tract S8.

problems.

335

Then ca : cp : : 1 \f\ and the three equations are rf — x,and vv = — 2 gfx, and tv = — x. Hence f — and v-0 =

——; the correct fluent of which gives v = \/( 2 tf X ■ - — )

= pd v'y = pd/-^-, the velocity at the point p; where pdand ce are perpendicular to ca. So that the velocity at anypoint p, is as the perpendicular pd at that point.

When the body arrives at c, then v = \/2gr = ^/(2g.Ac)— 25950 feet or 4‘9148 miles per second, which is the great-est velocity, or that at the centre c.

Again, for the time, t = — = V-rr X ,, . 1 t,> an( l th e

fluents give t == •/— x arc to cosine — = V-- X arc ad.° r 2gr

So that the time of descent to any point p, is as the corre-sponding arc ad.

When p arrives at c, the above becomes t =

X quadrant ae =— V— V510&V—- = 12671 se-

2gr 1 ac *2 g 4

conds = 21' 7"^-, for the time of falling to the centre c.

The time of falling to the centre is the same quantityl'5708y'-^, from whatever point in the radius ac the bodybegins to move. For let n be any given distance from c atwhich the motion commences: then, by correction, v =V\ — (n z — x l )'\- and hence t = V~ X .. , .,thefluents

of which give t — J— x arc to cosine —; which, when

x = 0, gives t — X quadrant = ] , 570Sv''-^; for the

time of descent to the centre c, the same as before.

As an equal force, acting in contrary directions, generatesor destroys an equal quantity of motion, in the same time ;it follows that, after passing the centre, the body will justascend to the opposite surface at b, in the same time inwhich it fell to the centre from a ; then from b it will returnagain in the same manner, through c to a ; and so vibratecontinually between a and b, the velocity being always equalat equal distances from c on both sides; and the whole time