Re: acceleration

[G.Lambert@unsw.edu.au (Geoff Lambert)]

On Mon, 8 May 2000 15:18:13 +1000, "geoff dawson"
<geoffrey.dawson@aph.gov.au> wrote:

>'Acceleration' can mean not only 'getting faster while travelling in a
>straight line' (what we think of mostly) but also 'moving at constant speed
>on a curve'.
>Does anyone know a formula to relate the radius of a curve, the speed of the
>train and the acceleration?
>Ie a train travelling at a constant speed x m/sec around curve radius y
>metres experiences an acceleration of how many m/sec/sec?
>Thank you.

Acceleration = r * omega^2

Where r = radius
omega = the rotational speed.

To make things come out correctly, omega should be expressed in
radians per sec. If r is expressed in metres, then the reultant
acceleration is in m/sec/sec.

If you have the linear velocity (v) and the radius of the curve (r),
the angular speed in radians/sec is  v/r. Therefore the acceleration
is rv^2/r^ = v^2/r.

Thus, for a velocity of 20 metres/sec on a curve of 400 metres radius
(about 20 chains):

a = 20*20/400 = 1 m/s/s (about one tenth g)

GL