(5) To find the radius of gyration of the area of the lemniscate r2 = a2 cos 20 about a tangent at its node.

The radius of gyration is equal to π3α.

(6) To find the moment of inertia of a triangle ABC about an axis drawn, in the plane of the triangle, through the angular point A.

If m be the mass of the triangle, and b, c, the lengths of the perpendiculars from B, C, upon the axis, the required moment of inertia is equal to

fm (b2 + bc+c2).

Messenger of Mathematics, Vol. IV. p. 115.

SECT. 4. A Plane Area about a Perpendicular Axis.

(1) To find the radius of gyration of a triangular lamina ABC, (fig. 177), about an axis through A at right angles to its plane.

Take two points P, p, indefinitely near to each other in the side AB, and draw PM, pm, parallel to BC. Take P', p', in PM, pm, and construct the indefinitely small parallelogram P'p', two of the sides of which are parallel to AC. Let AM=x, PM=y, P'M=y', Am=x+dx, p'm=y' + dy', ▲ ACB= C; let a, b, c, be the three sides of the triangle.

Then, Mk denoting the moment of inertia about 4, we have, 7 denoting the indefinitely small thickness, and p the density of the lamina,

Mk2

=

0

pr

(x2 + y2 — 2xy' cos C') pт sin C' dæ dý

=PT y3-xy

sin C' [* (x2y + }y3 — xy3 cos C) dx

рт

=12. pr ab sin C (632 + 2a2 — 6ab cos C)

=M {6b2+2a2 - 3 (a2 + b2 — c2)},

and therefore

k2 = 11⁄2 (362 + 3c2 — a2).

(2) To find the radius of gyration of a triangular lamina ABC about a perpendicular through its centre of gravity G.

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Let AG, BG, CG, be represented by a, ß, y; and BC, CA, AB, by a, b, c. Then, M denoting the mass of the whole triangle ABC, M will be the mass of each of the triangles BGC, CGA, AGB. Hence, by the preceding problem, the moment of inertia of these three triangles respectively about the axis through & will be

and therefore the moment of inertia of the whole triangle about G will be equal to

3% M {6 (a2 + ß2 + y2) − (a2 + b2 + c2)};

or, by a property of the centre of gravity of a triangle, to 3M {2 (a2 + b2 + c2) − (a2 + b2 + c3)}

= 3 M (a2 + b2+c2).

Hence the square of the radius of gyration will be equal to

3% (a2 + b2 + c3).

Euler; Theoria Motus Corporum Solidorum, cap. VI.

Prob. 32. Cor. 1.

(3) To find the radius of gyration of an elliptic area about a perpendicular axis through its centre.

If M be the mass of the area, the moment of inertia about the two axes of the ellipse will be

But the moment of inertia of a plane area, about any perpendicular axis, is equal to the sum of the moments of inertia about any two lines, at right angles to each other in the plane area, passing through the point at which the axis meets the area. Hence, in the present problem, the moment of inertia about the proposed axis is equal to

† M (a2 + b2),

and the square of the radius of gyration = (a2 +b2).

(4) To find the radius of gyration of a circular annulus about a perpendicular axis through the centre of the circle.

Let r be the distance of any point of the annular area from the centre of the circle, the angular co-ordinate, p the density, and the indefinitely small thickness of the area; then, a, b, being the radii of the two concentric circles,

(5) To determine the form of a uniform plane lamina, and the position of an axis at right angles to it, in order that the moment of inertia of the lamina about the axis may be a minimum, the mass, density, and thickness of the lamina being known.

Let mk denote its moment of inertia, m denoting its mass: then, p being its density, and T its thickness, x and y being polar co-ordinates,

= f(yʻ — ay1) dx,

V=y' - ay3.

ρτ

0

Hence, by the formula of the Calculus of Variations,

which shews that the disk is circular, its centre being in the axis of rotation.

(6) To find the radius of gyration of a parallelogram about an axis perpendicular to it through its centre of gravity.

If 2a, 2b, be the lengths of two adjoining sides of the parallelogram, then, whatever be the angle of their inclination,

k2 = } (a2 + b2).

Euler; Theoria Motus Corp. Solid. Cap. VI. Prob. 35.

(7) To find the radius of gyration of a regular polygon about an axis perpendicular to it through its centre. If n be the number of sides, and c the length of each,

(8) To find the radius of gyration of a portion of a parabola, bounded by a double ordinate to the axis, about a perpendicular line through its vertex.

If x, y, represent the extreme co-ordinates of the portion, k2 = ‡ x2 + }y2.

SECT. 5. Plane Area about an Oblique Axis.

Having given the greatest of the moments of inertia of any plane figure about the three principal axes, which have the same origin, to find the moment of inertia about an axis passing through the same origin and equally inclined to the three principal axes.

Let A, B, C, be the moments of inertia about the three principal axes: one of these will evidently be at right angles to the plane area; we will suppose C to correspond to this. Then, μ being the moment of inertia about the other axis, and a, ß, y, its inclinations to the principal axes,

μ= A cos2 a + B cos2 B + C cos2 y

C being evidently greater than either A or B.

cos2 a + cos2 B+ cos y = 1,

But, since

and a=ẞ=y, we see that cos2 α =

: hence,

μ = 3 C.

W. S.

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