On some moments of Planar convex bodies - 3
Copying questions as recorded at Mathoverflow:
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Consider any planar convex region C. A line l may be called an inertia bisector of C if it divides C into 2 pieces each of which has same moment of inertia with respect to l.
Which shape of C causes the envelope of all its inertia bisectors to enclose the maximum fraction of the area of C? Guess: a triangle. But will any triangle do?
If for a certain C, all inertia bisectors are concurrent, is C necessarily centrally symmetric? One would guess it is. Note: As in reference given above, one can ask about lines that cut C into two pieces with their moments of inertia in some specified ratio rather than equal to each other.
What happens if instead of the moment of inertia one tries to make the integral of some other function of the distance x to the line (other than x2) equal between the two pieces?
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The perpendicular axis theorem states that the moment of inertia of a planar lamina about an axis perpendicular to the plane of the lamina is equal to the sum of the moments of inertia of the lamina about the two axes at right angles to each other, in its own plane intersecting each other at the point where the perpendicular axis passes through it.
If we are given the moments of inertia M1 and M2 of a convex lamina C about two lines that are at some specified angle α, what could be said about the moment of inertia about an axis perpendicular to the plane through the point of intersection of the two lines? The MI about this perpendicular axis might lie in some range determined by M1, M2 and α. In particular, one could ask: if M1=M2 and α is specified, which shape of C maximizes (minimizes) the MI of C about the perpendicular axis?
Moving to 3D, if the moments of inertia of a convex body about the 3 axes Mx, My and Mz are given, how closely could one calculate the MI of the body about some other line through the origin?
Note: One can also consider integrals of other functions of the distance x to the line (other than x2).
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Consider any planar convex region C. A line l may be called an inertia bisector of C if it divides C into 2 pieces each of which has same moment of inertia with respect to l.
Which shape of C causes the envelope of all its inertia bisectors to enclose the maximum fraction of the area of C? Guess: a triangle. But will any triangle do?
If for a certain C, all inertia bisectors are concurrent, is C necessarily centrally symmetric? One would guess it is. Note: As in reference given above, one can ask about lines that cut C into two pieces with their moments of inertia in some specified ratio rather than equal to each other.
What happens if instead of the moment of inertia one tries to make the integral of some other function of the distance x to the line (other than x2) equal between the two pieces?
------------------
The perpendicular axis theorem states that the moment of inertia of a planar lamina about an axis perpendicular to the plane of the lamina is equal to the sum of the moments of inertia of the lamina about the two axes at right angles to each other, in its own plane intersecting each other at the point where the perpendicular axis passes through it.
If we are given the moments of inertia M1 and M2 of a convex lamina C about two lines that are at some specified angle α, what could be said about the moment of inertia about an axis perpendicular to the plane through the point of intersection of the two lines? The MI about this perpendicular axis might lie in some range determined by M1, M2 and α. In particular, one could ask: if M1=M2 and α is specified, which shape of C maximizes (minimizes) the MI of C about the perpendicular axis?
Moving to 3D, if the moments of inertia of a convex body about the 3 axes Mx, My and Mz are given, how closely could one calculate the MI of the body about some other line through the origin?
Note: One can also consider integrals of other functions of the distance x to the line (other than x2).
posted by R.Nandakumar @ 10:37 PM
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