Mathematics of Physics and Engineering

58 Kinematics and Dynamics of a Point Mass

dulum starts to swing in the Northern Hemisphere in the meridian plane,
then the Coriolis force will tend to turn the plane of the swing clockwise as
seen from above.
Let us perform a simplified analysis of the motion of the Foucault pen-
dulum in the Northern Hemisphere, away from the North Pole and the
equator. Assume that the pendulum starts to swing in a meridian plane.
Figure 2.1.7 presents a (grossly out-of-scale) illustration, with the Earth's
surface represented by the semi-circle. In reality, the length of the sup-
port and the height of the supporting point 0\ are much smaller than the
radius R of the Earth, so that \00\ « |0.Po| ^ R- Also, the ampli-
tude of the swing is small compared to the length of the support, so that
|OiPo| « \0\PN\ = |OiPs|, and the linear distance \PNPS\ is approxi-
mately equal to the length of the corresponding circular swing arc.

Fig. 2.1.7 Foucault Pendulum

The pendulum is suspended at the point 0\, and the points PN, PS are

the two extreme positions of the weight. The angle 9 is the latitude of the

support point 0\. Denote the distance \PNPS\ by 2r. As seen from the

picture, the point PN is closer to the axis ON of the rotation of the Earth

than the point 0\, and the amount of this difference is \QNPQ\ = rsm9.

Similarly, the point Ps is farther from the axis than 0\ by the same amount.

Since the Earth is rotating around the axis ON with angular speed u, the

points PN, 0\, and Ps will all move in the direction perpendicular to the

meridian plane. The point PN will move slower than Oi, and the point Ps,

faster, causing the plane of the swing to turn. The speed of Ps relative to

0\ and of 0\ relative to PN is ruj sin 9. If we assume that these relative