# Centripetal Acceleration of hard drive

• UMich1344
In summary: Looks like I'll need to revisit the section on linear velocity and angular velocity in order to get a better understanding of this problem.In summary, 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. At a point that is 0.050 m from the center of the disc, the centripetal acceleration is 72 m/s2.
UMich1344

## 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?

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.

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.

jc10
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.

## 1. What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It is directed towards the center of the circle and is always perpendicular to the object's velocity.

## 2. How does centripetal acceleration affect a hard drive?

Centripetal acceleration is responsible for keeping the hard drive's spinning disk in its circular path. It helps maintain the stability and rotation of the disk, allowing the read/write heads to access data accurately and efficiently.

## 3. What factors affect the centripetal acceleration of a hard drive?

The centripetal acceleration of a hard drive is affected by the rotational speed of the disk, the distance between the read/write heads and the center of the disk, and the mass of the disk. Higher rotational speeds and smaller distances result in higher centripetal acceleration, while a heavier disk will require more centripetal force to maintain its circular motion.

## 4. Can centripetal acceleration cause damage to a hard drive?

While centripetal acceleration is necessary for the proper functioning of a hard drive, excess or uneven forces can lead to damage. This is why it is important to handle hard drives carefully and ensure proper installation and maintenance.

## 5. How is centripetal acceleration measured in a hard drive?

The centripetal acceleration in a hard drive can be measured using a formula: a = v^2/r, where a is the centripetal acceleration, v is the linear velocity of the disk, and r is the radius of the disk. This can be measured using specialized equipment or calculated using known values.

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