Coriolis 1

# Coriolis 1 - The Centrifugal and Coriolis Forces The"proof...

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

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 04/29/2008 for the course PHYS 230 taught by Professor Harris during the Fall '07 term at McGill.

### Page1 / 2

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

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

View Full Document
Ask a homework question - tutors are online