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581. The moving force is

P-W.
Hence the accelerating force is

P-W
F-

P +W
That part of P's weight which is sustained is
P - W

2PW
P
P +W

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
2.4

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'
and if n be the number of degrees in this arc, we get

27in
360°

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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

=

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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

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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

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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+)

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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

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= surface of sphere x

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.. 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""""
L=

Mk+M'k'+M"K" +.xk"
Mka + M'k'a'

Mk + MK by the question. Now

M = id, M' = (1-2), M" E a

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2

d, a'r 3

3
Hence
wd3

(1 d) + all-d)2 + zdo
31 31
L=
W

(1 - d)2 + a(l d) + xd
21

22

+

W

+

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OTHERWISE.
Let L, L' be the lengths of the two pendulums. Then

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3

+

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Again,

d3
(1-0)

+ a(l-d2 + xd
3

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
2 P-31d +3d 2 l. (12 - 31d +3d*) +3al-d)*+ 3dox
3 1-2d

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
Hence

1-d 3(1-d)Q-2P
d

2P-3Qd
al.(1-d)

1-30
X
d

21 - 9ld + 124* as before.

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+k

587. Generally

S S + Mk S
L=
Mk
Mk

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 =
Mk

k
LEV(L? - 4r2)
.. k =

2
which indicates two axes of rotation.

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