# Rotational motion question

1. Nov 11, 2004

### jenavira

Q: What is the linear acceleration of a point on the rim of a 30-cm-diameter record rotating at a constant angular speed of 33.5 rev/min?

I seem to have all the variables and equations in hand -
r = .15m and $\omega$ = 3.49 radians/second;
$v = \omega r$ ;
(radial component of linear acceleration) $a_r = \frac {v^2} {r} \omega^2 r$

...but I get 3.29 $\frac {m} {s^2}$ (pointing inward), and the book's answer is 1.8 $\frac {m} {s^2}$. Am I wrong? Is the book wrong? Did I just convert something wrong somewhere? I'm very confused.

2. Nov 11, 2004

### Angie913

I got 1.8. You need to use the equation for centripetal acceleration (since the record is spinning) and that is ac= angular velocity^2 times radius. That should give you your answer!

3. Nov 11, 2004

### jenavira

*facepalm* Figured it was something simple like that. Thanks!