Cylinder vs cylinder of double the radius roll down an incline plane, which one wins? A solid cylinder and another solid cylinder with the same mass but double the radius start at the same height on an incline plane with height h and roll without slipping. Consider the cylinders as disks with moment of inertias I=(1/2)mr^2. Which one reaches the bottom of the incline plane first?
According to this, the velocity of any body rolling down the plane is
v=(2 g h/1 + c) ^½
where c is the constant in moment of inertia (for example, c=2/5 for a solid sphere). 
My thought process was that since the radius doubled, c=2. So, the velocity of the doubled cylinder would be less, therefore finishing later. Similarly, if it’s moment of inertia increases, it’s angular and linear acceleration decreases. However, my other peers and even my professor disagree, saying that radius and mass do not play a role in the velocity of the body, since both m and r will cancel in an actual calculation of the velocity.
Could anyone elaborate on whether I am right or wrong?
 A: Conservation of energy tells us potential energy becomes kinetic energy as the disks fall. If they roll without slipping, some energy goes into translational kinetic energy and some goes to rotational kinetic energy. 
The rolling without slipping condition requires that the velocity of the disk be equal to the rotatoinal velocity times the radius $v=\omega R$. 
Total kinetic energy = $\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$
So:
$$Mgh=\frac{1}{2}Mv_1^2+\frac{1}{2}I_1(\frac{v_1^2}{R^2})=\frac{1}{2}Mv_2^2+\frac{1}{2}I_2\frac{v_2^2}{4R^2}$$
$I_1=\frac{1}{2}MR^2$
$ I_2=\frac{1}{2}4MR^2=2MR^2$
We can take the ratio of the squared velocities:
$$\frac{1}{2}Mv_1^2+\frac{1}{2}(\frac{1}{2}MR^2)(\frac{v_1^2}{R^2})=\frac{1}{2}Mv_2^2+\frac{1}{2}(2MR^2)\frac{v_2^2}{4R^2}$$
$$\frac{1}{2}Mv_1^2+\frac{1}{4}(M)(v_1^2)=\frac{1}{2}Mv_2^2+\frac{1}{4}(M)v_2^2$$
$$\frac{v_1^2}{v_2^2}=\frac{1+\frac{I_2}{4MR^2}}{1+\frac{I_1}{MR^2}}=\frac{\frac{3}{2}}{\frac{3}{2}}=1$$
So I stand corrected. Using the proper radii consistently, the velocities end up the same. 
A: The following equation from @R. Romero's analysis is correct:
$$Mgh=\frac{1}{2}Mv^2+\frac{1}{2}I\left(\frac{v^2}{R^2}\right)\tag{1}$$But, the moment of inertia of a cylinder is given by:  $$I=M\frac{R^2}{2}\tag{2}$$ So, combining Eqns. 1 and 2 gives:$$Mgh=\frac{1}{2}Mv^2+\frac{1}{4}Mv^2\tag{3}$$Calcelling M from both sides of the equation yields:$$gh=\frac{1}{2}v^2+\frac{1}{4}v^2\tag{4}$$So, solving for v, we have:$$v=\sqrt{\frac{4gh}{3}}$$Note that this is independent of the radius of the cylinder.  So, both cylinders roll down the ramp in the same amount of time.
A: 
My thought process was that since the radius doubled, c=2

$c$ is not the moment of inertia itself, it's the constant in $I = cMR^2$.  For your two solid cylinders, the constant will be the same, even though $I$ will differ because $R$ will differ.

Similarly, if it’s moment of inertia increases, it’s angular and linear acceleration decreases.

You're correct that the angular acceleration decreases.  But that doesn't mean the linear acceleration decreases.  
If we put the same rotational energy into the cylinders, the larger one must spin slower.  How much slower?
$$ E = \frac{1}{2} I \omega^2$$
$$ \omega ^2 = \frac {2 E}{I}$$
$$ \omega = \sqrt{ \frac {2 E}{MR^2} }$$
Since mass and energy are constant here, we can replace them and the factor of two with a single constant $k$.
$$ \omega = \frac{k}{R}$$
So as I goes up (and energy and mass are constant) it has angular speed that is inversely proportional to R.  But because it's rolling, we know that $v = \omega R$.  
$$ v = \omega R$$
$$ v = \frac{k}{R} R = k$$
The radius has fallen out.  The rotational speed depends on the radius, but the linear speed does not.  
