3
$\begingroup$

I have read in available sources that product of inertia is just a term that is defined because it is useful in calculating the minimum and maximum moments of inertia of a body and also in finding the moments of inertia about a rotated system of axes with respect to the original system of axes. But does not there stand any physical significance of product of inertia?

$\endgroup$

marked as duplicate by ACuriousMind, Kyle Kanos, Qmechanic Sep 19 '14 at 14:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ Possibly helpful link: physicsforums.com/showthread.php?t=4485 $\endgroup$ – Floris Sep 19 '14 at 12:43
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/60843/2451 , physics.stackexchange.com/q/66350/2451 and links therein. $\endgroup$ – Qmechanic Sep 19 '14 at 12:45
  • $\begingroup$ @Floris It gives an basic idea...Can anybody explain why the factor of stability about that axes is defined by the expression of product of inertia?means why its physical significance is this? $\endgroup$ – Dvij Mankad Sep 19 '14 at 12:48
  • 1
    $\begingroup$ The "physical significance" is "degree of asymmetry" of the body; after that, simple math using the POI allows you to compute torque when you apply rotation along an axis other than a principal axis. That makes it useful, for example, in balancing tires: you compute the POI and then add mass to set the POI to zero along the axis of the wheel - resulting in a smooth ride... Put another way - when POI is non zero, there will be a net torque when you rotate about that axis. A simple diagram should convince you. See one at encyclopedia2.thefreedictionary.com/Product+of+Inertia $\endgroup$ – Floris Sep 19 '14 at 12:49
  • 1
    $\begingroup$ The physical significance of non-zero products of inertia is that the chosen axes are not the principal axes of the body. There might be good, solid engineering reason to use those non-principal axes, but that's a question of engineering rather than physics. $\endgroup$ – David Hammen Sep 19 '14 at 12:56

Browse other questions tagged or ask your own question.