A student attempts to spin a rubber stopper(m=0.5kg) in a horizontal circle with a radius of 0.75m.If the stopper completes 2.5 revolutions every second,determine the following; a)The centripetal acceleration. b)The centripetal force.
Question
Explanation:
a) ac = r(omega)^2
where omega = 2.5 rev/s
r = 0.75 m
We need to convert rev/s first into rad/s:
2.5 rev/s × (2pi rad/rev) = 5pi rad/s = omega
Therefore, the centripetal acceleration ac is
ac = (0.75 m)×(5pi rad/s)^2
= 185 m/s^2
b) Fc = mac
= (0.5 kg)(185 m/s^2)
= 92.5 N