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581. The moving force is
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
582. Generally the time of an oscillation through an arc, the versed sine of half of which is (h), is (see 571.)
+ &c.} 9
21 I being the radius or length of the pendulum.
Now A the are being small compared with the whole circumference, we have
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
52520 In the question, we have
1° 13° n = 2°, N = 2° 10' = 2
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
sin 0 X w
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
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 =
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 + MK by the question. Now
M = id, M' = (1-2), M" E a
d, a'r 3
(1 – d) + all-d)2 + zdo
(1 - d)2 + a(l – d) + xd
+ 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
21 - 9ld + 124* as before.
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