Why do objects roll down slopes at the same speed?

Question

I had to do an online simulation in class, and for some reason an object with twice as much as another rolled down a slope at the same speed of the other object. In other words, objects with a large difference in mass still rolled down the slope at the same rate. Is this correct, first off, or was the simulation wrong? If it is correct, why so?

Answer 1

Before dealing with rolling objects, assume it's a block sliding down a frictionless slope with a starting height h

Conservation of energy gives:

$$ mgh = \frac 1 2 mv^2$$

Notice that the mass cancels out leaving: $$v= \sqrt{2gh}$$ which is independent of mass.

For rolling objects:

$$mgh =\frac 1 2 mv^2 + \frac 1 2 I\omega^2$$ where I is the moment of Inertia of the object and $\omega$ is the angular velocity of the rolling object.

The moment of Inertia is analogous to mass but is also dependent on the size and shape of the rolling object. For example, the moment of Inertia of a sphere is

$$I = \frac 2 5 mR^2$$ where R is the radius of sphere. Plugging in to the equation above and solving for v gives: $$v = \sqrt{2gh - \frac 2 5 R^2\omega^2}$$ Edit: For pure rolling: $$\omega = \frac v R$$ and

$$v = \sqrt{\frac{2gh}{1+\frac 2 5 }}$$

Notice the velocity is still independent of mass but the moment of inertia has an effect. So as long as they are spheres, they will roll with the same velocity despite having different masses and sizes.

Answer 2

Hmm, for sliding, we know:

$$ mgh \propto T \propto m v^2 $$

so

$$ v^2 \propto gh $$

mass plops out of the equation. It doesn't matter, hence: general relativity (but that's a long story).

Now if you look at rolling, where:

$$ T' \propto I \omega^2 $$

you have $I \propto MR^2$ and $\omega \propto v/R$, so:

$$ T' \propto \Big(MR^2\Big)\Big(v/R\Big)^2$$

$$ T'\propto Mv^2 \propto gh $$

so mass just doesn't matter, as long as the shape doesn't change.

The mass distribution does, since mass is the zero-th movement of $\vec r$ w.r.t. to the density distribution:

$$ M = \langle \rho(\vec r) ||r||^0\rangle$$

Center-of-mass is the first moment:

$$ \vec R = \langle \rho(\vec r) \vec r\rangle$$

(unless $\vec g(\vec r) \ne g$ is non-uniform--then forget it),

while the moment-of-interia is the variance (and when variance is computed from a vector, it's a dyadic):

$$ I = \langle \rho(\vec r)[r^2{\bf I} - \vec r \vec r] \rangle$$

Basically, it introduces a length-scale (difference) that can't be canceled with $M$ and $g$.

Answer 3

The simulation is correct. The explanation is deep, and arguably beyond Newtonian mechanics. In General Relativity, the Earth warps spacetime around itself, and other objects fall along so-called "geodesics" which do not depend on the object's mass. Hence the speed at which something falls does not depend on its mass.

In Newtonian mechanics, there is:

$$F = GMm/r^2 = ma$$

The first equality is Newton's law of gravitation. The second is Newton's second law. The two $m$'s cancel, which leaves:

$$GM/r^2 = a$$

Hence the acceleration of an object does not depend on its mass. Once the acceleration is the same, from kinematics, the speed is also the same.

(The subtlety here is that we have asserted that the two $m$'s cancel, but there's no a priori reason to expect that they will cancel.)