A computer is reading data from a rotating CD-ROM. At a pointthat is 0.030 m fro

Question

A computer is reading data from a rotating CD-ROM. At a pointthat is 0.030 m from the center of the disc, the centripetalacceleration is 120 m/s2. What is the centripetalacceleration at a point that is 0.050 m from the center of thedisc?

Attempt:
v = [(120 m/s2)(0.030 m)] = 1.897366596m/s

ac = (1.897366596 m/s)2 / (0.050 m)= 72 m/s2


I feel this is correct because this question appears in the sectionof problems that deals only with the following twotopics:
- Uniform Circular Motion
- Centripetal Acceleration

I've seen this question asked before on this site and severalothers, and there is definitely inconsistency in how it isanswered. Many people that respond consider angular velocity,which is NOT covered in these two sections of the textbook. If that was, indeed, part of the reasoning for the solution, then Iwould imagine the problem would have been moved to a differentsection by now (since this is the 7th edition of the textbook andhas gone through plenty of revisions). Based on the followinglink, my answer is correct: http://www.physics.unomaha.edu/Sowell/Phys1110/quizzes/QuizzesSpr06/quiz7.html. This seems to be the only reliable source I have been able to findas it is from a university and not a forum. This sitesuggests that we are to assume that linear velocity is constantwhen solving this problem. I agree with it.

Could someone evaluate my work and check to see if I have done itcorrectly? Please do not include any information aboutangular velocity because that concept has not been discussedneither in lecture for my course or in the textbook.

Explanation / Answer

The ROTATIONAL PERIODS are the same for both points, NOT THEVELOCITIES. Both points will make one revolution in the same amountof time. However, velocity is not the same at both points. Sinceboth points make a complete rotation in the same amount oftime, the larger point will have covered a much largercircumference in that same time. The .050m radius makes a largercircle than the .030m radius and so the velocity is greaterthe further out the point is. So, using the centripetal acceleration formula ac =(42r/T2)... 120 = (42*0.030/T2) T2 = (42*0.030/120) T2 = 0.0098696044s2 Now, plug that T2 into the same equation forthe other point on the disk: ac = (42r/T2) ac = (42*0.050/0.0098696044) ac = 200m/s2, the finalanswer

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