Circular Motion - RPM and Radius question

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SUMMARY

The discussion focuses on calculating the linear speed and acceleration of data on a CD-ROM spinning at 1200 revolutions per minute (RPM) with the outer row located 5.6 cm from the center. The correct approach involves converting RPM to radians per second, allowing the use of the formula v = ωr for speed and a = V²/r for acceleration. The user initially miscalculated the period (T) but received guidance on the correct conversion method, emphasizing the importance of understanding angular velocity in circular motion.

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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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