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

`a_2 = 114.28\ m/s^2`

Step-by-step 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 = (v^2)/(r_1)`

`v= (2\pi r_1)/(T)`

`a_1 = (( (2\pi r_1)/(T))^2)/(r_1)`

`a_1 = (4\pi^2 r_1)/(T^2)`

similarly

`a_2= (4\pi^2 r_2)/(T^2)`

now,

`(a_2)/(a_1)= (r_1)/(r_2)`

`a_2 = a_2* (r_1)/(r_2)`

`a_2 = 339* (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`