Help in understanding sum of torque equation

AI Thread Summary
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
Messages
45
Reaction score
2
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.
 

Attachments

  • torque.png
    torque.png
    4.9 KB · Views: 560
Physics news on Phys.org
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.
 
  • Like
Likes Elias Waranoi
Compare with lifting a weight: your force acts at some point, but the weight moves in its entirety
 
Thread 'Is 'Velocity of Transport' a Recognized Term in English Mechanics Literature?'
Here are two fragments from Banach's monograph in Mechanics I have never seen the term <<velocity of transport>> in English texts. Actually I have never seen this term being named somehow in English. This term has a name in Russian books. I looked through the original Banach's text in Polish and there is a Polish name for this term. It is a little bit surprising that the Polish name differs from the Russian one and also differs from this English translation. My question is: Is there...
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...
Back
Top