An object with velocity $ v$ has kinetic energy:

$\displaystyle E_{\textrm{kin}}=\frac{1}{2}M v^2$ (1)

If the object is also rotating around an axis going through its center of mass (CM in the figure), then it also has rotational energy:

$\displaystyle E_{\textrm{rot}}=\frac{1}{2}I_{CM} \omega^2$ (2)

Here $ I_{\textrm{CM}}$ is the moment of inertia of the cylinder around an axis going through the center of mass, and $ \omega$ is called the angular velocity.

Click here to learn more about the moment of inertia and about angular velocity.

We take a hollow cylinder with outer radius $ R$, inner radius $ R_i$, length $ l$ and mass $ M$ and we place this cylinder on top of an inclined plane with inclination angle $ \alpha$ and height $ h$ . If we release the cylinder it shall have no initial velocity, because all of its energy is still in potential energy:

$\displaystyle E_{\textrm{pot}}=M g h$ (3)

where $ g$ is the acceleration of gravity (on earth about $ g=9.81m/s^2$). When the cylinder starts rolling, it will loose potential energy and gain kinetic and rotational energy. At the end of the inclined plane all potential energy will have been transfered into kinetic and rotational energy. In this process there is no loss or gain of energy (if we neglect friction), so we can write down:

$\displaystyle E_{\textrm{init}}$ $\displaystyle = E_{\textrm{final}}$    
$\displaystyle Mgh$ $\displaystyle = \frac{1}{2}M v^2 + \frac{1}{2}I_{\textrm{CM}} \omega^2$ (4)

To solve these equations we need a relationship between $ v$ and $ \omega$. If the cylinder may not slip during its rolling then:

$\displaystyle v = \omega R \Rightarrow \omega =\frac{v}{R}$ (5)

Substituting into 4 gives:

$\displaystyle Mgh$ $\displaystyle = \frac{1}{2}M v^2 + \frac{1}{2}I_{\textrm{CM}} \frac{v^2}{R^2}$    
$\displaystyle 2gh$ $\displaystyle = v^2 + \frac{I_{\textrm{CM}}}{M}\frac{v^2}{R^2}$    
$\displaystyle v$ $\displaystyle = \sqrt{ \frac{2gh}{1+I_{\textrm{CM}}/(MR^2) }}$ (6)

Now all we need to know is the $ I_{\textrm{CM}}$ of the hollow cylinder:

$\displaystyle I_{\textrm{CM}}=\frac{1}{2}M \left(R_i^2+R^2\right)$ (7)

The moment of inertia of the solid cylinder is found by taking $ R_i=0$.

Read here how you determine the moment of inertia of a cylinder around its symmetry axis.

Substituting the moment of inertia into equation 6:

$\displaystyle v$ $\displaystyle = \sqrt{ \frac{2gh}{1+\frac{1}{2}M \left(R_i^2+R^2\right)/(MR^2) }}$    
  $\displaystyle = \sqrt{ \frac{2gh}{1+\frac{1}{2} \left(R_i^2/R^2+1\right) }}$    
  $\displaystyle = \sqrt{ \frac{4gh}{R_i^2/R^2+3 }}$    

It's clear that when $ R_i\neq0$ the velocity $ v$ will decrease. A hollow cylinder will therefore be slower than a solid one.