on unchanged. A pulley with mass, or a rough one, does not. Never write one \(T\) for both sides of a pulley whose mass is given.
Blocks of \(5\,\text{kg}\) and \(3\,\text{kg}\) hang over a pulley that is a uniform disc of mass \(4\,\text{kg}\). The string does not slip. Take \(g = 10\,\text{m/s}^2\). Find the acceleration of the blocks, in m/s².
In an Atwood machine, the pulley is changed for a heavier disc of the same radius. What happens to the difference between the two tensions?
In an Atwood machine with a heavy pulley, explain why the tension is larger on the side of the heavier block.
The pulley starts from rest and turns faster and faster towards the heavier block, so it has an angular acceleration in that sense. By \(\tau = I\alpha\) it needs a net torque in that sense. The only forces with a moment arm about the axle are the two tensions, at the same radius. So the tension on the side of the heavier block must be the larger one, and \((T_1 - T_2)R = I\alpha\).
Above most lift shafts sits a machine room with a grooved steel wheel, the sheave. The lift's cables pass over it, with the cabin on one side and a heavy counterweight on the other. The motor turns the sheave, and the grip of the grooves moves the cables. The sheave is heavy. So while the lift gains speed, the motor must also supply the torque \(I\alpha\) that turns the heavy sheave. The sheave is the heavy pulley of an Atwood machine, driven by a motor.
Now return to the well. Suppose the bucket of water has a mass of \(10\,\text{kg}\) and the drum is a solid wooden cylinder of \(20\,\text{kg}\). For a solid cylinder \(I/R^2 = M/2 = 10\,\text{kg}\). So \(a = 100/(10 + 10) = 5\,\text{m/s}^2\), half of \(g\), and the rope keeps a tension of \(10 \times 5 = 50\,\text{N}\). A fall of \(10\,\text{m}\) takes \(2\,\text{s}\), against about \(1.4\,\text{s}\) for a stone. The drum's moment of inertia holds the bucket back.
- Writing the same tension on both sides of a pulley with mass.Name \(T_1\) and \(T_2\). They differ by \(I\alpha/R\).
- Setting the tension equal to the weight of a falling block.The block accelerates, so \(T = m(g - a)\), less than \(mg\).
- Dividing \(\alpha\) by \(R\) to find \(a\).A string that does not slip moves with the rim: \(a = \alpha R\).
- Forgetting that the weight of a pulley and the push of its axle give no torque.Both act at the axle, with zero moment arm. Only the tensions turn the pulley.