Centripetal Acceleration of hard drive

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Homework Help Overview

The problem involves calculating the centripetal acceleration of a point on a rotating CD-ROM at different distances from the center. The original poster presents an initial calculation based on the assumption of constant linear velocity, while others introduce concepts of angular velocity and period to explore the relationship between centripetal acceleration and radius.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to solve for centripetal acceleration using a direct calculation based on linear velocity, while others question this approach by suggesting the use of angular velocity and the period of rotation.

Discussion Status

Participants are exploring different interpretations of the problem, with some suggesting that the original poster's method may not be appropriate due to the nature of rotational motion. Guidance has been offered regarding the use of angular velocity and period, which has led to a revised understanding by the original poster.

Contextual Notes

The original poster emphasizes that angular velocity has not been covered in their course or textbook, which influences their approach to the problem. There is a noted inconsistency in how similar problems are addressed in various sources.

UMich1344
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Homework Statement



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

Homework Equations



ac = v2 / r

The Attempt at a Solution



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

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


I feel this is correct because this question appears in the section of problems that deals only with the following two topics:
- Uniform Circular Motion
- Centripetal Acceleration

I've seen this question asked before on this site and several others, and there is definitely inconsistency in how it is answered. 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 I would imagine the problem would have been moved to a different section by now (since this is the 7th edition of the textbook and has gone through plenty of revisions). Based on the following link, 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 find as it is from a university (not saying it's infallible) and not a forum (which often contains more false information than a site directly from a university... as I'm sure most of you would agree). This site suggests that we are to assume that linear velocity is constant when solving this problem. I agree with it.

Could someone evaluate my work and check to see if I have done it correctly? Please do not include any information about angular velocity because that concept has not been discussed neither in lecture for my course or in the textbook.
 
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You have the relationship that

ac = ω²r

This implies directly that with ω constant

ac1 / ac2 = r1 / r2

Or

ac1 * r2 / r1 = ac2

120*5/3 = 200

Your solution is seemingly based on holding v constant which for a spinning disk is not the case, it's the ω that's constant.
 
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If you aren't into angular velocity yet, just consider the period. T will be the same for all radii, so it is a useful thing to work with. You can find T from your velocity at the first radius using v = 2(pi)r/T
Once you have T, which applies everywhere, you can find v at the 2nd radius using the same formula. Put that in your centripetal force formula and you'll get the correct answer!
 
Thank you very much. I solved for T in terms of the point that is 0.030 m from the center of the disc, then plugged that value into find the centripetal acceleration at the point that is 0.050 m from the center of the disc.

The numbers worked out very nicely and it came out to be 200 m/s2, as the first response suggested.

I knew there had to be more to it, but I wasn't seeing it.
 

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