A CD-ROM drive in a computer spins the 12-cm-diameter disks at 9400 rpm .
1. What is the disk's period (in s) ?
2. What is the disk's frequency (in rev/s)?
3. What would be the speed of a speck of dust on the outside edge of this disk?
4.. What is the acceleration in units of g that this speck of dust experiences?
+37084 Period is time for one revolution = 60 sec/min / 9400 r/m = 6.383 x 10^(-3) Seconds
f=frequence = RPM = 9400 rpm
Find the circumference
pi x d = circumference = 3.1417 x 12 = 37.699 cm
The speck will travel this far 9400 times/min or a distance of 37.699 x 9400 cm per minute
= 354371.65 cm per minute this is 3543.7165 meters per minute or (divide by 60) = 59.062 m/s
Constant speed circular motion acceleration is given by v^2 / r and is directed toward the center
v^2 / r = (59.062m/s ) ^2 / .06 m (make sure the units are consistent! And r = 1/2 d = .06m)
58138.54 m/sec^2
+37084 Sorry....question asked for rev/sec
revs/sec = revs/min / 60 sec/min = 9400/60 = 156.667 revs/sec
ALSO the acceleration in units of g
58138.54 m/s^2 / 9.8 m/s^2 = 5932.5 g's
