Question

A computer is reading data from a rotating CD-ROM. At a point that is 0.0268 m from the center of the disk, the centripetal acceleration is 339 m/s2. What is the centripetal acceleration at a point that is 0.0795 m from the center of the disc?

Answer 1

Answer:

$a_2 = 114.28\ m/s^2$

Explanation:

Given,

centripetal acceleration, $a_1 =339\ m/s^2$

Distance from the center,$r_1 = 0.0268\ m$

Centripetal acceleration,$a_2 = ?$

Distance, $r_2 = 0.0795\ m$

$a_1 = \frac{v^2}{r_1}$

$v= \dfrac{2\pi r_1}{T}$

$a_1 = \frac{( \dfrac{2\pi r_1}{T})^2}{r_1}$

$a_1 = \dfrac{4\pi^2 r_1}{T^2}$

similarly

$a_2= \dfrac{4\pi^2 r_2}{T^2}$

now,

$\dfrac{a_2}{a_1}= \dfrac{r_1}{r_2}$

$a_2 = a_2\times \dfrac{r_1}{r_2}$

$a_2 = 339\times \dfrac{0.0268}{0.0795}$

$a_2 = 114.28\ m/s^2$

Hence, the acceleration of disc at 0.0795 m is equal to $a_2 = 114.28\ m/s^2$