= Solution
Apply the <rotating-frame derivative formula> first to $r$, then to $v'= [dr/dt]_{S'}$. Since the <angular velocity> is constant in either frame,
$$
\left[\frac{dr}{dt}\right]_S=v'+\omega\times r,
$$
$$
\boxed{\left[\frac{d^2r}{dt^2}\right]_S=\left[\frac{d^2r}{dt^2}\right]_{S'}+2\omega\times v'+\omega\times(\omega\times r).}
$$
The two apparent <forces> in the rotating equation are the <Coriolis force> and the <centrifugal acceleration> multiplied by <mass>. Thus
$$
m\left[\frac{dv'}{dt}\right]_{S'}=-\nabla\phi+G-2m\omega\times v'-m\omega\times(\omega\times r).
$$
Set $\phi_{\rm eff}=\phi-\tfrac m2|\omega\times r|^2$. The supplied gradient identity gives $\nabla\phi_{\rm eff}=\nabla\phi+m\omega\times(\omega\times r)$. Taking the scalar product with $v'$ removes $G$, since it is orthogonal to $v'$, and removes the <Coriolis force>. Therefore the <energy balance in a uniformly rotating frame> is
$$
\boxed{\frac{d}{dt}\left(\frac m2|v'|^2+\phi-\frac m2|\omega\times r|^2\right)=\left(\frac{\partial\phi}{\partial t}\right)_{r'}.}
$$
\b[The displayed quantity is conserved when the potential is stationary in the rotating coordinates.] This includes a static inertial potential invariant under rotations about the chosen axis, such as vertical gravity in part (b). If the word conservative only means a time-independent inertial potential, that extra rotational invariance is needed; a conservative <force> alone does not imply the claimed conservation.
For an explicit counterexample to that unrestricted interpretation, take unit <mass>, an inertial <force> $F=e_x$, $\phi=-x$, no $G$, rotation $\omega=\omega e_z$, and inertial trajectory $r(t)=(t^2/2,1,0)$ with initial <velocity> zero. It satisfies <Newton's second law>. Substituting $v'=\dot r-\omega\times r$ into the proposed energy gives $E=\omega t$, which is not constant for nonzero rotation. The qualified conservation law above is the one used for the hoop.
Back to article page
