Help in understanding sum of torque equation

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Discussion Overview

The discussion revolves around the equation for the sum of torque, ∑τ = Iα, in the context of rigid body dynamics. Participants explore the implications of torque contributions from different particles within a rigid body when an external force is applied.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant describes the derivation of the torque equation and questions the relevance of torque contributions from particles not directly experiencing an external force.
  • Another participant suggests that the blue dot in the example is a point on the disk rather than a particle, implying that the simplification to a single term is valid for rigid bodies.
  • A different participant clarifies the distinction between 'external to the body' and 'external to the particle,' noting that the external torque is distributed internally, maintaining the same angular acceleration across the rigid body.
  • A comparison is made to lifting a weight, indicating that while the force acts at a specific point, the entire weight moves, suggesting a similar principle may apply to torque in rigid bodies.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of torque contributions from various particles in a rigid body, indicating that multiple competing perspectives remain without a clear consensus.

Contextual Notes

Some assumptions about the nature of forces and torques in rigid bodies are not fully articulated, and there may be unresolved nuances regarding the distribution of internal forces and their effects on angular acceleration.

Elias Waranoi
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I've been reading my physics book and there they derived the formula ∑τ = Iα where τ is torque, I is moment of inertia of a rigid body and α is the angular acceleration. They did by taking an arbitrary particle on the rigid body with an applied external force tangent to the rotation. τ1 = Ftan * r1 = m1 * a1 * r1 = m1 * r12 * α where m is the particles mass and r is the particles distance from the center of rotation and a is the particles linear acceleration. They sum this into ∑τ = α∑miri2 = Iα

What I don't understand is why the torque beside the first particle matters. In the image I attached an external force is applied to the blue particle so its torque is Ftan * r1 but if you take any other particle there is no external force applied there so Ftan * r1 should be zero? I mean the internal force of particles in a rigid body cancel out each other right? And the force that the blue particle exert on other particles is an internal force right? I understand that for an infinitesemal mass m1 the angular acceleration must be super high to become Ftan so I am quite puzzled.
 

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I think the blue dot isn't a particle, but a point on the disk. And if it is rigid, then: yes, the sum simplifies to this one single term.
 
Elias Waranoi said:
if you take any other particle there is no external force applied
You might be confusing 'external to the body' and 'external to the particle'. The external torque is distributed internally, under the constraint that all the parts have to undergo the same angular acceleration.
 
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Compare with lifting a weight: your force acts at some point, but the weight moves in its entirety
 

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