= Solution
A straight diameter at nonzero <angle> $\chi$ to the rotation axis moves the body toward that axis during the inward half of its trip. With no external axial <torque>, the total axial <angular momentum> of the <Earth> and body is conserved. The body's contribution to the <moment of inertia> decreases toward the centre, so the <Earth> spins slightly faster; the reverse happens on the outward half, restoring the original rotation rate at the far surface.
For a body of <mass> $m$ constrained to follow a tunnel fixed in the <Earth>, an illustrative angular-momentum balance is
$$
\boxed{\Omega(x)=\Omega_0\frac{I_E+mr_0^2\sin^2\chi}{I_E+mx^2\sin^2\chi}.}
$$
Here $I_E$ is the <Earth>'s axial <moment of inertia>, treated as constant, and $x$ is the signed distance along the diameter. A frictionless guide can still exert normal forces: these supply the transverse acceleration and exchange axial <angular momentum> with the rotating <Earth>. Simply allowing an unconstrained body to fall would not keep it inside a narrow nonpolar straight tunnel. The effect vanishes for $\chi=0$ and is largest for an equatorial diameter.
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