Question

A coin completes 18 spins in 12 seconds. The centripetal acceleration of the edge of the coin is 2.2 m/s2. The radius of the coin is?

Answer 1

In order to find the radius of the coin, we need to use:

Ac = V^2 r

In which,  
Ac = acceleration of the coin = 2.2 m/s^2 
V= rotational Speed  =  (18/12) * 2πr
r= Radius

so,
2.2 = 9(π^2)(r^2) / r 
= 9(π^2)r

r = 2.2 / 9(π^2) = 0.02476740044 m . . .or you can rounded it up to 0.025 m

Answer 2

Answer: 24.8 mm

Explanation:

The centripetal acceleration of the edge of the coin is given by:

$a=\omega^2 r$ (1)

where

$\omega$ is the angular frequency

r is the radius of the coin

The frequency of revolution of the coin is equal to the number of spins divided by the time taken:

$f=\frac{18}{12 s}=1.5 Hz$

And the angular frequency is given by

$\omega=2 \pi f=2 \pi (1.5 Hz)=9.42 rad/s$

Since we know the centripetal acceleration, $a=2.2 m/s^2$, we can re-arrange eq.(1) to find the radius of the coin:

$r=\frac{a}{\omega^2}=\frac{2.2 m/s^2}{(9.42 rad/s)^2}=0.0248 m=24.8 mm$