# Moment of Inertia Race [closed]

Wikipedia shows an example of how to the moment of inertia determines how fast a rotating object rolls down an incline plane. the race Includes: spherical shell, solid sphere, cylindrical ring. solid cylinder. https://en.wikipedia.org/wiki/Moment_of_inertia#/media/File:Rolling_Racers_-_Moment_of_inertia.gif (No-Friction). My questions are; How is it that the solid sphere hits the bottom first, while the cylindrical ring, finishes last? I would like a conceptual insight on to why this is,(gifs or animations are welcome.),of why the velocity of the center of masses are different because of different rotational inertia. What would happen if the race included a solid cube that slid down the same path. Would the solid cube finish last?

• Conversion od potential energy to kinetic energy is you friend here.
– Gert
Commented Mar 23, 2019 at 3:21
• Your question about the solid cube is not answerable without knowing the properties of friction between the cube and the incline. But I would edit this to be one question. You should avoid asking multiple questions in a single post. Commented Mar 23, 2019 at 3:29

A cylindrical ring rolls the slowest because its mass is farthest from the axis of rotation. A solid sphere rolls the fastest because it has a lot of its mass close to the axis of rotation. When mass is far from the axis of rotation, the object is harder to rotate; when close, easier. The moment of inertia simply measures the spatial distribution of the mass.

The short answer: The more of mass is distributed at the object circumference, the more energy is spent on the object rotation and less on the object linear speed. So the speed decreases in the order of the list just below, as with higher momentum of inertia, more object segments move by higher speeds, up to $$2v$$.

The moment of inertia $$I$$ increases in order

• solid sphere $$I = 2/5.m.r^2$$
• solid cylinder $$I = 1/2.m.r^2$$
• spherical shell $$I = 2/3.m.r^2$$
• cylindrical shell $$I = m.r^2$$

because more and more mass shifts toward the radius distance from the rotation axis and more and more energy is needed to bring them to the same rotation frequency $$f$$.

Their kinetic energy provided by their potential energy is distributed between the translational energy and rotation energy.

$$E_p = E_t + E_r$$ $$m.g.h = 0.5.m.v^2 + 0.5.I.\omega^2$$ $$\omega = 2.pi.f = v/r$$ $$m.g.h = 0.5.m.v^2 + 0.5.I.v^2/r^2$$ $$v = \sqrt{ \frac{2.m.g.h}{ m + I/r^2 }}$$

If we substitute $$I$$ by the expression for the particular shape,

the absolute winner is :

The zero friction sliding cube.

The winner in the rolling category is:

The solid sphere

• zero friction cube $$v = \sqrt{ 2.g.h }$$
• solid sphere $$v = \sqrt{10/7.g.h }$$
• solid cylinder $$v = \sqrt{ 4/3.g.h}$$
• spherical shell $$v = \sqrt{6/5.g.h }$$
• cylindrical shell $$v = \sqrt{g.h }$$

You can notice that the speed does not depend on the mass nor the radius.

You can see the higher the moment of inertia $$I$$ is, the lower is the speed, as there is lower translational energy and higher rotational energy.

If a sliding cube with zero friction was involved, it would be the fastest, as it would not rotate, so all energy goes translational.

Edit: With real nonzero friction, the cube can have a wide range of accelerations and final speeds, so it can still be first, or it can even never leave the start. Similarly, things would get more complicated with low friction, when rolling is combined with sliding. But the order should remain the same, unless a coefficient of friction strongly depends on pressure.

• "spherical cylinder" Cylindrical shell? Commented Mar 23, 2019 at 3:59
• Hehe, sure. :-) Commented Mar 23, 2019 at 4:03

Since each object has the same diameter, each must turn the same number of times to make it down the hill. It takes kinetic energy to increase the rotation of an object. This energy must come from the kinetic energy of its fall. Since the sphere has the least mass about its radius, it takes the least energy from its fall to rotate. Conversely, the cylindrical ring has the most mass about its radius, so it takes the most energy from its fall to rotate, and thus falls the slowest.

• Alas for a rather nice argument the race comes out the same when the various balls and wheels have different sizes—so long as the multiplier on their moments of inertia remain the same. That happens be because of the no-slipping condition and the consequent coupling of translational and angular velocity. Commented Mar 23, 2019 at 3:58