Yes, "straight" lines on a manifold are geodesics, which obey the geodesic equation $$\frac{\text{d}^2x^\rho}{\text{d}\tau^2} = -\Gamma^\rho_{\mu\nu}\frac{\text{d}x^\mu}{\text{d}\tau}\frac{\text{d}x^\nu}{\text{d}\tau}$$

If we take the surface of a sphere as our manifold, the geodesics are great circles. A frictionless track will only be able to exert a perpendicular force on the object, so any object traveling along it will always remain at constant speed. The first such force exerted by the track is the normal force which points upwards. If the track is not a geodesic, it will exert an additional force of constraint on the object.

Yes, "straight" lines on a manifold are geodesics, which obey the geodesic equation $$\frac{\text{d}^2x^\rho}{\text{d}\tau^2} = -\Gamma^\rho_{\mu\nu}\frac{\text{d}x^\mu}{\text{d}\tau}\frac{\text{d}x^\nu}{\text{d}\tau}$$

If we take the surface of a sphere as our manifold, the geodesics are great circles. A frictionless track will only be able to exert a perpendicular force on the object, so any object traveling along it will always remain at constant speed and will circle the Earth forever. 

The track exerts two forces on the object. The first is the normal force which points radially. If the track is not a great circle, it will exert an additional force of constraint on the object which is tangential and also perpendicular to the track.

Yes, "straight" lines on a manifold are geodesics, which obey the geodesic equation $$\frac{\text{d}^2x^\rho}{\text{d}\tau^2} = -\Gamma^\rho_{\mu\nu}\frac{\text{d}x^\mu}{\text{d}\tau}\frac{\text{d}x^\nu}{\text{d}\tau}$$

On the surface of a sphere, geodesics are great circles. If the frictionless track is a great circle, the object will travel along it without feeling any force (except the normal force). Otherwise, the track will apply a force in order to keep the object moving along it.

Note that the geodesics on the surface of the Earth (great circles) are not the same as geodesics in spacetime, which correspond to test particles in free fall.

Yes, "straight" lines on a manifold are geodesics, which obey the geodesic equation $$\frac{\text{d}^2x^\rho}{\text{d}\tau^2} = -\Gamma^\rho_{\mu\nu}\frac{\text{d}x^\mu}{\text{d}\tau}\frac{\text{d}x^\nu}{\text{d}\tau}$$

If we take the surface of a sphere as our manifold, the geodesics are great circles. A frictionless track will only be able to exert a perpendicular force on the object, so any object traveling along it will always remain at constant speed. The first such force exerted by the track is the normal force which points upwards. If the track is not a geodesic, it will exert an additional force of constraint on the object.