Help in understanding sum of torque equation

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The discussion focuses on the equation for the sum of torque, ∑τ = Iα, which relates torque (τ), moment of inertia (I), and angular acceleration (α). A participant questions why the torque from particles without external forces is considered, given that internal forces cancel out. It is clarified that the torque is distributed internally across the rigid body, ensuring all parts experience the same angular acceleration. The analogy of lifting a weight illustrates that while force acts at a specific point, the entire body moves together. Understanding the distinction between external forces acting on the body versus on individual particles is crucial for grasping the concept.
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
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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