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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)

 Sep 21, 2016
 #1
avatar+129849 
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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

 

 

cool cool cool

 Sep 21, 2016
 #2
avatar+223 
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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.

 Sep 21, 2016

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