What is the centripetal acceleration of the end of the minute hand if its length is 2 cm?

Turnover period of the minute hand (T):
T = 60sec
Centripetal acceleration a = V ^ 2 / r
V when moving in a circle:
V = 2pi * r * nu
Let’s substitute:
a = 4 * pi * pi * r / T ^ 2 = 0.0002m / s ^ 2

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