# Masteringphysics question?

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A CD-ROM drive in a computer spins the 12-cm-diameter disks at 1.02×104rpm .

T = 5.9×10−3 s

f = 170 rev/s

v= 64m/s

What is the acceleration in units of g that this speck of dust experiences?

I have no idea what this question wants me to do. Isn’t the velocity constant so there is no acceleration? Please explain.

• a=v^2/r .[centrifugal acceleration]

a=64^2/0.06=68266,67 m/s^2

a=6966g

• There is centripetal acceleration at constant rotational speeds.

Acceleration = (v^2/r).

You need to work out the radius of the speck of dust’s path. The disc is doing 170 rps., and has a tangential V of 64m/sec.

(64/170) = 0.376470588 metre per 1 revolution.

Divide by pi, then halve the result for radius, = 0.059917155 metre. Looks like the dust speck is on the rim! Maybe you were told that?

Acceleration = (64^2/0.06) = 68,266.67m/sec^2. Divide by 9.8, = 6,966g.

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