Question

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.​

Answer 1

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