Calculate the magnitude of centripetal acceleration of the tip a the second hand, the minute hand, and the hour hand of a standard clock. The radius's are:
second hand - 9.8 cm
minute hand - 8.0 cm
hour hand - 6.0 cm
I know that the equation I need to use is Ac=v^2/r. However, how would I find the velocity if they only the give us the radius and are looking for the centritpetal accleration?

This is a clock and it has its own velocity!

Remember that v = r\omega

So, a = \frac{v^2}{r} = vr\omega = r\omega^2

Use the latter formula.

Omega is the angular velocity, or \omega = 2\pi f = \frac{2\pi}{T}

Where f is the frequency and T is the periodic time (time to complete one turn).

Can you solve it now? Post what you get!