A rotating merry-go-round makes one complete revolution in 4.0s a) what is the linear speed of a child seated 1.2 meters from the center? b) what is her acceleration (give components)
So for part a I did (2Pi/4) =1.57
1.57x1.2 (radius) =2.96 m/s???
Am I on the right track? And I don't know how to set up part b. Help!

Question

A rotating merry-go-round makes one complete revolution in 4.0s a) what is the linear speed of a child seated 1.2 meters from the center? b) what is her acceleration (give components)
So for part a I did (2Pi/4) =1.57
1.57x1.2 (radius) =2.96 m/s??? 
Am I on the right track? And I don't know how to set up part b. Help!

anonymous · Answer

First part looks good. 

She is not accelerating in the tangential direction as her time of rotation is constant. She is, however, experiencing centripetal acceleration, which is required to keep her rotational course. This is expressed as$$a_c = {v^2 \over r}$$ this is directed towards the center of her rotation.

anonymous · Answer

Thank you so much for your input! So just to make sure I have this, I just want to clarify: V=2.96 and r=1.2? So the set up is:
(2.96$$^{2}$$ m/s) / 1.2
which would be about 7.30?

anonymous · Answer

* (2.96^2m/s/1.2m) = 7.30s

anonymous · Answer

It is 7.30 m/s^2.