The position of a point moving on a
circular path is given by the equation
s= 2t^3 + 6t^2
where s and t are measured in feet and seconds,
respectively. The magnitude of the acceleration
of the point is 60 fps2 when t = 2 sec.
Determine the radius of the circle.
The speed of the point is
V = ds/dt = 6 t^2 + 12 t
= 48 ft/s when t = 2
At t = 2, there is a tangential acceleration of d^2s/dt^2 = 12 t + 12 = 36 ft/s^2
If the magnitude of acceleration is 60 ft/s^2 at that time, the centripetal acceleration at that time must be
sqrt(60^2 - 36^2) = 48 ft/s^2
Since the centripetal acceleration is V^2/R,
48 = (48)^2/R
R = 48 feet