Question

A manufacturer of CD-ROM drives claims that the player can spin the disc as frequently as 1200 revolutions per minute.

a. If spinning at this rate, what is the speed of the outer row of data on the disc?


This row is located 5.6 cm from the center of the disc.




b. What is the acceleration of the outer row of data?

Answer 1

a. 7.0 m/s

First of all, we need to convert the angular speed (1200 rpm) from rpm to rad/s:

`\omega = 1200 (rev)/(min) \cdot (2\pi rad/rev)/(60 s/min)=125.6 rad/s`

Now we know that the row is located 5.6 cm from the centre of the disc:

r = 5.6 cm = 0.056 m

So we can find the tangential speed of the row as the product between the angular speed and the distance of the row from the centre of the circle:

`v=\omega r = (125.6 rad/s)(0.056 m)=7.0 m/s`

b.
`875 m/s^2`

The acceleration of the row of data (centripetal acceleration) is given by

`a=(v^2)/(r)`

where we have

v = 7.0 m/s is the tangential speed

r = 0.056 m is the distance of the row from the centre of the trajectory

Substituting numbers into the formula, we find

`a=((7.0 m/s)^2)/(0.056 m)=875 m/s^2`