R is determined by the supposition of inelasticity, i. e. that an inelastic body would after the impact have no velocity perpendicular to the plane; These determine the direction of motion and the velocity after impact. It appears from them that the ball rebounds on the opposite side of the normal to that on which it impinges, the direction of its motion making a greater acute angle with it after the impact than before; its velocity is diminished by the impact. In the case of perfect elasticity, the velocities before and after impact are equal, and in directions equally inclined to the normal on opposite sides of it. APPENDIX. OF THE CYCLOID. 1. IF a circle roll on a straight line, any point in its circumference will trace out a cycloid. It is evident that the form of the curve will be such as that in the figure, P being the point in the generating circle which traces it, and CC' the line on which the circle rolls. The curve may be continued to an unlimited extent, as indicated in the figure, where the beginning of the portion beyond C' is shewn; but we shall only consider the one portion CAC', with which all the others will be identical. If AB bisect CC' at right angles, the curve will be symmetrical with respect to AB, which is called the axis of the cycloid. The point A is called the vertex and C, C' are cusps. 2. In the generating circle, if P be joined with the extremities of the diameter ROH, which passes through the point of contact R of the circle with the straight line CC', the line PH will be perpendicular to PR. And at the instant of time when the circle is in the position HPR, since it rolls along CBC' the point R is at rest, and therefore P is moving at that instant as if describing a circle about R; therefore the motion of P is for the instant perpendicular to PR, and therefore is in the direction PH. Therefore PH is a tangent to the cycloid. 3. The arc of a cycloid measured from the vertex to any point is equal to twice the length of the portion of the tangent at that point intercepted by the generating circle passing through that point. (AP = 2.PH). P' P M N Ρ Take a point P' very near to P. Draw P'p parallel to CB and join Hp. Then ultimately Hp is parallel to the tangent at P', and HP is the tangent at P; therefore if PM be drawn parallel to P'p, the figure PMPP' is ultimately a parallelogram*, and pM= PP'. Also ultimately pH is perpendicular to PR, i. e. the angles PNP, PNM are ultimately right angles. C And the chord PR of the circle makes equal angles with the tangents at its extremities, therefore the angles pPR, CRP are ultimately equal; i. e. pPN, NPM are ultimately equal. Therefore ultimately the triangles pPN, MPN have two angles equal each to each, and the side PN common; therefore ultimately pN= NM, or pM=2.pN. And ultimately HN=HP; therefore ultimately PP' =2.pN=2 (Hp – HP), i. e. the increment of the arc AP is twice the corresponding increment of the chord HP. This arc and chord begin simultaneously from zero; therefore the arc AP twice the chord HP. = 4. If two equal cycloids be placed with their axes parallel and a cusp of the one coincident with the vertex of the other, their concavities being turned towards the same parts; and if a * The chords PP', Pp (which are ultimately coincident with the arcs PP', Pp) have not been drawn, to avoid complicating the figure. MODE OF DESCRIBING A CYCLOID BY A STRING. 97 tangent drawn at any point of the one within the concavity of the other be produced to meet it, it will be a normal to it, and will be equal in length to the arc of the first cycloid from the vertex to the point of contact. Take the generating circle in the position HPR: then PH is a tangent to the cycloid AC. Produce RH to K, make KH=RH, and describe the semicircle KQH, which will = RPH. Produce PH to meet the semicircle KQH in Q, and join QK. Then the angles PHR, KHQ are equal: therefore, since the circles are equal, arc HQ arc PH = =arc HPR-arc PR C K П A B R = BC – RC, by the mode of generation of the cycloid, = AH. Therefore Q is in the cycloid A'A: and HQK is its generating circle; therefore KQ is the tangent at Q, and consequently HQ is the normal. From this it is evident that if a string fastened at C equal in length to CA be unwrapped from the cycloid CA, its extremity will trace out an equal cycloid in the position AA'. A. 5. This may be investigated by means of the differential calculus. Take the vertex of the cycloid for origin, and its axis for the axis of x; and for any point P(x, y) let the angle POH=0, HPR being the generating circle. Let the radius of this circle be a, L. |