Why does mass not affect the time and maximum steepness in which an uphill roller rolls? While investigating the classic mechanical problem of an uphill roller, I realized that the mass of the uphill roller does no affect the steepness of how high the roller can roll and also the time it takes to roll up a certain height. Why is this so?

https://youtu.be/SUjY0sdtGus
 A: When the conically shaped objects 'climb' uphill, this is really a kind of  an illusion, because the Centre of Gravity (CoG) is actually lowered during the climb, due to the objects' shape and how the roll bars are placed. This is of course not true of the cylindrical object and explains why it can only roll truly downhill.
Let's assume the system is free of friction, airdrag and other non-conservative forces.
As the conical object 'climbs' the height of its CoG is lowered by say, $\Delta h$, potential energy $U$ is released:
$$\Delta U=-mg\Delta h$$
As the object starts rolling (assumed without slipping) it gains both translational and rotational kinetic energy $K$:
$$\Delta K=E_{trans}+E_{rot}=\frac12 mv^2+\frac12 I\omega^2$$
$I$ is the inertia moment about the axis of rotation, which for the double sided regular conical object equates to:
$$I=\frac{6}{10} mR^2$$
where $R$ is the centre radius of the conical obbject, $\omega$ is the angular velocity of rotation and for rolling without slipping we have:
$$v=\omega R\rightarrow \omega=\frac{v}{R}$$
Inserting gives:
$$\Delta K=\frac12 mv^2+\frac12 \frac{6}{10}mv^2=\frac45 mv^2$$
Conservation of energy tells us:
$$\Delta U+\Delta K=0$$
So:
$$mgh=\frac45 mv^2$$
$$gh=\frac45 v^2$$
$$v=\frac12 \sqrt{5gh}$$
So the velocity gained for a given decrease in the height of the CoG of the conical object is independent of the mass $m$. of the object. The objects 'climbs' the rolling bars without defying gravity.
This is also true of the second 'conical' but without knowing the mathematical shape we can't carry out the conservation of energy calculation.

At the request of @Alex in the comments, here's the equation of motion.

Note that $\alpha$ may not be constant here, due to the peculiar geometry of the problem. I've assumed it to be constant for the sake of simplicity.
$$F-F_f=ma_x$$
where $F_f$ is the friction force, needed for the no slipping equirement.
$$F=mg\sin\alpha$$
$$F_f=\mu N=\mu mg\cos\alpha$$
with $\mu$ the minimum friction coefficient required for rolling without slipping.
$$a_x=g\sin\alpha-\mu g\cos\alpha\tag{1}$$
$\mu$ now needs to be determined.
Torque causes rotation:
$$\tau(x)=R(x)F_f=\dot{\omega}I$$
Note that due to the geometry of the problem, $R(x)$ is a function of $x$ (not determined here).
Rotation without slipping means:
$$v_x=R(x)\omega$$
$$a_x=\frac{\partial \omega}{\partial t}=R(x)\dot{\omega}$$
$$\dot{\omega}=\frac{a_x}{R(x)}=\frac{\mu mg\cos\alpha R(x)}{I}$$
$$a_x=\frac{\mu mg\cos\alpha R^2(x)}{I}\tag{2}$$
From $(1)$ and $(2)$, $\mu$ could then be determined.
