Deriving Moment of Inetia using just linear dynamics

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SUMMARY

The moment of inertia can be derived using linear dynamics and calculus, specifically through the relationship between force, torque, and angular acceleration. The derivation begins with the equation F=ma, which is modified to τ=rma=r²mα, where α represents angular acceleration. By summing the contributions of all mass elements, the total torque can be expressed as τ=∫r²dm(dω/dt), providing a clear alternative to the traditional energy-based approach.

PREREQUISITES
  • Understanding of linear dynamics principles, specifically F=ma.
  • Familiarity with calculus, particularly integration techniques.
  • Knowledge of angular motion concepts, including torque and angular acceleration.
  • Basic grasp of moment of inertia and its significance in rotational dynamics.
NEXT STEPS
  • Study the derivation of moment of inertia using energy equations for comparison.
  • Explore advanced calculus techniques relevant to physics applications.
  • Investigate the relationship between linear and angular dynamics in greater detail.
  • Learn about practical applications of moment of inertia in engineering and physics.
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in advanced dynamics and rotational motion analysis.

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Can moment of inertia be derived using just linear dynamics and calculus. Textbooks usually derive moment of inertia using energy equation and and analogy of 1/2mr^2w^2 with 1/2mv^2. I would like to know if it can be approached in a different manner using just linear dynamics.
 
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Yes, If you consider a mass being accelerated and rotates in a circle.
Then the acceleration is:

[itex]F=ma[/itex]
multiply both sides by r:

[itex]\tau=rma=r^{2}m\alpha[/itex]
where [itex]\alpha[/itex] is the angular acceleration.
Take this sum of all masses:

[itex]\sum r^{2}dm[/itex]

Or another way:

The force on a small element dm is:
[itex]dF=r\frac{d\omega}{dt}dm[/itex]
then the torque on this small mass dm is:
[itex]d\tau= rdF=r^{2}\frac{d\omega}{dt}dm[/itex]
integrating this over the total mass gives the total torque:
[itex]\tau=\int r^{2}dm\frac{d\omega}{dt}[/itex]

Hope it helps
 

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