Acd-rom drive in a computer spins the 12-cm-diameter disks at 9500 rpm. find acceleration in units of g that a speck of dust on the outside edge of the disk experiences

The net force on an object is:
F = ma
The only force acting on the speck of dust as it is lays on the CD is centripetal force given by:
F = (mv²)/r
Equating the two
ma = (mv²)/r
We get:
a = v²/r
v is the linear velocity in this case; however, we can calculate only the angular velocity with the given data. Therefore, we must use:
v = ωr; where
ω = 2πf; f is the rotations per second
f = 9500 / 60
f = 950/6
ω = 2π(950/6)
ω = (950π)/3
v = (950π)/3 × 0.12
v = 38π
a = (38π)²/0.12
a = 1.2 × 10⁵ m/s²
To express this in terms of gravitational acceleration, we divide by the value of gravitational acceleration, 9.81
a = 1.2 × 10⁴ g