Finding the orbits from the focus given

PROPOSITION 50 PROBLEM 33

Cause a pendulous body to oscillate in a given cycloid

Let:

• Globe QVS have a center C
• cycloid QRS, bisected in R, and meeting the superficies of the and S on either globe with its extreme points Q Let there be drawn CR birxcting the arc in such in O, and let it be produced to to CR. as to sort that may be hand. QS CO CA About the centre C, with the A CO interval CA, let and there be described an exterior globe DAF is diameter whose wheel a this within globe, by ; AO, let there be described two semi-cycloids AQ,, and meeting the exterior globe and S, AS, touching the interior globe in Q, thread a with From that point A, in A. APT AR, let the body T depend, and oscillate in in length equal to the line such manner between the twoSlCC. OF NATURAL PHILOSOPHY. X.J 187 that, as often as the pendulum parts from the per the may be applied to that pendicular AR, upper part of the thread the motion and fold itself round which APS towards tends, semi-cycloid that curve line, as if it were some solid obstacle, the remaining part of the semi-cycloids AQ, AS, AP same thread PT Then straight. which has not yet touched the semi-cycloid continuing will the T weight oscillate in the given cycloid QRS. Q.E.F. PT For QRS QOS PT in T, and the circle let the thread meet the cycloid and to the rectilinear part of the thread be drawn V, and let let there be erected the perpendiculars from the extreme points P and in B and W. It is evident, from the the line W, meeting BP, right m 0V j T T CV construction and generation of the similar figures AS, SR, that those per pendiculars PB, TVV, cut off from CV the lengths VB, VVV equal the diameters of the wheels OA, OR. Therefore is to VP (which is dou TP VBP when ^BV is radius) as BYV to BV, or AO -f-OR to AO, that is (since CA and CO, CO and CR and by division AO and OR are proportional), as CA + CO to CA, or, if BV be bisected in E, ble the sine of the angle ; as 2CE to CB. Therefore (by Cor. 1, Prop. XLIX), the length of the of the cycloid of the cycloid PT is always equal to the arc APT is always equal to the half rectilinear part of the thread PS, and the whole thread that is (by Cor. 2, Prop. XLIX), to the length AR. And there fore contrariwise, if the string remain always equal to the length AR, the APS, T will always move in the given cycloid QRS. Q.E.D. COR. The string AR is equal to the semi-cycloid AS, and therefore has the same ratio to AC the semi-diameter of the exterior globe as the like point semi-cycloid SR has to CO the semi-diameter of the interior globe.