M014 · Rope constraints and work

Atwood Machine

In an Atwood machine, why are the rope tensions on the two sides unequal? Because the pulley has rotational inertia and the rope has weight. Changing the pulley's spin takes a torque, and that torque can only come from different tensions on the two sides. As the rope runs over, its own weight moves from one side to the other and changes the imbalance.

Run an Atwood machine whose rope follows its real path over the wheel, and compare an ideal setup with a heavy pulley, a heavy rope and a damped axle. Read the tension at both masses and at both sides of the rim, sample the position, and fit an acceleration to the measurements.

Left rim tension $0\,\mathrm N$

Right rim tension $0\,\mathrm N$

Recorded time $0\,\mathrm s$

Load coordinate, upward positive $0\,\mathrm m$

Signed load velocity $0\,\mathrm{m/s}$

Instantaneous load acceleration $0\,\mathrm{m/s^2}$

Generalized inertial mass $0\,\mathrm{kg}$

Translational kinetic energy $0\,\mathrm J$

Sheave rotation energy $0\,\mathrm J$

Rope kinetic energy $0\,\mathrm J$

Gravitational energy change $0\,\mathrm J$

Signed input travel $0\,\mathrm m$

Work at the free end $0\,\mathrm J$

Viscous bearing work $0\,\mathrm J$

Integrated energy residual $0\,\mathrm J$

Acquired positions $1$

Whole-record quadratic acceleration $\text{Need 3 samples}$

Quadratic position fit RMS $\text{Need 3 samples}$

Record end time $0\,\mathrm s$

First record boundary $\text{Travel boundary}$

Left mass rope force $0\,\mathrm N$

Right mass rope force $0\,\mathrm N$

Rope, motion and work

In an Atwood machine, why are the rope tensions on the two sides unequal?

Physical forcesForce arrows keep one fixed scale for each trial.Left mass

m_1

1.00\,\mathrm{kg}

Right mass

m_2

1.50\,\mathrm{kg}

Pulley radius

R

0.22\,\mathrm m

Rope linear density

\lambda

0.00\,\mathrm{kg/m}

Pulley moment of inertia

I

0.04\,\mathrm{kg\,m^2}

Viscous axle coefficient

B

0.00\,\mathrm{N\,m\,s}

Initial load coordinate

q_0

0.00\,\mathrm m

Position sensor sample period

\Delta t

0.10\,\mathrm s

Bounded position noise amplitude

\varepsilon

0.00\,\mathrm{mm}

Inspect recorded time

t/t_{\mathrm{end}}

0.00\,\%

Positive motion raises the left mass and lowers the right one. The two vertical hanging lengths of rope always add up to four metres. The rope's mass changes both the weight imbalance and the inertia. The tension at the pulley rim includes the weight of the hanging rope; each block's arrow shows only the rope force on that block.

The sensor records only up to the cursor time. The noise is bounded and reproducible, and editing any setting starts a fresh trial. The quadratic position fit needs three samples and assumes constant acceleration. Rope weight and viscous drag can break that assumption, so look at the fit residual and the instantaneous acceleration. The CSV holds the recorded positions and their fit, kept apart from the model's true values.

The pulley is rigid and fixed. The rope stays taut, does not stretch and does not slip on the rim. The heavy rope has uniform mass per length. Axle drag is viscous, with no static friction. The run ends at the limit of travel, before anything hits. Shapes and rope thickness are schematic, and elasticity, air drag and material calibration are left out.

Start with the essentials

Focus question: How much acceleration does the ideal prediction miss?
One-sentence intuition: Both the weight imbalance that drives the motion and the inertia that resists it must include the rope. A spinning pulley needs a torque, and unequal tensions at the rim supply it.

Core mathematical model

One coordinate for the whole rope

\ell_1=\ell_0-q,\quad\ell_2=\ell_0+q,\quad\ell_0=2\,\mathrm m,\quad\omega=-\dot q/R

The coordinate is positive when the left side moves up. Both hanging lengths stay positive throughout the run, and the rope over the top semicircle keeps a constant length.

Inertia and driving force

\left[m_1+m_2+\lambda(2\ell_0+\pi R)+\frac I{R^2}\right]\ddot q+\frac B{R^2}\dot q=(m_2-m_1)g+2\lambda gq

A heavy rope adds translational inertia, and its weight imbalance changes with position. The viscous axle torque opposes the spin; it is not Coulomb friction at the axle.

Rope forces on the masses

T_1=m_1(g+\ddot q),\qquad T_2=m_2(g-\ddot q)

These are the upward arrows on the two masses. Each mass is treated on its own, without any of the rope.

Tension at the rim includes the hanging rope

T_{1r}=(m_1+\lambda\ell_1)(g+\ddot q),\quad T_{2r}=(m_2+\lambda\ell_2)(g-\ddot q)

The difference between the two rim forces turns the wheel, accelerates the rope over the top arc and overcomes the axle loss. The rim forces are equal only when the wheel is massless and frictionless and the rope is massless too.

Energy balance with the rope included

\Delta K+\Delta U=-\int_0^t\frac B{R^2}\dot q^2\,\mathrm dt,\quad \Delta U=(m_1-m_2)g\Delta q-\lambda g(q^2-q_0^2)

Kinetic energy includes both masses, the whole rope and the spinning pulley. The power lost in the bearing is integrated on its own, not worked out from the energy difference.

Fit the measurements, not a theoretical answer

q_i\simeq c_0+c_1t_i+\tfrac12a_{\mathrm{fit}}t_i^2

The fit needs at least three samples. It returns a single acceleration, and that number depends on the model: with a heavy rope or damping the motion is not exactly quadratic. The RMS value measures how far the positions sit from the fit; it does not guarantee an accurate acceleration.

Common difficulties

One rope, different tensions

Typical misconceptionThe rope has the same tension everywhere.

Better mental modelGive the pulley some inertia or the rope some mass, then compare the four tension readings.

Equal masses can still move

Typical misconceptionEqual masses always stay still.

Better mental modelWith a heavy rope, release the masses away from the center. The side with more rope hanging is also heavier.

A good-looking fit can still be biased

Typical misconceptionA small position RMS proves the acceleration is correct.

Better mental modelChange the time span and look at the instantaneous acceleration. Sensor noise and a changing acceleration are different kinds of error.

Run the experiment

01
Reproduce the ideal result
Choose Equal tension. Open Position sensor and more readings, set the sensor noise to zero, and press Complete record.
What to observe: The quadratic fit matches the constant acceleration, and the rope forces on the two masses agree.
02
Add a heavy pulley
Choose Spin needs torque and compare the end times.
What to observe: The acceleration drops, the two rim tensions move apart, and the rotational energy grows.
03
Unbalance the rope
Choose Longer side wins with equal masses, then reverse the release offset.
What to observe: The motion reverses direction even though the masses are still equal.
04
Check a measurement
Add axle drag, change the sample period and noise, then export the positions.
What to observe: Tell apart three effects: sensor noise, the error of assuming constant acceleration, and energy lost in the bearing.