568. Let a, b, be the lengths of the arms of the lever ; A, B their weights, and a = the 2 between them. Then since when in equilibrio, the vertical line passing through the point of suspension also passes through their centre of gravity, we easily get the distance (k) of this centre of gravity from the axis of suspension, viz. a sin. ß xv{1 + (aA + bB cos. a)"} 26B sin. where ß denotes the angle contained by a, and the line joining the extremities of the lever. Again, the mass is M = A + B. Again, if x denote the distance of any point in the arm a from the point of suspension, then the momentum of inertia of that point is dx x x sin.' ? bB cos. a + aA bB sin. a Hence the whole momentum of inertia of « is and that of the arm a is sin. d. 3 sin." (« + ). is the momentum of inertia of the other arm. Hence the distance of the centre of oscillation from the point of suspension is (Venturoli). AS B: sin. + sin." (+ a) S 3 3 Mk a sin. B (A + B) XN {1+(aA+bB cos.c)"} 26B sin.a the length of the simple pendulum. But the time of an oscillation in the simple pendulum whose length is 1, (see Venturoli) is 569. Let w be the given weight of the rod, a its length, and T the given time of oscillation; then the distance of the centre of oscillation from that of suspension is being the distance of the centre of gravity of the rod from the point of suspension. Hence ce a lx = 12 570. Let w be the given weight of the cylinder, r the radius of its base, and b its length; also let w' be the weight of the chain, l its whole length, and a that of the part thereof un wound at the commencement of the motion; then since the inertia of the cylinder is 1 wr, wr2 that of the chain wr2 and the moving force after the space x has been described wa w'w wa + 2 therefore the accelerating force is gw'(a + x) g2w'(a + x) 2w + w) 2 But if v be the velocity acquired vdu = Fdx 2w'g x{(a + x) – ao} l(2w'+w) which gives the velocity for any descent x. Hence dix 1x (2w'+w) dx Х ♡ (2ar - 2) 1.2w' +w atx+(2ax +- x*) hyp. log. 2gw' Let x=1- a; then the time of unwinding the length la is 11.2w + w 1+(1 a") hyp. log. 2gw' a the time required. 571. If v, udenote the velocities of the oscillating and revolving bodies at the distance x from the lowest point of the circle, then we have o= 2g. (h-x), v = v 29.(h' — «). Again, let t, t' denote the times in which the arc 2ml sured from the highest point, s being that part of the semicircle which corresponds to the abscissa x; then -ds -ldx s mea dt = C x{1+ + 2.4 and dť = -ldx v 2g.(h'-x). (21x — 2") which being integrated between = h and o x = 2land o will give t and t. 2 dx Х 21 1.3 + + &c.} 21 2.4 (21) 1.3 x2 and dt = X + 2.4 - 2ndx Х 12 rdx X h 2m - 1dx ♡ 3o) (hx - x) and if we write H, H.-1, &c. for the integrals of 20 d.c &c. (hx - x2) 7(hx - x*)' when x = h, we get H. = 2m-1 1h. H 2 (hx m m m &c. = &c. 23 h h 1.3.... (2m-3) (2m - 1) 1.2.3......m 쯤 (n). Hence the time of the semi-oscillation is 2 h X* X{1+ + 4g 21 Similarly it may be shewn that 2 21 1.3 + 2.4 m* To find the velocity V and periodic time T in the circle, not considering the accelerations due to gravity, we have V = v 29.(h' – 21) 4ť -21 gl X(21—h) h 271 Hence T = V W g.(21-h) Hence moreover since the periodic time in a circle is generally (440.) R F R being the radius and F the centripetal force, the centripetal force in this case, or the tension of the string is, 49 (21—"); h 27 572. Let A, B, be the weights of bodies, B being that of the one projected; also let a be the distance of B from the ring before projection, and its distance after it has been in motion during the time t, and make A = the angle between ę and a, and let B be the velocity of projection. Then, if u denote the velocity |