(2)
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 |
(3)
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
GO · GP = κ2.
(4)
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,
| 1⁄2 | l1 + l2 | + 1⁄2 | l1 − l2 | = 1, |
| h1 + h2 | h1 − h2 |