# Centripetal Acceleration again

1. Oct 11, 2005

### rockmorg

Hey all -

I've got a problem that's rather encompassing...

The second hand and the minute hand on one type of clock are the same length. What is the period T of the motion for the second hand and for the minute hand?

It goes on to say the centripetal acceleration is given by a_c = v^2/r where v is speed and r is the radius. How is centripetal acceleration related to the period?

And then finally it asks for the ratio a_c, second / a_c, minute for the tips of the second hand and the minute hand.

I'm not even sure where to start really...

Any help would be great, thanks!

2. Oct 11, 2005

### amcavoy

Basically if you take the parametric equations:

$$x=r\cos{\omega t}$$
$$y=r\sin{\omega t}$$

...and note that the second derivatives (acceleration) are:

$$x''=-\omega^2 r\cos{\omega t}$$
$$y''=-\omega^2 r\sin{\omega t}$$

The magnitude is therefore:

$$\left|\vec{a}\right|=\omega^2 r=\frac{v^2}{r}$$

because

$$\omega=\frac{d\theta}{dt}=\frac{v}{r}$$