Proving $\tau = I\alpha$ for Continous Mass Distribution

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SUMMARY

The proof of the equation τ = Iα for continuous mass distribution involves transitioning from a discrete mass distribution to an integral representation. The net external torque (τ) is related to the moment of inertia (I) and angular acceleration (α). The integral formulation arises from the Riemann sum, which converges to the Riemann integral as the mesh value decreases. This approach effectively incorporates the continuous nature of mass distribution, validating the relationship in a broader context.

PREREQUISITES
  • Understanding of torque (τ) and its physical significance.
  • Familiarity with moment of inertia (I) and its calculation for continuous bodies.
  • Knowledge of angular acceleration (α) and its relation to rotational motion.
  • Basic concepts of calculus, specifically Riemann sums and integrals.
NEXT STEPS
  • Study the derivation of moment of inertia for various shapes in continuous mass distributions.
  • Explore the application of Riemann sums in physics and engineering contexts.
  • Investigate the relationship between torque and angular momentum in rotational dynamics.
  • Learn about advanced calculus techniques for evaluating integrals in physics problems.
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in the principles of rotational dynamics and continuous mass distributions.

pardesi
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how does one prove [tex]\tau=I\alpha[/tex] for continious mass distribution where [tex]\tau[/tex] is the net external torque along the axis of rotation [tex]I[/tex] is the moment of inertia,and [tex]\alpha[/tex] is the angular accelaration ...
i know the proof when the mass distribution is discrete...
 
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look at your discrete version and see how you can turn the sum into an integral ... eventually, i think the integral get absorbed in the definition of I (moment of inertia)
 
well taht doesn't happen ...because the discrete version necissates the existence of point like particles...what does ahppen that thsi turns out to be a very good approximation...using the fact that as the mesh value of the riemann sum decrease it converges to the riemann integral; for a closed bounded function
 

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