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An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s). If a particular disk is spun at 646.1 rad/s while it is being read, and then is allowed to come to rest over 0.569 seconds, what is the magnitude of the average angular acceleration of the disk?

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Answer:

Angular acceleration will be
1135.5rad/sec^2

Step-by-step explanation:

We have given initial angular speed of a particular disk
\omega _i=646.1rad/sec

And it finally comes to rest so final angular speed
\omega _f=0rad/sec

Time is given as t = 0.569 sec

From third equation of motion we know that


\omega _f=\omega _i+\alpha t

So angular acceleration
\alpha =(\omega _f-\omega _i)/(t)=(0-646.1)/(0.569)=1135.5rad/sec^2

User Trixo
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