CHAPTER VII.

MOTION OF RIGID BODIES ABOUT FIXED AXES.

SECT. 1. Various Problems.

LET F denote the resolved part of any one of a system of forces acting on a rigid body, at right angles to a fixed axis, r being the perpendicular distance between the fixed axis and the direction of F. Then Fr will be the moment of this force about the axis, and, if Σ (Fr) denote the sum of the moments of all the forces affected by their appropriate signs, we shall have, for the determination of the motion of the body, the general formula

where Mk2

=

the angular velocity of the body after a time t, and its moment of inertia about the fixed axis.

(1) A straight uniform rod, moveable about its upper end, hangs vertically: to find the least angular velocity with which it must begin to move in order that it may perform complete revolutions in a vertical plane.

=

Let OA (fig. 182) be the rod in any position; let its inclination to the vertical line Ox at any time t. Let G be the centre of gravity: draw GH at right angles to Ox. Let ОА = a, m = the mass of the rod. Then, for the motion,

let the initial angular velocity of the rod: then

(2) A straight rod AB (fig. 183) is freely moveable about its lower end A, which is fixed, while the other end B is suspended by a fine string BC attached to a fixed point C: when the system is slightly displaced from its position of equilibrium, so as to keep the string at full stretch, to find the time of a small oscillation.

Let AB= a; join CA; let a = the inclination of CA to the horizon, <BAC=e, 0 = the inclination of the plane BAC to the vertical plane through AC, mk the moment of inertia of AB about AC.

=

The component of the weight mg of the rod at right angles to AC is mg cos a, and the arm of the moment of this component about AC is a sin e. sin : hence, for the motion,

Hence the time of an oscillation is equal to T

2a sin e

3g cos

(3) Supposing the force which acts on the crank of a steamengine to be vertical, and to vary as the sine of the angle through which the crank has revolved at any time from a ver

tical position; to find the angular velocity of the crank in any position, the moment of the resistance being always equal to half the greatest moment of the force, and the moment of the weight of the crank being regarded as inconsiderable.

Let AO (fig. 184) be the crank, O being the fixed extremity; draw Ox vertical; let ‹ AOx = 0 at any time t; F= the force acting at the extremity A; OA=a: assume Fμ sin 0; let mk2 denote the moment of inertia of the crank about 0.

=

Then, the moment of the resistance about O being pa, we have, for the motion of the crank,

ω

let o denote the angular velocity of the crank when 0=0; then

which gives the angular velocity of the crank in any position: from this result we see that the angular velocity is always w when the crank is in either a horizontal or a vertical position.

(4) The extremities of a uniform rod, moveable about its middle point, are connected with a fixed point by elastic strings, the natural length of each string being equal to the distance of the fixed point from the middle point of the rod to find the period of the rod's oscillations, when it has been slightly displaced from its position of equilibrium in the plane through the rod and the fixed point.

Let b be the distance of the fixed point from the end of the rod in the position of equilibrium, c the distance between the fixed point and the middle point of the rod, m the mass of the

rod, and λ the modulus of elasticity of either string: then the

time of oscillation is equal to

πb (mb) +

SECT. 2. Uniform Revolution.

(1) An isosceles right-angled triangle ABC (fig. 185) is suspended at the right angle A, and its side AB is kept in a vertical position by a ring at B: an angular velocity w being communicated to the triangle round AB, to determine the magnitude of w in order that there may be no pressure at B.

Bisect BC in L, join AL, and take AG=AL; then G will be the centre of gravity of the triangle; draw GH at right angles to AC. Take P any point in the area of the triangle, and draw PM at right angles to AB. Let AM=x, PM= [= y, AC=a= AB; m= the mass of a unit of area of the triangle. Then

π

π

2

=

=

AH= AG cos - 3AL cos = ŝa (cos 7) – ja.

4

=

4

4

Also the area of the triangle is equal to a2, and therefore its mass to ma: hence the moment of the triangle about an axis through A at right angles to its plane at any instant, in consequence of gravity, is

ma3g. }a = {ma3g.

Again, the moment about the same axis due to centrifugal force is equal to

Now, since there is no pressure on the ring at B, the moments of gravity and of centrifugal force about the axis through A must be equal; hence we have

=

=

(2) A string lying in the form of a circle on a smooth table is revolving like a wheel: to find the tension of the string. Let m the mass of a unit of length of the string, mds the mass of the element Pp (fig. 186): the moving force on the element due to rotation is equal to mds.wr, w being the angular velocity and r the radius.

L

Let t be the tension at P, the tension at p being accordingly t + dt. Resolving tangentially we have, ‹ POp being denoted by 0, for the equilibrium of Pp,

To find this constant value we have, resolving normally, mds. w'r = (t + dt) sin 0 = t0, in the limit:

whence

t = mr2w2,

or the tension varies as the square of the angular velocity.

(3) Two equal uniform rods AB, AC (fig. 187) are connected at one extremity A by a fixed hinge, the other extremities being connected by a fine string BC: they are whirled round with a given angular velocity, so that the axis of the isosceles triangle formed by the string and rods is always vertical: to find the tension of the string.

Let AB=2a, T=the tension of the string, W = the weight of either rod, ∞ = the angular velocity about the vertical axis AE of the triangle, BAE= a. Take Pany point in AB; let

AP = r.

w

Then, taking moments of the forces acting upon AB, about the point A, we have

= Wa sin a +

Wo2 sin a cos a 8

as