The force you experience is of the form $\vec{F} = - Gmr\vec{u_r}$, and we also know that in the surface, $r=R$, it is $\vec{F}=- gm\vec{u_r}$, so
$$\vec{F} = -gm\frac{r}{R}\vec{u_r}$$
This is a conservative force that can be derived from a potential
$$U = \frac{1}{2}gm\frac{r^2}{R}$$
Because this is a central force, angular momentum is conserved, so $r^2 \dot{\theta} = L$, and if $\Omega$ is the rotational velocity of earth,
$$r^2 \dot{\theta} = R^2 \Omega$$
And of course we have conservation of energy,
$$\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)+\frac{1}{2}gm\frac{r^2}{R} = E$$
but we also know the initial conditions, $r=R$, $\dot{\theta}=\Omega$, $\dot{r}=0$, so
$$E = \frac{1}{2}m(R^2\Omega^2)+\frac{1}{2}gmR$$
and conservation of energy can be rewritten as
$$\dot{r}^2+r^2\dot{\theta}^2+g\frac{r^2}{R} = R^2\Omega^2+gR$$
and including conservation of angular momentum as
$$\dot{r}^2+ \frac{R^4 \Omega^2}{r^2}+g\frac{r^2}{R} = R^2\Omega^2+gR$$
If you set $\dot{r} = 0$ and solve for $r$, there are two solutions, marking the annular region in which motion will happen. One is the obvious $r=R$, the other comes out to
$$r = \Omega R \sqrt{\frac{R}{g}}$$
which with the Earth parameters at the equator, comes out to $r=3740\ \mathrm{km}$.
You can rearrange the equation of energy as
$$\frac{dr}{\sqrt{R^2\Omega^2+gR - \frac{R^4 \Omega^2}{r^2}-g\frac{r^2}{R}}} = dt$$
which you could integrate to get a probably implicit relation between $r$ and $t$, which you could use in the conservation of angular momentum to get $\theta$ as a function of $r$ and/or $t$.
I have done that numerically, and again, for the point on the Equator, it would take about 21 minutes to reach the point closest to the Earth's center, and 21 more to get back up at the surface.
One neat result I don't fully understand where it comes from is that, at the minimum point, the angle $\theta$ has changed by $\pi / 2$, independently of what the rotation speed is, so that you always emerge at a point opposite where you went down. Since the Earth is rotating, you wouldn't actually come out at the antipodal point, but some $1175\ \mathrm{km}$ from it.
Away from the equator you would have a reduced $\Omega$, and the movement will happen in a plane perpendicular to the meridian going through that point.