Why do non-external particles contribute to total torque in ∑τ = Iα?

[Elias Waranoi]

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 = F_tan * 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∑m_ir_i² = 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 F_tan * r₁ but if you take any other particle there is no external force applied there so F_tan * 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 F_tan so I am quite puzzled.

 

[BvU]

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.

 

[A.T.]

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.

 

[BvU]

Compare with lifting a weight: your force acts at some point, but the weight moves in its entirety