Circular Motion - RPM and Radius question

AI Thread Summary
The discussion focuses on calculating the speed and acceleration of the outer row of data on a CD-ROM spinning at 1200 RPM, located 5.6 cm from the center. To find the speed, the correct approach involves converting RPM to radians per second and then applying the formula v = ωr. The acceleration can be determined using the formula a = V^2/r. There is confusion regarding the calculation of the period (T), with incorrect initial attempts noted. Clear guidance is provided on the proper method for conversion and calculation.
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Homework Statement



A manufacturer of CD-ROM drives claims that the player can spin the disc as frequently as 1200 revolutions per minute.
a. If spinning at this rate, what is the speed of the outer row of data on the disc; this row is located 5.6 cm from the center of the disc?

b. What is the acceleration of the outer row of data?

Homework Equations



f = 1/T
V = (2)(pi)(r)/T
a = V^2/r

The Attempt at a Solution



I know that rpm = T/60s, so I did rpm x 60 = T and I got 72,000 s = T (72,000 s for 1 rotation?) That is wrong, I'm not sure how to solve for T...
 
Last edited:
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Hi PhysicsNoob76! :smile:
PhysicsNoob76 said:
I know that rpm = T/60s, so I did rpm x 60 = T and I got 72,000 s = T (72,000 s for 1 rotation?) That is wrong, I'm not sure how to solve for T...

if you must do it this way, use 1200 rev per minute = 1 minute per 1200 rev,

so T = time for 1 rev = 1/1200 minutes :wink:

but the better way is to convert rev per minute to radians per minute (and then radians per second) …

then you can use speed = radians per second times radius (v = ωr) :smile:
 
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