Recent content by SaraM

OK, got it. Thank you so much for your help! Cleared things up.

Thank you for your reply! Following your hints, here's what I ended up with: In ##r##: ##~m( \ddot r -r \dot \theta^2)=- k(r-a)## In ##\theta##: ##~\dot\theta=\frac{v}{r}=-\frac{k}{mv}(r-a) ## (which I got from the centripetal force ##m\frac{v^2}{r}=-k(r-a)##) What's left to do is plug in...

1. Homework Statement A particle of mass m is attached to the end of a light spring of equilibrium length a, whose other end is fixed, so that the spring is free to rotate in a horizontal plane. The tension in the spring is k times its extension. Initially the system is at rest and the...