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.
2 Answers
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a=v^2/r .[centrifugal acceleration]
a=64^2/0.06=68266,67 m/s^2
a=6966g
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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.
