By Malcolm S. Longair

This moment version of a well-liked textual content demonstrates how physics could be favored via a chain of case reviews spanning an entire collage physics path. Highlights contain 3 new chapters on Newton's legislation (one on old astronomy as much as the time of Galileo, one on Galileo, and one on Newton); a brand new bankruptcy on dimensional equipment, chaos and self-organized criticality; and a brand new bankruptcy at the expertise of cosmology. even supposing the entire arguments are offered accurately in actual and mathematical terminology, the publication is written in a comparatively non-technical demeanour, meant to exhibit a deep figuring out and appreciation of the excellent achievements of theorists in growing the constitution of recent physics. First version Hb (1984): 0-521-25550-3 First variation Pb (1984): 0-521-27553-9

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The potential at the surface of the rotating sphere is constant by definition: V (r) = − γ mM + V0 . R According to the formulation of the problem, the earth’s surface is an equipotential surface. From this it follows that the attractive force acts normal to this surface. Because of the constancy of the potential along the surface, no tangential force can arise. V (r) = − ! =− γ mM r 1− R R − r m 2 2 ω R 1+2 2 R sin2 ϑ γ mM + V0 . R From this it follows that V0 = γ mM R2 r− m 2 2 2 ω R sin ϑ − mω2 R r sin2 ϑ.

11, 1868. In 1851, Foucault performed his famous pendulum experiment in the Panthéon in Paris as a proof of the earth’s rotation. In the same year he proved by means of a rotating mirror that light propagates in water more slowly than in air, which was important for confirming the wave theory of light. F. L. Fizeau. W. 1007/978-3-642-03434-3_3, © Springer-Verlag Berlin Heidelberg 2010 23 24 3 Foucault’s Pendulum the linear forces and the centripetal forces can be neglected, because for the earth’s rotation dω/dt = 0 and t · |ω| 1, t 2 ω2 ≈ 0 (t ≈ pendulum period).

Solution. Three forces act on the body, viz. (1) the gravitation (weight): Fg = −mge3 ; (2) the centrifugal force: Fz = −mω × (ω × r); (3) the chain tension force: T = −T sin ϕe1 + T cos ϕe3 . Fig. 5. e1 , e2 , e3 are the unit vectors of a rectangular coordinate system rotating with the bar; T is the chain tension force; Fg is the weight of the mass m; Fz is the centrifugal force Since the angular velocity has only one component in the e3 -direction, ω = ωe3 , and r = l(sin ϕe1 + (1 − cos ϕ)e3 ), we find for the centrifugal force Fz = −m(ω × (ω × r)) the expression Fz = +mω2 l sin ϕ e1 .