A CD-ROM drive in a computer spins the 12-cm-diameter disks at 9400 rpm .
1. What would be the speed of a speck of dust on the outside edge of this disk? (Make sure answer is in dimension of velocity)
+129852 A CD-ROM drive in a computer spins the 12-cm-diameter disks at 9400 rpm .
1. What would be the speed of a speck of dust on the outside edge of this disk?
9400rpm / 60sec = 470/3 revs per second
In one revolution, the particle would travel 2pi(12/2) = 12 pi cm
So...the velocity = [ 470/3 ] * [ 12 pi] cm = 1880pi cm / sec
+223 First you want to find the circumference of the drive by using the circumference formula:
2*pi*r = 2 * pi * .06 = 0.38 meters
This will be the distance the speck of dust travels in one rotation.
Now multiply that number by the number of rotations occuring in one minute (9400).
9400 * .38 = 3544 meters/minute
That number represents the distance the speck of dust travels in one minute. Now you can divide that number by the number of seconds in a minute to get the speed in m/s.
3544/60 = 59 m/s
I used rounded numbers so that the page wasn't cluttered, so be sure to use unrounded numbers until the end so you get an accurate answer.
