581. The moving force is P-W. P-W P +W 2PW P +W And it is evident that the same part of W is sustained. Hence the whole pressure on the axis is 4PW P +W XP: + 582. Generally the time of an oscillation through an arc, the versed sine of half of which is (h), is (see 571.) 2 h 1.3 x{1+ + &c.} 9 21 21 I being the radius or length of the pendulum. Now A the are being small compared with the whole circumference, we have h = 2l' 27in Hence, if the pendulum oscillating through n degrees keeps true time, the error arising from its vibrating through N degrees is No -n2 t' x nearly 52520 In the question, we have 1° 13° n = 2°, N = 2° 10' = 2 = W 583. The accelerating force down the plane is sin. 0, and 3 the mass + inertia = w + w, being the inclination 2 of the plane. Hence the moving force is 3 sin 0 X w 2 to counteract which by means P, we must have 2P i. sin 0 = 3w which gives the ratio of the height of the plane to its length. (If P = as in the question, then w 10 584. Let a be the length of the rod, then S being the moment of inertia, M the mass, and k the distance of the centre of gravity from the axis of suspension, the length of the first pendulum (see Venturoli) is a ? 1 a? S L= Mk = + 3 2 9 a 4 (1+) 585. Take any zone contained between two sections indefinitely near to each other, made by planes I axis of rotation, and distant from the centre by x and x + dx; then the momentum of inertia of this zone is = surface of sphere x .. the distance of the centre of gyration from the axis of rotation, ist ✓ ' (Venturoli.) 586. Let x be the weight required, a, k, M; a, k, M'the distances of the centres of oscillation from the axis of rotation, those of the centres of gravity and the masses of the respective parts d, 1 – d. Also, let a“, k”, M", 2", k", è be those of (the weights added to the lower and upper extremities of the rod; then the length of the pendulum is Mka+M'k'X' + M"k"2" + ak"""" Mk+M'k'+M"K" +.xk" Mk + MK by the question. Now M = id, M' = (1-2), M" E a 2 d, a'r 3 3 (1 – d) + all-d)2 + zdo (1 - d)2 + a(l – d) + xd 22 + W + OTHERWISE. 3 + Again, d3 + a(l-d2 + xd 3 L' = (1 + a + x) k But k' x (1 + a + x) = l. - d) + a (l – d) + xd by property of the centre of gravity. :. since by the question L = L', we have 3 1.(1-2d)+2a(l-d) + 2x and putting 1 – 3ld + 3d = P, and 1 – 2d=Q P IP + 3a (1-d): + 3xd: Q IQ + 2a (1 – d) + 2xd 1-d 3(1-d)Q-2P 2P-3Qd 1-30 21 - 9ld + 124* as before. +k 587. Generally S S + Mk S Mk see Venturoli,' p. 120, where S denotes the momentum of inertia referred to the axis passing through the centre of gravity, M the mass, and k the distance of the centre of gravity from the axis of rotation. But S = Mrs Mr? go?+ ko :: L tk = k 2 |