This reasoning may be repeated for the other vertices A, B, C, and we see that the mass-centre is the same as that for four equal particles placed at the vertices (§ 74, Example iv.). Hence the mass-centre of any tetrahedron lies on the line joining the vertex to the mass-centre of the base, the way up from the base. 96. This result can be readily extended to the case of Fig. 67. any pyramid. For if V be vertex, ABCD... the polygon forming the base, and if O be the mass-centre of the base, then by joining O to ABC... we divide the base into a number of triangles and by joining VO, we divide the pyramid into a number of tetrahedra. The mass-centre of each of the latter lies on E the line joining V to the mass-centre of its base the way up, hence the masscentre of each and all lies at the same distance from the of base. If we take any section parallel to the base, abcd..., then the polygon abc... is similar to ABC... and therefore VO will cut the area at its mass-centre. Hence the masscentre of the whole must lie in VO the way up from the base. This proof, being independent of the number of sides to the base, is true if we suppose the number of sides to increase indefinitely and their magnitude to decrease indefinitely. But the limit to such a polygon is a curve and the limit of such a pyramid is a cone. Hence the mass-centre of a cone is on the line joining the vertex to the mass-centre of the base of the way up from the base to the vertex. 97. This method of proof can be applied to find the mass-centre of the surface of a pyramid or cone (neglecting base), its position will be one-third up the line joining the mass-centre of the sides of the base to the vertex. 98. Mass-centre of a spherical surface between two parallel planes. Let O be the centre of the sphere, LCD, M the circumscribing cylinder having OT as axis and let the planes bounding the surface be AB and CD, both 1 to OT. Then it is known that the area of the surface ABCD of the sphere is the same as that of the cylinder A1 B1C1D1. The same is true for any very thin slice PQ. The mass of the strip PQ is therefore the same as the mass of P1Q1, and it has the same mass-centre. This is true for all the strips into which the surface may be divided, hence the mass-centre of the spherical surface is the same as that of the corresponding part of the cylinder. Hence for ABCD, the mass-centre is half-way between R and S. A special case of this is the hemispherical surface, its mass-centre would be half-way between 0 and T. If a semicircle be revolved about its base as axis, it will generate a sphere. If a sector of a circle be revolved about one of its sides it will generate a sector of a sphere. Thus if OAB be a sector of a circle (ABA' being a semicircle) and it is revolved about AA', the volume swept out is a sector of the sphere generated by ABA'. A B A' Fig. 68 a. If BC be drawn to OA, then OBC will generate a right circular cone, AB a spherical surface and ACB a spherical cap or segment. A sector of a sphere may thus be considered as made up of a cone and segment. A sector may be divided into a great number of very small cones or pyramids, having a common vertex at the centre, and their bases on the surface. All the mass-centres of these are at the same distance from the centre, viz. 4ths of the radius. Hence we may replace the solid sector by a spherical surface of its radius. The mass-centre of this surface can be found as in § 98. For a hemisphere, the corresponding surface is hemispherical of the radius, the mass-centre of this is half-way up. Hence for a hemisphere, the mass-centre is radius from base along the axis of symmetry. 100. Segment or Cap of a sphere. This can be treated as a positive sector and a negative cone. 101. Determination of Mass-centres by the Projection of Vectors. If we have a number of points and a line all coplanar and we draw through the points lines in the same direction to cut the given line, then the points so determined are called the projections of the given points and the parallel lines are called the projecting rays or lines. Thus in fig. 69, A1, B1, C1, D1, are the projections of ABCD and AĂ ̧, BÂ1, etc. are the projecting rays. If the projecting rays are perpendicular to the given line, the projection is called orthogonal projection. B A A B Fig. 69. If a point moves from A to B, then the projection of the point moves from A, to B1, and the projection of the step AB is A,B1; calling these a and a', the vector a' is the projection of a. 1 Similarly B,C1 = B' is the projection of BC (= ß), and C1D, the projection of CD (=y). Now AD=a+B+y=σ (say) and the projection of this is A1D1 = σ' = a' + B' + y'. Hence the projection of a sum of vectors is the same as the sum of the projections of the vectors. It is evident that the projection of any vector a is independent of the position of a, hence the projections of any number of vectors on a line are the same as on any parallel line. It will be noticed that finding the projection of a vector on a given line is exactly the same as finding the component of the vector in the direction of that line. 102. Since the sum of the projections of a number of vectors is the same as the projection of the sum, it follows at once that if the vectors form a closed polygon, the sum of the projections = 0. Theorem :-If a number of points and their masscentre be projected on a line, then the mass-centre of the projections is the projection of the mass-centre of the points. Proof. Let m1, m2... be the masses, a, ẞ, y... the position vectors of the points relative to the mass-centre, then ma means a vector m, times as long as a, mß a vector m2 times as long as ẞ and since the origin is the mass-centre, then Em,a=0 (i.e. the vectors form a closed polygon). Let a', B', y',... be the projections of a, ß, y, and M' the projection of M, then the mass-centre of the mass-points so determined is given by μm Em,a'. But Σm, a' is the projection of a closed polygon and therefore = 0 ; .'. μ = 0 and M' is the mass-centre. = 103. The coordinates of the mass-centre may be deduced from the last theorem, for calling the lengths of the projecting rays y1, y2, Yз... and the ort in their direction, we have, if y be the length of the projecting ray from mass-centre get Multiply the equations by m1, m,,... and add, and we Similarly by projecting in some other direction we should get Em1x1 = xΣm. |