As a first example, suppose we have a flywheel free to rotate about a horizontal axis, and that a weight m hangs by a vertical string from the circumferences of an axle of radius b (fig. 72).
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| Fig. 72. | Fig. 73. |
Neglecting frictional resistance we have, if R be the tension of the string,
whence
| bω. = | mb2 | g. |
| 1 + mb2 |
This gives the acceleration of m as modified by the inertia of the wheel.
A “compound pendulum” is a body of any form which is free to rotate about a fixed horizontal axis, the only extraneous force (other than the pressures of the axis) being that of gravity. If M be the total mass, k the radius of gyration (§ 11) about the axis, we have
| d | ( Mk2 | dθ | ) = −Mgh sin θ, |
| dt | dt |
where θ is the angle which the plane containing the axis and the centre of gravity G makes with the vertical, and h is the distance of G from the axis. This coincides with the equation of motion of a simple pendulum [§ 13 (15)] of length l, provided l = k2/h. The plane of the diagram (fig. 73) is supposed to be a plane through G perpendicular to the axis, which it meets in O. If we produce OG to P, making OP = l, the point P is called the centre of oscillation; the bob of a simple pendulum of length OP suspended from O will keep step with the motion of P, if properly started. If κ be the radius of gyration about a parallel axis through G, we have k2 = κ2 + h2 by § 11 (16), and therefore l = h + κ2/h, whence
This shows that if the body were swung from a parallel axis through P the new centre of oscillation would be at O. For different parallel axes, the period of a small oscillation varies as √l, or √(GO + OP); this is least, subject to the condition (4), when GO = GP = κ. The reciprocal relation between the centres of suspension and oscillation is the basis of Kater’s method of determining g experimentally. A pendulum is constructed with two parallel knife-edges as nearly as possible in the same plane with G, the position of one of them being adjustable. If it could be arranged that the period of a small oscillation should be exactly the same about either edge, the two knife-edges would in general occupy the positions of conjugate centres of suspension and oscillation; and the distances between them would be the length l of the equivalent simple pendulum. For if h1 + κ2/h1 = h2 + κ2/h2, then unless h1 = h2, we must have κ2 = h1h2, l = h1 + h2. Exact equality of the two observed periods (τ1, τ2, say) cannot of course be secured in practice, and a modification is necessary. If we write l1 = h1 + κ2/h1, l2 = h2 + κ2/h2, we find, on elimination of κ,
| 12 | l1 + l2 | + 12 | l1 − l2 | = 1, |
| h1 + h2 | h1 − h2 |
whence
| 4π2 | = | 12 (τ12 + τ22) | + | 12 (τ12 − τ22) | . |
| g | h1 + h2 | h1 − h2 |
The distance h1 + h2, which occurs in the first term on the right hand
can be measured directly. For the second term we require the values
of h1, h2 separately, but if τ1, τ2 are nearly equal whilst h1, h2 are
distinctly unequal this term will be relatively small, so that an
approximate knowledge of h1, h2 is sufficient.
As a final example we may note the arrangement, often employed in physical measurements, where a body performs small oscillations about a vertical axis through its mass-centre G, under the influence of a couple whose moment varies as the angle of rotation from the equilibrium position. The equation of motion is of the type
and the period is therefore τ = 2π√(I/K). If by the attachment of another body of known moment of inertia I′, the period is altered from τ to τ′, we have τ′ = 2π√{ (I + I′)/K }. We are thus enabled to determine both I and K, viz.
The couple may be due to the earth’s magnetism, or to the torsion of a suspending wire, or to a “bifilar” suspension. In the latter case, the body hangs by two vertical threads of equal length l in a plane through G. The motion being assumed to be small, the tensions of the two strings may be taken to have their statical values Mgb/(a + b), Mga/(a + b), where a, b are the distances of G from the two threads. When the body is twisted through an angle θ the threads make angles aθ/l, bθ/l with the vertical, and the moment of the tensions about the vertical through G is accordingly −Kθ, where K = M gab/l.
For the determination of the motion it has only been necessary to use one of the dynamical equations. The remaining equations serve to determine the reactions of the rotating body on its bearings. Suppose, for example, that there are no extraneous forces. Take rectangular axes, of which Oz coincides with the axis of rotation. The angular velocity being constant, the effective force on a particle m at a distance r from Oz is mω2r towards this axis, and its components are accordingly −ω2mx, −ω2my, O. Since the reactions on the bearings must be statically equivalent to the whole system of effective forces, they will reduce to a force (X Y Z) at O and a couple (L M N) given by
L = ω2Σ(myz), M = −ω2Σ(mzx), N = 0,
where x̄, ȳ refer to the mass-centre G. The reactions do not therefore
reduce to a single force at O unless Σ(myz) = 0, Σ(msx) = 0,
i.e. unless the axis of rotation be a principal axis of inertia
(§ 11) at O. In order that the force may vanish we must also
have x̄, ȳ = 0, i.e. the mass-centre must lie in the axis of rotation.
These considerations are important in the “balancing” of
machinery. We note further that if a body be free to turn
about a fixed point O, there are three mutually perpendicular
lines through this point about which it can rotate steadily,
without further constraint. The theory of principal or “permanent”
axes was first investigated from this point of view
by J. A. Segner (1755). The origin of the name “deviation
moment” sometimes applied to a product of inertia is also
now apparent.
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| Fig. 74. |
Proceeding to the general motion of a rigid body in two dimensions we may take as the three co-ordinates of the body the rectangular Cartesian co-ordinates x, y of the mass-centre G and the angle θ through which the body has turned from some standard position. The components of linear momentum are then Mẋ, Mẏ, and the angular momentum relative to G as base is Iθ., where M is the mass and I the moment of inertia about G. If the extraneous forces be reduced to a force (X, Y) at G and a couple N, we have
If the extraneous forces have zero moment about G the angular velocity θ. is constant. Thus a circular disk projected under gravity in a vertical plane spins with constant angular velocity, whilst its centre describes a parabola.
We may apply the equations (9) to the case of a solid of revolution rolling with its axis horizontal on a plane of inclination α. If the axis of x be taken parallel to the slope of the plane, with x increasing downwards, we have
where κ is the radius of gyration about the axis of symmetry, a is the constant distance of G from the plane, and R, F are the normal and tangential components of the reaction of the plane, as shown in fig. 74. We have also the kinematical relation ẋ = aθ.. Hence
| ẍ = | a2 | g sin α, R = Mg cos α, F = | κ2 | Mg sin α. |
| κ2 + a2 | κ2 + a2 |
The acceleration of G is therefore less than in the case of frictionless
sliding in the ratio a2/(κ2 + a2). For a homogeneous sphere this
ratio is 57, for a uniform circular cylinder or disk 23, for a circular
hoop or a thin cylindrical shell 12.
The equation of energy for a rigid body has already been stated (in effect) as a corollary from fundamental assumptions.
