Question

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 $\mathrm{m} / \mathrm{s}^{2}$ What is the centripetal acceleration at a point that is 0.050 m from the center of the disc?

          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 $\mathrm{m} / \mathrm{s}^{2}$  What is the centripetal acceleration at a point that is 0.050 m from the center of the disc?

Added by Patrick C.

Physics

John D. Cutnell, Kenneth W. Johnson 9th Edition

Chapter 5

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03/22/2020

Step 1

In this case, we are given the centripetal acceleration at a point that is 0.030 m from the center of the disc, which is 120 $\mathrm{m} / \mathrm{s}^{2}$. Let's call this $a_{c1}$ and $r_{1}$. $a_{c1} = \frac{v_{1}^{2}}{r_{1}}$ Show more…

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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 $\mathrm{m} / \mathrm{s}^{2}$ What is the centripetal acceleration at a point that is 0.050 m from the center of the disc?