Does a larger ball of the same mass roll faster downhill than a smaller ball? My intuition tells me that a larger ball will cover more ground on each revolution and thus it will go faster downhill than a smaller ball of the same mass.
Assuming both balls are the same mass and the mass is uniformly distributed, what is the answer?
 A: Interestingly, the only thing that matters is the shape (or distribution of mass) of the ball.  Consider the total energy for an object rolling down an incline:
$$E = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2 + mgy$$
Now note that the rotational inertia of any rolling object can be written in the form $I = \beta mr^2$, where $\beta$ is some unites constant.  For a solid sphere, $\beta = 2/5$, for a hollow sphere, $\beta = 2/3$, for a hollow cylinder, $\beta = 1$, while for a solid cylinder, $\beta = 1/2$.  You can look these up in any physics textbook, or calculate them by performing an integral.
Also note that the "rolling condition" means that the angular speed is fixed by the speed of the center of mass and the radius: $\omega = v/r$.
Making these two substitutions, we can write:
$$E = \frac{1}{2}(1+\beta)mv^2 + mgy$$
Now imagine that we start from rest at the top at some initial height $y=h$, and roll down to $y=0$ at some final time.  Imposing conservation of energy:
$$mgh = \frac{1}{2}(1+\beta)mv_f^2$$
$$v_f = \sqrt{\frac{2gh}{1+\beta}}$$
In other words, the final velocity is independent of the mass and the radius!  It only depends on the factor $\beta$ which is determined by the shape of the rolling object (and the initial height, of course)!
EDIT:
I thought I'd add this edit about the acceleration. One can determine the acceleration also by conservation of energy.  Start with:
$$E = \frac{1}{2}(1+\beta)mv^2 + mgy$$
and take $\frac{dE}{dt}$. Be sure to impose the chain rule since $y=y(t)$, and $v=v(t)$.  Then, conservation of energy implies
$$ \frac{dE}{dt}=(1+\beta)mv \frac{dv}{dt} + mg \frac{dy}{dt} = 0$$.
Now $y$ is the height, which we can express in terms of the distance $\ell$ along the incline (from the bottom) as $y = \ell \sin \theta$.
But, $\frac{dv}{dt}=a$ and $\frac{dy}{dt}=\frac{d\ell \sin \theta}{dt}=v\sin{\theta}$, assuming $\theta$ is the constant angle of the incline and $v=\frac{d\ell}{dt}$.
$$ \frac{dE}{dt}=(1+\beta)mv a + mg v \sin\theta = 0$$.  So the acceleration is:
$$a=\frac{-g\sin\theta}{1+\beta}$$
Note that it reduces to the acceleration $a=-g\sin\theta$ of a sliding mass when $\beta \rightarrow 0$, and to "free fall" when $\beta \rightarrow 0$ and $\theta \rightarrow \pi/2$.
A: Moment of inertia of a sphere is $I=\frac 25 mr^2$. Using energy conservation, $$\text{initial potential energy = final kinetic energy}$$ $$mgh=\frac 12mv^2+\frac 12  I\omega ^2$$ After some calculations (using equations of motion as well) you will get, $$a=\frac {5g\sin\alpha}{7}\tag{1}$$ This is the acceleration of a sphere along a inclined plane whereas $\alpha$ is the inclined angle. Thus the acceleration of two objects with the same shape are equal (because it only depends on the constant in front of $mr^2$[moment of inertia] which is $\frac25$ here) So you see, the time for the two balls to reach the bottom of the hill is equal!

 More general form of the equation(1) is $a=\frac{g\sin\alpha}{k+1}$ , whereas $k$ is the constant mentioned above. This is valid for any shape.
