To find the centre of a sector of a circle. chord of arc X radius of circle length of arc 11 distance from centre of circle, on radius perpendicular to the chord, and bisecting the sector. (chord of arc2+ (radius versed sine - distance from centre of circle)2) distance from both extremities of the arc. Radius distance from centre of circle = distance from vertex. = To find the centre of a parabola. On the abscissa at the length from the middle of the base. (base distance from base2) ities of the curve. = distance from both extrem Abscissa distance from base = distance from vertex. In an Ellipse, the centre is at a right section of the figure, as in the parallelogram. In a Zone, the centre is at a right section of the figure, as in a trapezium. SOLIDS. The centre of gravity of a Cube is its geometrical centre, and the same is true for a Right Prism, Parallelopiped, Cylinder, Sphere, Spheroid, Ellipsoid, and any Spindle. It is also true for the Middle Frustum of any spindle, the Middle Frustum of a spheroid, the two Equal Frustums of a paraboloid, and the two Equal Frustums of a cone. In a Cone and Pyramid, the centre of gravity is in the axis, at the length of the axis from the base. In a Paraboloid, the centre of gravity is in the axis, at the length of the axis from the base. To find the centre of gravity of a frustum of a cone or frustum of a regular pyramid. height (D + d)2 + 2 D2 = distance on the axis from the centre of the less base; D being the diameter of the greater base, and d the diameter of the less, in a frustum of a cone; or, D being a side of the greater base, and d a side of the less, in a frustum of a regular pyramid. To find the centre of gravity of a prismoid. (WA+Na)2+2 A X height = distance on the axis (WA+Na)2 — ✔AX Na from the less base; A being the area of the greater base, and a the area of the less. And this rule is applicable to the frustum of a pyramid, of any number of equal sides, and to the frustum of a cone. To find the centre of gravity of a frustum of a paraboloid. D2 × 2+ď2 D2 + d2 X height = distance on the axis from the less base; D being the diameter of the greater base, and d the diameter of the less. To find the centre of gravity of a spherical segment. 3.1416 height of segment2 × (radius of sphere - height of segment)? solid contents of segment =dis tance from the centre of the sphere, on the axis of the segment; the solid contents and lines being in the same denomination of measure. Radius of sphere gravity of segment segment's vertex. NOTE. (Radius = distance from centre of sphere to centre of distance on the axis of the segment from the Radius of base of segment2 + height of segment2 height of segment X 2 radius of sphere. To find the centre of gravity of a spherical sector. 4 = distance on bisecting radius from centre of sphere. To find the centre of gravity of a system of bodies. ax bax b' + a" × b", &c. a+a+a", &c. centre required; a, a', a", &c., being the solid contents or weights, and b, b', b", &c., the distances of their respective centres of gravity from the given plane. OF CENTRES OF OSCILLATION AND PERCUSSION. The centre of oscillation applies to bodies fixed at one end and vibrating in space; and the centre of percussion applies to bodies revolving around a fixed axis. The two centres, in the same body, are at the same point. The centre of oscillation or percussion is that point in a body, under one or the other of the supposed motions, in which the whole motion, tendency to motion and force in the body, may be supposed collected. It is that point, therefore, in the body under motion, that would strike any obstacle with the greatest effect, or, if met by a staying force, would serve to stop the motion and tendency to motion, of the whole mass, at the same instant. In Pendulums, or rods of uniform diameter and density, having one end fixed, and a ball or weight appended to the other: Weight of rod X length of rod = momentum of rod. Weight of ball x (length of rod + radius of ball) = momentum of ball. 3 force of rod. Weight of ball X (length of rod + radius of ball)2= force of ball. Force of rod force of ball momentum of rod +- momentum of ball distance from the point of suspension or axis of motion, to the centre of oscillation or percussion. And in a rod of uniform diameter and density, having one end fixed, and two or more balls or weights attached : Weight of 1st ball X distance from axis of motion momentum of 1st ball. By another Rule. — Suspend the body freely by a fixed point, and cause it to vibrate in small arcs, and note the number of vibrations it makes per minute. Then - Number of vibrations per minute2: 602: length of pendulum that vibrates seconds in the respective locality: distance from point of suspension to centre of oscillation. EXAMPLE. -The rod is 10 feet in length, and weighs 21 pounds; the weight of a ball, having its centre at the lower extremity of the rod, is 16 pounds, and that of another, affixed to the rod 4 feet from the point of suspension, is 9 pounds; required the centre of oscillation in the system. (21 × 102) + (9 × 42) + (16 × 102) 21 X 5+ (9 × 4) + (16 × 10) point of suspension. Ans. = 8.12 feet from the EXAMPLE.—If, on suspending the body supposed in the last example by its unloaded end, it is found to vibrate 38 times in a minute, what is the distance from the point of suspension to the centre of oscillation? 382 602: 3.25846: 8.12 feet. Ans. In a Right Line, Cylinder or Equilateral Rectangular Prism, one end being in the axis of motion, the centre of oscillation or percussion is distant from that end the length of the line, cylinder or prism. The axis of motion being in the vertex of the figure, and the figure moving flatwise, the centre of oscillation or percussion is distant from the vertex, In an isosceles triangle, its height; In a circle, its radius; In a parabola, its height. But the axis of motion being in the vertex of the figure, and the figure moving sidewise, the centre of oscillation or percussion is distant from the vertex, radius2 X 2 radius 5+ radius ; In a rectangle, suspended by one angle, the diagonal ; In a parabola, suspended by the middle of its base, axis + parameter; radius2 X 2 In a sphere, suspended by a thread, 5X (radius + length of thread) +radius length of thread. EXAMPLE.Required the length of a rod of uniform diameter and density, that, being suspended by one end, will vibrate seconds in the latitude of New York. 2:3::39.10153 : 58.6523 inches. Ans. EXAMPLE.-A ball of 4 inches radius is suspended by a string 24 inches in length; required the centre of oscillation or percussion in the system. 32 = .2286+4+24= 28.2286 inches from 42 × 2 (424) X 5140 the point of suspension. Ans. EXAMPLE.Required the centre of oscillation in a parabola whose height is 10 inches and base 8 inches, supposing the parabola to be vibrating sidewise. 10 X 5 + 42 10 X 3 = 7.676 inches from vertex. Ans. The centre of oscillation in a sphere suspended by a point in its circumference being given, to find thè radius of the sphere. h 2h = radius; h being the distance from the point of suspension to centre of oscillation. EXAMPLE.Required the radius of a sphere, that, being suspended by a point in its surface, will vibrate seconds at 23° of latitude. 39.01206 39.01206 X 2 = 27.865 inches. Ans. The radius of the ball or bob of a pendulum being given, to find the length of the rod or string by which that ball must be suspended, in order that the pendulum may vibrate seconds at a given locality. P r2 X 2 PX )+r= PX5+r=1; P being the length of a pendulum that vibrates seconds in the respective locality, r the radius of the ball, and 7 the length of the string or rod; the last supposed to be without weight. EXAMPLE.The radius of a ball being 4 inches, required the length of a thread, (supposed without weight,) whereby to form a pendulum with that ball, that will vibrate seconds in the latitude of New York. 39.10153 42 X 2 39.10153 X 5 +1) 34.937 inches. Ans. |