| Fig. 67. |
the equation (21) would assume the same form as § 12 (5). The motion along the arc would then be accurately simple-harmonic, and the period 2π √(k/g) would be the same for all amplitudes. Now equation (22) is the intrinsic equation of a cycloid; viz. the curve is that traced by a point on the circumference of a circle of radius 1⁄4k which rolls on the under side of a horizontal straight line. Since the evolute of a cycloid is an equal cycloid the object is attained by means of two metal cheeks, having the form of the evolute near the cusp, on which the string wraps itself alternately as the pendulum swings. The device has long been abandoned, the difficulty being met in other ways, but the problem, originally investigated by C. Huygens, is important in the history of mathematics.
The component accelerations of a point describing a tortuous curve, in the directions of the tangent, the principal normal, and the binormal, respectively, are found as follows. If OV>, OV′> be vectors representing the velocities at two consecutive points P, P′ of the path, the plane VOV′ is ultimately parallel to the osculating plane of the path at P; the resultant acceleration is therefore in the osculating plane. Also, the projections of VV′> on OV and on a perpendicular to OV in the plane VOV′ are δv and vδε, where δε is the angle between the directions of the tangents at P, P′. Since δε = δs/ρ, where δs = PP′ = vδt and ρ is the radius of principal curvature at P, the component accelerations along the tangent and principal normal are dv/dt and vdε/dt, respectively, or vdv/ds and v2/ρ. For example, if a particle moves on a smooth surface, under no forces except the reaction of the surface, v is constant, and the principal normal to the path will coincide with the normal to the surface. Hence the path is a “geodesic” on the surface.
If we resolve along the tangent to the path (whether plane or tortuous), the equation of motion of a particle may be written
| mv | dv | = T, |
| ds |
(23)
where T is the tangential component of the force. Integrating with respect to s we find
1⁄2mv12 − 1⁄2mv02 = ∫s1s0 T ds;
(24)
i.e. the increase of kinetic energy between any two positions is equal to the work done by the forces. The result follows also from the Cartesian equations (2); viz. we have