be the axis of the cycloid, then the velocity g.x* y = x tan. 450 2lg cos. 450 1 the horizontal range is x = l. Hence the time of flight is 21 lg cos. 45° g Again, the time of an oscillation is t, 425. Since, generally, the time of an oscillation is 426. Let l be the length of the pendulum ; then 21 = length of the whole cycloid, and, since the time of an oscillation is 1 and that down 2l is we have 427. Let the body descend from rest through the arc FA (Fig. 90); then the greatest velocity will be at the lowest point A; and, since s oc = v*, if BR = BA, the velocity of a body de a scending from rest through BA at R will = velocity at A. Hence, through R drawing RM 1 AB, and meeting the curve in M, the point M is determined in which the velocity of the body descending down the cycloid = the greatest velocity. 2 428. Since generally, vo = 2gs, the velocity due to 2r (r being the radius of the generating of the cycloid), is v 4gr = Vgr. Now the velocity V of the moving point in the cycloid : velocity of the moving point in the circle :: ds : dz, s and z denoting the cycloidal and circular arcs respectively. But 2 chord of supp 12 S4r of z z 4 cos. x being the abscissa corresponding to s, measured from the vertex along the axis of the cycloid Hence V = ✓ 29 (2r velocity due to 2r - x = that of a body oscillating in the cycloid. (Venturoli, p. 102.) 429, In the generating O AB (Fig. 90,) inflect AP= its radius, and draw R'PM' I AB; then since AM' = 2AP= AB AF, AF is bisected in M'. Now (Translat. of Venturoli, p. 103.) the time through any arc of the cycloid, whose abscissa, measured from the lowest point is x, is d - d d cos. -1 But when z= 0, the time (t)) down FA is d X 180° 29 430. In Newton's Construction (Prop. 50, Princep.), we have CA: CO :: CO : CR .. AO : CO :: OR: CO - OR. But in the common cycloid, SOQ is a straight line, and :: CO .. CO= CO – OR, and AO = OR; Also BW is parallel to RA, and BV = AO = OR = VW ;. PT = 2PV = PS, &c. &c. 431. Let R be the radius of the base, r that of the wheel; then (by Prop. 49, Princip. Newt.) it easily appears that AP = 4r (R—-). vers. AB R 2 vers. AB 4r (R – 9) to rad. 2 R Also by Prop. 52 of that work, we learn that the time through AL AP o arc vers. AP to rad. AB 4r (R QAB, 2 ( Hence s = 9T= 1000 , the space required. 433. Let l be the length of either pendulum ; then the distance descended vertically from the highest to the lowest points of the O and cycloid, will be and the velocities in the curves (v, d) will be those due to such distances. :: v::: 2gl : 2g :: N 2:1. Now the chord (C) of the quadrant = 1 2 the arc (A) of 1 the cycloid = 2. 2 :: C:A :: N2 : 1 viu' :: C:A 434. The semi-cycloidal arc FA (Fig. 90) is bisected by AB inflecting AP = &c. &c. (4.29.) |