Formula for centripetal acceleration and force
Consider an object of mass m moving with constant angular velocity (w) and constant speed (v) in a circle of radius r with centre O (see Figure
1).
It moves from P to Q in a time t.
Change in velocity parallel to PO = vsinq - 0
Change in velocity perpendicular to PO = vcosq - v
When q becomes small (that is when
Q is very close to P) sinq is close to q in radians and cosq tends to 1.
The equations then become:
Change in velocity
along PO = vq - 0 = vq
Change in
velocity perpendicular to PO = v - v = 0
Therefore acceleration along PO = vq/t = vw = v2/r = w2r
Centripetal acceleration (a) = v2/r = w2r
Applying Newton's Second Law (F = ma)
gives:
Centripetal force (F) = mv2/r = mw2r
Example problems
1. A space station has a radius of 100 m and is rotated with an angular velocity of 0.3 radians per second.
(i) which side of a "room" at the rim is the floor
(ii) what is the artificial gravity produced at the rim
(i) the floor is the outer rim of the space station
(ii) a = g = p2r = 0.32x100 = 9 ms-2
2. Calculate the rate of rotation for a space station of radius 65 m so that astronauts at the outer edge experience artificial gravity equal to 9.8 ms2.
But T = 2pr/v and so T = 16.18 s giving the rotation rate (1/T) as 0.062 Hz.
