SCIENCE · INVESTIGATE AND EXPLAIN

Couple translation and rotation through a massive pulley

Recover unequal tensions from the rotational equation instead of assuming an ideal massless pulley.

Learning outcome: At m 2=4 compare I=.5,1,2. Solve acceleration, both tensions and torque independently; retain the no-slip conversion and compare the massless limit. Explain which observations would challenge the assumed constraint.

Starting recap

Declare the fixed axis, integrate the actual mass or load distribution, and test constraint feasibility before interpreting a numerical result.

All lessons are open directly. Your saved experiments remain when you leave and return.

Try

m 2=4,I=1 yields a 2,alpha 4,T 1=24,T 2=32 and net pulley torque 4 in SI units. Doubling I increases effective rotational resistance to acceleration.

Open the experiment

Predict

Must the tensions on both sides of an accelerating inertial pulley be equal?

Your prediction and reason

Experiment

At m 2=4 compare I=.5,1,2. Solve acceleration, both tensions and torque independently; retain the no-slip conversion and compare the massless limit. Explain which observations would challenge the assumed constraint.

Couple translational forces to pulley inertia

Worked teaching model · synthetic values · Virtual masses 2 kg and selected heavier mass joined by ideal massless nonstretching rope on fixed radius 0.5 m pulley. g=10 m/s², no slip or axle friction. Pulley inertia prevents equal tensions. No physical hanging loads are requested.

Descending mass: 4 kg Pulley rotational inertia: 1 kg m²

Reset keeps your prediction, captures and practice history.

Acceleration versus pulley inertia

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

Acceleration versus pulley inertia: numerical coordinates

inertia (kg m²)	acceleration (m/s²)
0	3.33333
0.05	3.22581
0.1	3.125
0.15	3.0303
0.2	2.94118
0.25	2.85714
0.3	2.77778
0.35	2.7027
0.4	2.63158
0.45	2.5641
0.5	2.5
0.55	2.43902
0.6	2.38095
0.65	2.32558
0.7	2.27273
0.75	2.22222
0.8	2.17391
0.85	2.12766
0.9	2.08333
0.95	2.04082
1	2
1.05	1.96078
1.1	1.92308
1.15	1.88679
1.2	1.85185
1.25	1.81818
1.3	1.78571
1.35	1.75439
1.4	1.72414
1.45	1.69492
1.5	1.66667
1.55	1.63934
1.6	1.6129
1.65	1.5873
1.7	1.5625
1.75	1.53846
1.8	1.51515
1.85	1.49254
1.9	1.47059
1.95	1.44928
2	1.42857

Calculated values under the declared model
Quantity	Value	Unit
Linear acceleration magnitude	2	m/s²
Pulley angular acceleration	4	rad/s²
Rising-side tension	24	N
Descending-side tension	32	N
Net pulley torque	4	N m
Rotational I alpha	4	N m
Displacement after 1 s from rest	1	m

An inertial pulley needs unequal tensions to produce the torque for its angular acceleration.

For the rising mass T1−m1g=m1a; for the descending mass m2g−T2=m2a; and (T2−T1)R=I(a/R). Eliminating tensions gives the denominator m1+m2+I/R². The zero-inertia plotted comparison is an ideal limit, not a claim every pulley is massless.

Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.

An inertial pulley needs unequal tensions to produce the torque for its angular acceleration.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Declare the coupled constraints

Two virtual masses 2 kg and m 2 are joined by a massless nonstretching rope over a fixed pulley of radius.5 m and inertia I. There is no slip or axle friction. The heavier mass descends and the lighter rises with common acceleration magnitude a. No-slip gives angular acceleration alpha=a/R; it does not require equal tensions when the pulley has inertia.

Write recipient-specific equations

For the rising mass, T 1−m 1g=m 1a. For the descending mass, m 2g−T 2=m 2a. On the pulley the two tensions create opposing torques, so(T 2−T 1)R=I alpha. The fixed axle can exert a force but has zero moment about that axis in this ideal model. Keep the pulley as a separate recipient before combining equations.

Eliminate tensions carefully

Substitute alpha=a/R and the two tension expressions. The result is a=(m 2−m 1)g/(m 1+m 2+I/R²). The extra I/R² has mass units and reduces acceleration compared with the zero-inertia limit. Replacing both tensions by one common value would give zero pulley torque despite a nonzero angular acceleration.

Verify an original case

For m 2=4,I=1,R=.5,g=10, acceleration is 2 m/s², alpha 4 rad/s², T 1=24 N and T 2=32 N. Their torque difference times radius is 4 N m, exactly I alpha. From rest after 1 s, each mass travels 1 m in its own direction. The force and torque equations independently check the same constraint.

Keep the feasibility and physical gates visible

The supported controls give positive tensions and a below g. A negative tension or incompatible no-slip requirement would invalidate the assumed rope branch. Real ropes stretch, pulleys have friction and finite strength, and axles move; none is tested here. The zero-inertia curve is an ideal comparison, not a statement that actual pulley mass never matters. No physical suspended loads or lifting experiment is requested.

Worked original case

m 2=4,I=1 yields a 2,alpha 4,T 1=24,T 2=32 and net pulley torque 4 in SI units. Doubling I increases effective rotational resistance to acceleration.

Investigation

Public numerical transfer practice

Try a new practice case

For m 1=2,m 2=4,g=10 and a zero-inertia pulley limit, what is acceleration magnitude in m/s²?

Numerical result in the requested units

This public, unsaved check is not a secure assessment. Explain the derivative, integral or boundary in the retained course evidence workspace.

Sources and evidence boundary

Original teaching and numerical cases checked 30 October 2026 against the official AP Physics C Mechanics framework and OpenStax fixed-axis rotation. Supplied/generated evidence is not physical measurement, a load rating or laboratory participation. No rotating machinery, hanging weights or mechanical construction is requested. Public feedback and retained self-reports do not certify independent learner mastery.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Why does the pulley inertia enter the translational denominator as I/R²?

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Integrate a rotational torque law with its initial condition Next: Integrate rotational energy without double counting

Save a science backup or import one

Science work is stored locally in this browser, separately from mathematics and AI. It does not sync automatically. Use this science-only backup; this checkout's older all-subject export does not include science.

A legacy file is available only when the complete snapshot fits the 36 MB limit. For larger snapshots, prepare a complete file set below; never discard notebooks to make a smaller file.

Choose a science backup (maximum 36 MB)

Matching imported notebook drafts replace their current drafts. Omitted lessons and known first attempts remain. Export your current work first if you want to retain both drafts.

Complete file sets and course backups

A file set can hold more work than one science file. Keep its manifest and every part together. Each file is limited to 36 MB. This format needs this version of LearnAI or a compatible newer version; older versions can use only the legacy single-file backup above.

Export scope

The set includes every saved science course and global continuation at this moment. Later edits need a new export. A conflicting tab exports its own visible work, which may differ from saved work in another tab.

Choose the manifest and all parts together

0 file(s) selected. File order and filenames do not determine identity. Missing, changed or conflicting parts block the entire restore; no good part is applied by itself.

Science storage: loading. No account or remote computation is required.