M014 · Rope constraints and work

Atwood Machine

An Atwood rig with a real rope path compares ideal, inertial, heavy-rope and damped trials. Resolve the tension at both hanging masses and both rims, sample the position and fit an acceleration from the observations.

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 $0\,\mathrm N$

PREDICT / SAMPLE / AUDIT

Why are the two rope forces unequal?

Physical forcesOne fixed force scale per trialLeft 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

0.00\,\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 load motion lifts the left mass while lowering the right. Two vertical hanging lengths sum to four metres. Rope mass changes both the gravitational imbalance and inertia. Rim tension includes the hanging rope weight; block arrows show only the rope force on that block.

The sensor acquires only through the selected cursor time. Noise is bounded and reproducible; editing creates a fresh trial. A quadratic position fit needs three samples and assumes constant acceleration. Rope weight and viscous drag can violate that assumption; inspect the fit residual and instantaneous acceleration. CSV exports the acquired positions and their fit, separately from model truth.

Rigid fixed pulley, taut inextensible rope and no rim slip. The heavy rope has uniform linear density. Axle drag is viscous, with no static friction. The record ends at the travel boundary before impact. Shapes and rope thickness are schematic; elasticity, air drag and material calibration are omitted.

Start with the essentials

Focus question: How much acceleration does the ideal prediction miss?
One-sentence intuition: The driving imbalance and the inertia must both include the rope. A rotating pulley needs a torque, supplied by unequal rim tensions.

Core mathematical model

Define the same rope coordinate

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

The left coordinate is upward positive. Both hanging lengths stay positive over the finite record. The top semicircle has constant length.

Inertia and drive

\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

The heavy rope adds translational inertia and a position-dependent gravitational imbalance. Viscous torque opposes angular velocity; it is not Coulomb axle friction.

Forces at the masses

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

These are the two upward arrows on the masses. The rope is not included in either block system.

Rim forces include 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 rim difference supplies the wheel torque, the top arc rope inertia and axle loss. A massless frictionless wheel gives equal rim forces only when the rope is also massless.

Energy includes the rope

\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 full rope and pulley rotation. Bearing power is integrated independently of the energy difference.

Fit observations, not a theoretical answer

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

Three samples are needed. A single acceleration estimate is model-dependent; heavy-rope or damped trajectories are not exactly quadratic. RMS measures position residuals, not guaranteed acceleration accuracy.

Common difficulties

Constraint versus force

Typical misconceptionThe rope has one tension everywhere.

Better mental modelCompare the four sampled tension traces with nonzero inertia or rope density.

Equal masses do not settle a heavy rope

Typical misconceptionEqual masses always remain still.

Better mental modelRelease the heavy-rope preset away from the center. The longer hanging side also weighs more.

A good-looking fit can be biased

Typical misconceptionA small position RMS proves the acceleration is correct.

Better mental modelChange the time span and inspect instantaneous acceleration. Noise and acceleration variation are different errors.

Run the experiment

01
Recover the ideal result
Use ideal masses and complete the record with zero sensor noise.
What to observe: The quadratic fit equals the constant acceleration and the mass-end tensions agree.
02
Spin a flywheel
Choose the inertial pulley and compare end times.
What to observe: Acceleration falls while rim tensions separate and rotational energy rises.
03
Unbalance the rope
Choose equal masses with a heavy rope; reverse the release offset.
What to observe: The direction reverses even though the masses remain equal.
04
Audit a measurement
Add axle drag, change sample period and noise, then export positions.
What to observe: Separate sensor perturbations from the constant-acceleration approximation and bearing energy loss.