Posted by **katy** on Monday, November 8, 2010 at 10:32pm.

a wheel of radius R rotates making f revolutions per second. the quantity f is know as the frecuency of the motion. show tha 1/f is the time to complete one revolution, called the period T of the motion. show that the centripetal acceleration of a point at the rim of the wheel is a=4(3.14)^2Rf^2

3.14=pi

- physics help!! -
**bobpursley**, Monday, November 8, 2010 at 10:45pm
The distance around the wheel is 2PI*r

that is the distance it goes on one rev.

so on one rev, it goes 2PI*r, now in one second, it does this f times,

so the time for one revolution then is 1/f.

centripetal acceleration= v^2/r

but v= 2PI*r/period=2PI*r*f

v^2=4PI^2 r^2 f^2

so v^2/r= 4PI^2 f^2 r

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