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Find cable tension as bar and cylinder accelerate from rest.

Source: homework.study.com

TL;DR

Physics problem on a uniform bar pinned at O, linked by cable over pulley C to hanging cylinder m1.
System released from rest at 23° incline; find cable tension T with given masses and bar length L.
Uses Newton's laws for translation and rotation to link cylinder acceleration and bar angular acceleration.

The story at a glance

This homework problem describes a mechanics setup where a uniform bar of mass m, pinned smoothly at O, connects via a light cable over pulley C to a hanging cylinder of mass m1. The system starts from rest in a shown position—bar inclined at 23° to horizontal, marks at L/3 and 2L/3—and asks for the cable tension right after release. It provides sample values like m=40 kg, m1=12 kg, L=4.5 m.[[1]](https://www.numerade.com/ask/question/the-uniform-bar-of-mass-m-is-supported-by-the-smooth-pin-at-o-and-is-connected-to-the-cylinder-of-mass-m1-by-the-light-cable-which-passes-over-the-light-pulley-at-c-if-the-system-is-released-49097)

Key points

Bar is uniform, mass m, length L, supported by smooth pin at O (no friction, reacts normal and axial forces).
Cable is light, passes over light pulley at C, connects bar (likely at end A) to cylinder mass m1 hanging vertically.
Initial position: bar at 23° to horizontal, segments shown as OL/3 and 2L/3 along bar, points labeled O, B, A, C.
System released from rest; determine instantaneous tension T in cable.
Sample values vary across sources: e.g., m=40 kg, m1=12 kg, L=4.5 m; or m=30 kg, m1=20 kg, L=6 m.[[2]](https://www.chegg.com/homework-help/questions-and-answers/uniform-bar-mass-m-supported-smooth-pin-o-connected-cylinder-mass-m1-light-cable-passes-li-q19891406)[[3]](https://www.numerade.com/questions/the-uniform-bar-of-mass-m-is-supported-by-the-smooth-pin-at-o-and-is-connected-to-the-cylinder-of-ma)
Goal requires dynamics: acceleration of m1 downward equals distance along bar times angular acceleration α of bar.

Details and context

The smooth pin at O allows rotation without friction, so the bar pivots there under gravity and cable tension. The center of mass of the uniform bar is at L/2 from O, creating a torque from its weight mg (component perpendicular to bar).

Cable tension T pulls the bar (reducing clockwise torque from weight) and accelerates m1 downward via T = m1 a, where a is linear acceleration. Kinematic link: if cable attaches at distance d from O (possibly 2L/3 or L), then a = d α.

To solve, draw free-body diagrams—for cylinder: vertical forces; for bar: pin reactions at O, weight at center, tension T tangent at attachment. Use ΣF=ma for cylinder, ΣM_O = I_O α for bar (I_O = (1/3)m L^2 for uniform bar about end), and relate a and α. Position marks (L/3, 2L/3) likely indicate attachment or pulley location relative to O.[[1]](https://www.numerade.com/ask/question/the-uniform-bar-of-mass-m-is-supported-by-the-smooth-pin-at-o-and-is-connected-to-the-cylinder-of-mass-m1-by-the-light-cable-which-passes-over-the-light-pulley-at-c-if-the-system-is-released-49097)

This is a standard rigid body dynamics problem, testing coupled translation-rotation. No friction means pure gravitational drive; light cable/pulley neglects their inertia.

Key quotes

None; this is a textbook-style homework problem without sourced quotes.

Why it matters

Tests core engineering/physics principles of statics-to-dynamics transition in linked systems.
Helps students practice free-body diagrams, torque equations, and acceleration constraints for exams or real designs like cranes, linkages.
Watch worked solutions on Chegg/Numerade for method confirmation, as numerical T depends on exact attachment point (not fully visible here).[[2]](https://www.chegg.com/homework-help/questions-and-answers/uniform-bar-mass-m-supported-smooth-pin-o-connected-cylinder-mass-m1-light-cable-passes-li-q19891406)