= Solution
The starting radius is $r_0=\sqrt2\,c$. In uniform <circular motion>, the required <centripetal acceleration> has magnitude $\omega^2r_0=kr_0/m$. The initial <velocity> must be tangent to the starting radius, with the orientation chosen anticlockwise:
$$
\boxed{\mathbf v(0)=\omega\widehat{\mathbf z}\times(c,c,0)=(-\omega c,\omega c).}
$$
The resulting trajectory is
$$
x=c(\cos\omega t-\sin\omega t),\qquad y=c(\cos\omega t+\sin\omega t).
$$
It has constant $x^2+y^2=2c^2$, so
$$
\boxed{U(t)=kc^2\quad\text{for all }t.}
$$
This is the horizontal line in the right panel of the figure. The <kinetic energy> also equals $kc^2$, giving total <mechanical energy> $2kc^2$, twice that for release from rest at the same point.
Back to article page
