Coriolis 1

Coriolis 1 - The Centrifugal and Coriolis Forces The...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
The Centrifugal and Coriolis Forces The “proof” given in class of the general expression (in two dimensions) is my own, and different from what is in the textbook, or in other books. Here it is: First, look at motion in a straight line as measured in polar coordinates. Suppose that the velocity of a point mass m is ±v = v ˆ i as shown in the diagram, superposed on the drawing of a stationary disk. Imagine that the motion is along a straight line drawn on the disk parallel to the x -axis, not through the centre. r w ¡ ¡ @ @ @ @I φ v ¡ ¡ ¡± @ @ @I ˆ φ ˆ r Then, the velocity of the mass has components v r = v cos φ and v φ = - v sin φ so that ±v = v cos φ ˆ r - v sin φ ˆ φ The angle φ is written instead of the usual θ , because we will need θ for when the disk is rotating. The unit vectors ˆ r and ˆ φ are as indicated. Notice that the acceleration corresponding to this ±v is zero, because ±a = - v sin φ ˙ φ ˆ r + v cos φ ˙ φ ˆ φ - v cos φ ˙ φ ˆ φ + v sin
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 2

Coriolis 1 - The Centrifugal and Coriolis Forces The...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online