The reasoning based on conservation of energy does not yield directly the time dependence of the motion. In order to simulate the motion by means of a computer, we need to use the equation of motion.

The equation of motion for rotating bodies is:

$\displaystyle r \times F = I \dot{\omega}$

(8)

where

is the distance between the point of application of the force and the axis of rotation, $\dot{\omega}$ is the angular acceleration and

is the moment of inertia around the axis.

The cylinder rotates around the line where the cylinder and the plane touch each other, so

in 9. We already determined the moment of inertia around the symmetry axis. The rotation axis is parallel to the symmetry axis, therefore we can use the parallel axis rule to determine the moment of inertia as:

$\displaystyle I=I_{\textrm{CM}}+MD^2,$

(9)

where

denotes the distance between both axes,

in this case.

in equation 9 is the component of the gravitational force parallel to the plane:

$\displaystyle F_{\textrm{netto}}=M g \sin(\alpha),$

(10)

where $\alpha$ is the angle of inclination.

The rotational axis and the force are perpendicular, so the crossproduct in equation 9 is simply:

$\displaystyle Rmg\sin\alpha$	$\displaystyle = (I_{\textrm{CM}}+MR^2) \dot{\omega}$
	$\displaystyle = \left(\frac{1}{2}M\left(R_i^2+R^2\right)+MR^2\right) \dot{\omega}$
	$\displaystyle = \frac{1}{2}M\left(R_i^2+3R^2\right) \dot{\omega}$
$\displaystyle \dot{\omega}$	$\displaystyle =\frac{2Rg\sin\alpha}{R_i^2+3R^2}$	(11)

Since the angular acceleration is not time dependent:

$\displaystyle \omega(t) = \int_0^t\dot\omega \d t'= \dot\omega \int_0^t \d t'=\frac{2Rg\sin\alpha}{ R_i^2+3R^2}t$

(12)

Using the fact that the cilinder may not slip:

$\displaystyle v(t)$	$\displaystyle =\omega(t)R$	(13)
	$\displaystyle =\frac{2R^2g\sin\alpha}{ \left(R_i^2+3R^2\right)}t$	(14)
	$\displaystyle =\frac{2g\sin\alpha}{ \left(R_i^2/R^2+3 \right)}t$	(15)

So for larger $R_i\neq0$ the velocity will decrease. So a hollow cylinder rolls more slowly than a solid one.