A computer is reading data from a rotating CD-ROM. At a point that is 0.0302 m from the center of the disk, the centripetal acceleration is 284 m/s2. What is the centripetal acceleration at a point that is 0.0803 m from the center of the disc?

Answer 1

omega = v/R is the same for each
Ac = omega^2 R = v^2/R

284 = omega^2 (.0302)

SO
Ac = [284/.0302](.0803)