82. Mass-centre of a circular arc. Let AB be the given circular arc, and suppose that is the centre. Let OC be the radius to AB, then OC is an axis of symmetry and the mass-centre must obviously lie on OC; all we require then is to determine the distance of the mass-centre from 0.

Divide the arc BC into a number of equal parts say n parts, let BE, EF,... be these parts, and draw the chords.

Consider the chord FE, draw ON to it, through F, N and E draw lines to OC, and through E draw EH || to BD.

Denote the length of the chord FE by a.

To find the distance of the mass-centre of all the chords like EF, we suppose each concentrated at its midpoint, and take moments of the masses about OB, (the radius | to DB).

The moment of FE is a. NN1.

Now NON, and FEH are similar triangles, for the sides of the one are to those of the other each to

each.

.'.

ON: NN, FE: HE,

=

or

ON.FE, FE.NN1 = a.NN1 = moment of EF.

=

Now all the chords are equal, ... ON is the same for all.

.. moment of sum = ON.B2O.

However great the number of parts taken this equation will remain true. If n be very great indeed then the chords become indistinguishable from the arcs, and hence, in the limit

But in this case ON becomes the radius of the circle r (say), .. moment of arc BC=r. B10.

If length of arc BC is s, and semi-chord BD is c and y is the distance of mass-centre from B10, then

If the angle BOC = 0, then s = r0, and BD = r sin 0.

83. The following is an easy graphical construction for determining the position of the mass-centre.

From C draw a line || to DB, and with a pair of

compasses (dividers) step off along this line CL equal (approximately) to the arc BC. (The chords must be taken so small that on the drawing it is difficult to distinguish the chord from the arc.) Join OL, draw BN to OC and MN || to BD. Then M is the masscentre.

or

For from figure OM: MN- OC : CL,

=

y.s= c.r.

MASS-CENTRES OF AREAS.

84. If we have a distribution of points with equal masses over an area, such that for any bit of the area, however small, the mass is proportional to the area, then there is a uniform distribution of mass, and the area is said to be evenly covered (or loaded).

The mass-centres of such areas can often be easily found.

85. If the area be such that it has an axis which bisects all lines (bounded by the area) drawn parallel to a fixed direction, the axis is called (in general) an axis of skew symmetry. The fixed direction is called the conjugate

direction. In the special case for which the axis is perpendicular to the conjugate direction, it is simply called an axis of symmetry.

An area having such an axis, its mass-centre must lie on it, for to any point with mass m at a distance x from the axis (measured in the conjugate direction) must correspond a point with mass m at a distance

Emx=0 or the mass-centre is on the axis.

x. Hence

If the area has two such axes of symmetry, the masscentre must lie at their intersection, if more than two they must all pass through a point.

The medians of a triangle are examples of axes of skew symmetry; there are three medians and therefore the mass-centre of a triangle is the same as for three equal mass-points at the vertices and the three medians must intersect in the same point.

86. Mass-centre of a trapezium.

Let ABCD be the trapezium, where AB || CD.

Let AB=2a, and CD=2b. Then if E and F the mid-points of AB and CD, be joined, EF is an axis of skew symmetry and the mass-centre must lie on it.

Draw the diagonal BD, dividing the trapezium into two triangles ABD and BCD. These triangles have a common altitude, viz. the distance p between AB and CD, and

area ABD=pa,

area BCD=pb.

The mass-centre of a triangle being the same as the mass-centre of three equal mass-points at its vertices, we may replace ABD by masses pa at A, B and D, and BCD by masses pb at B, C and D.

Multiply each of these masses by the distance (measured || to EF) from AB, then if M is the mass-centre

To find M graphically:-set off from D in the opposite sense to DC, DQ= 2a, and from B set off in opposite sense to BA, BP = 2b, join PQ, the point where PQ cuts EF is the mass-centre M.

For

EM

==

EP

=

a +26

EF EP + FQ ̄ ̄ 3 (a + b) *

87. Mass-centre of any quadrilateral.

Theorem: If we have any triangle ABC, and produce the base BC both ways equal distances to B, and C1, then the mass-centre of AB,C, is the same as that of ABC.

For since ABB, ABC and ACC, have the same altitude, the mass-centre of each is at the same distance