A video of this demonstration is available at this link.

You simultaneously release three cylinders of equal mass (~2.7 kg) and equal diameter (~21 cm) but different moments of inertia to roll down an incline. In which order do they reach the end?

Please note that a second, smaller version of this demonstration, which is described below, is available.

If the gate, at left in the photograph, is not set, raise the lever and insert the pin to hold it in place. Set the cylinders against the gate in whichever lanes you would like them. When you pull the pin, this opens the gate, releasing the cylinders simultaneously. Because the cylinders roll down the incline, gravity must not only translate them, but also rotate them. The cylinders all have the same radius, so gravity has the same moment for all of them. The moments of inertia, however, are all different, so their rotational accelerations (and, thus, also their linear accelerations) are different. The cylinder with the greatest moment of inertia travels most slowly down the incline, while the one with the least moment of inertia travels fastest. (The one with the intermediate moment of inertia travels at a rate intermediate between those of the other two.) Thus, the cylinder with its mass concentrated near the center (at front in the photograph) gets to the end first, followed by the cylinder with uniform mass distribution (in the middle), with the hoop (at rear) coming in last. You can have the class try to predict the outcome of the race before you open the gate.

We can analyze this system in two ways. If we consider the dynamics, the total force on the cylinder is that of gravity, acting at the center of mass (which is the center of the cylinder) to accelerate the cylinder down the incline, minus the friction between the contact point of the cylinder and the incline, which opposes the force due to gravity and causes the cylinder to rotate. If we call the angle of the incline above the horizontal

θ, and the mass of the cylinderM, then this givesMa=Mgsinθ-f. The contact point of the cylinder is the full radius,R, from the center, so the torque exerted by the friction isfR=Iα, whereαis the rotational acceleration. The linear acceleration,a, that corresponds to this rotational acceleration equalsαR. , soα=a/R. Substituting and rearranging, we getf=Ia/R^{2}. Once we know whatais, this gives the minimum friction necessary to keep the cylinder rolling without slipping. Substituting this into our first equation givesMa=Mgsinθ-Ia/R^{2}, ora=gsinθ-Ia/MR^{2}, which rearranges toa(1 +I/MR^{2}) =gsinθ, which then givesa=gsinθ/(1 +I/MR^{2}). From this we can see that the linear acceleration of the cylinder depends on its moment of inertia, but not on its mass (theMin the moment of inertia cancels with theMin theMR^{2}of the denominator), and that the greater the moment of inertia, the smaller the acceleration.If we call the height difference between the starting point and the end point of the travel of the cylinder

h, and the distance over which it travelss, then sinθ=h/s, and we can rewrite the equation for the acceleration asa=gh/(1 +I/MR^{2})s. From the equations of linear motion, we know that after the cylinder has rolled over the distances, the square of its (linear) speed,v^{2}, equals 2as, orv= √(2as). This givesv= √[2gh/(1 +I/MR^{2})]. As we should expect, the greater the moment of inertia, the smaller the final velocity of the cylinder.We can also analyze the system in terms of energy. At the beginning, the cylinders have a potential energy of

mgh(and no kinetic energy). If we allow them to roll the entire distance of the ramp (from the gate), so the entire heighth, this potential energy is converted to both linear kinetic energy and rotational kinetic energy, thus:Mgh= (1/2)Mv^{2}+ (1/2)Iω^{2}.ω=v/R, so we haveMgh= (1/2)Mv^{2}+ (1/2)Iv^{2}/R^{2}. Dividing through byMand rearranging a bit givesgh=v^{2}[(1/2)+I/2MR^{2})], orgh=v^{2}[(2MR^{2}+ 2I)/4MR^{2}]. Canceling the 2s in the numerator on the right with the 4 in the denominator and rearranging givesv^{2}= [2ghMR^{2}/(MR^{2}+I)], orv= √[2gh/(1 +I/MR^{2})], which agrees with what we arrived at above, as it should.The cylinder with the greatest acceleration, and thus velocity, at any given point, will win the race. We could also calculate the time,

