A CD-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