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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?

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    $\begingroup$ twice as much as another Twice as much what? $\endgroup$
    – Ghoster
    Commented May 2 at 2:52
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    $\begingroup$ Try rolling a solid cylinder and a hollow cylinder of the same mass and radius. They won’t roll at the same speed because they have different moments of inertia. $\endgroup$
    – Ghoster
    Commented May 2 at 3:14
  • $\begingroup$ They don't, for a general object. Why do people ask questions 'why does something happen?', when it doesn't happen? For specifically similar objects - same moment of inertia when rolling, neglecting friction and air resistance - they do have the same acceleration when rolling down a slope. $\endgroup$
    – Neil_UK
    Commented May 2 at 10:48

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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.

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    $\begingroup$ $v = \sqrt{2gh - \frac 2 5 R\omega^2}$ doesn’t take into account that $\omega$ and $v$ have a second relationship. $\endgroup$
    – Ghoster
    Commented May 2 at 4:50
  • $\begingroup$ Yes. You are correct: $\omega = \frac v R $ for the pure rolling condition. Since it's independent of mass I left it out but I'll add it. $\endgroup$ Commented May 2 at 4:56
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    $\begingroup$ Your equation $I = \frac 2 5 mR$ is dimensionally incorrect and thus your equations for $v$ are as well. You can’t add $1$ and $\frac 2 5 R$. $\endgroup$
    – Ghoster
    Commented May 2 at 5:16
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    $\begingroup$ $1+\frac 2 5 R$ is dimensionally incorrect, which is too bad, because I don't want to solve the problem, but I am interesting the scaling symmetry of the problem. $\endgroup$
    – JEB
    Commented May 2 at 5:18
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    $\begingroup$ $\omega=\frac{v}{r}$ should be $\omega=\frac{v}{R}$. $\endgroup$
    – Ghoster
    Commented May 2 at 5:54
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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 $g$.

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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.)

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  • $\begingroup$ Don't objects with greater mass have greater gravitational potential energy, and therefore if they are rolled from the same height, they will gain more kinetic energy? I heard this somewhere, but is kinetic energy completely related to the speed of an object then? $\endgroup$
    – user386598
    Commented May 2 at 2:42
  • $\begingroup$ @user386598 they do gain more kinetic energy, but you can have more kinetic energy without also having more speed - see physics.stackexchange.com/questions/812729/… $\endgroup$
    – Allure
    Commented May 2 at 2:43
  • $\begingroup$ Okay, thank you for the help! $\endgroup$
    – user386598
    Commented May 2 at 3:48
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    $\begingroup$ This answer does not analyze the effect of rolling. $\endgroup$
    – my2cts
    Commented May 2 at 4:31
  • $\begingroup$ @my2cts yes, but rolling adds unneeded complexity w.r.t the original question. $\endgroup$
    – Allure
    Commented May 2 at 5:31

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