a CD-ROM drive in a computer spins the 12-cm diameter disks at 8600rpm. Part A W

Question

a CD-ROM drive in a computer spins the 12-cm diameter disks at 8600rpm.
Part A What is a disks period (in s)?
Part B What is a disks frequency in rev/s?
Part C What would be the speed of a speck of dust on the edge of this disk?
Part D What is the acceleration in units of g that this speck of dust experiences? a CD-ROM drive in a computer spins the 12-cm diameter disks at 8600rpm.
Part A What is a disks period (in s)?
Part B What is a disks frequency in rev/s?
Part C What would be the speed of a speck of dust on the edge of this disk?
Part D What is the acceleration in units of g that this speck of dust experiences? a CD-ROM drive in a computer spins the 12-cm diameter disks at 8600rpm.
Part A What is a disks period (in s)?
Part B What is a disks frequency in rev/s?
Part C What would be the speed of a speck of dust on the edge of this disk?
Part D What is the acceleration in units of g that this speck of dust experiences?

Explanation / Answer

given

w = 8600 rpm ( 2 pi/ 60) = 900.13 rad/s

w = 2 pi/T

T = 2 pi/ w = 2 pi/ 900.13 = 0.0069 s

(b)

f = 1/T = 1/ 0.0069 s=143.33 Hz

(c)

v = rw = 0.06 m ( 900.13) = 54 m/s

(D)

a = v^2/ r = ( 54)^2/ 0.06 = 48614.041014 m/s^2 ( g/ 9.8 m/s^2)

in g

a= 4960.61 g

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