t, it takes the cylinders to go down the ramp. The equations of motion gives= (1/2)at^{2}, andt= √(2s/a). They also gives= (1/2)(v+_{i}v)_{f}t, and sincev= 0, dropping the subscript, we also have_{i}t= 2s/v. So we could put the expression we obtained earlier for the acceleration,a, intot= √(2s/a) to findt. In all of the following, we find expressions forv; if we would like to knowt, we could divide 2sby the expression we obtain forv.The cylinder in front consists of a 6.25-cm-diameter copper cylinder mounted in the center of two plastic discs, which ride inside an outer rim of plastic. If we call the radius of the copper cylinder

r, then its moment of inertia is (1/2)Mr^{2}. If we treat the (large) cylinder as if most of its mass is in the copper hub, and neglect the mass of the plastic, then this is also the moment of inertia of the whole cylinder, whose radius,R, is 21 cm, as noted above. The equation forvthen becomesv= √[2gh/(1 +Mr^{2}/2MR^{2})], orv= √[2gh/(1 +r^{2}/2R^{2})].r^{2}/R^{2}equals 3.12^{2}/10.5^{2}, or 0.0883, sor^{2}/2R^{2}= 0.0441, andv= √(2gh/1.044^{}), orv= √(1.92gh), which equals 1.38√(gh). If we took the copper cylinder as a point mass,rwould go to zero, and the second term in the denominator would go to zero, givingv= √(2gh), or 1.41√(gh), a difference of about 2 percent. This is a limiting case. If the moment of inertia goes to zero, then all of the energy goes into translation, and none into rotation. Thev= √(2gh) result is what we would obtain for an object sliding down a ramp through heighthwithout any friction between itself and the ramp. While it is not really possible to have an object with all of its mass at the center, it is possible to make one in which most of the mass is near the center, which reduces the moment of inertia enough that most of the potential energy goes into translation and only a small portion goes into rotation, and the result we obtain is fairly close to this limiting case.The cylinder in the middle is a plastic disc of uniform mass, so its moment of inertia is (1/2)

MR^{2}, and we havev= √[2gh/(1 +MR^{2}/2MR^{2})], orv= √[2gh/(3/2^{})], which equals √(4gh/3) or 1.15√(gh).The cylinder at the rear is essentially a hoop. If we consider the difference between the inner and outer radius to be small enough that we may treat them as equal, its moment of inertia is

MR^{2}, and our equation becomesv= √[2gh/(1 +MR^{2}/MR^{2})], orv= √(gh). In the case of our hoop, this difference is about 0.6 cm, the outer radius being 10.5 cm and the inner one 9.9. So the full-blown ratio in the denominator should be (1/2)(M(R_{1}^{2}+R_{2}^{2})/MR_{2}^{2}, or (R_{1}^{2}+R_{2}^{2})/2R_{2}^{2}, which gives (9.9^{2}+ 10.5^{2})/2(10.5^{2}), or (98.0 + 110)/220 = 0.95. This givesv= √(2gh/1.95), orv= √(1.02gh), orv= 1.01√(gh). So the error introduced by not accounting for this is about 1 percent. A hoop with all its mass at radiusR, for which the final velocity is √(gh), represents the opposite limiting case to the one with all the mass at the center. In this case, the second term in the denominator goes to one, making both terms equal, and the energy is essentially equally split between translation and rotation. (If we write the energy expressions explicitly, we find that linear kinetic energy equals (1/2)Mv^{2}, and rotational kinetic energy equals (1/2)Iω^{2}= (1/2)MR^{2}(v/R)^{2}= (1/2)Mv^{2}.) Again, such an object could not really exist, but we can make one in which all the mass is concentrated in a region close enough to the outer diameter that we can come fairly close to this limiting case.So we see that the cylinder with its mass concentrated at its center wins the race, followed by the disc, with the hoop coming last, with their final velocities in the ratio of about 1.4:1.2:1. We also see that our results are independent of the masses of the cylinders. What they do depend on, however, is the way the mass of the cylinders is distributed – mostly concentrated at the center, uniformly distributed, or mostly concentrated at the outside. In addition, if the moment of inertia and friction act at the same radius, since radius cancels out of the equation above, the rates of acceleration and the final velocities of the cylinders do not depend on their radius and, hence, on their size.

A perhaps more traditional version of this demonstration uses a sphere, disc and hoop, all having the same radius. We have a version of this demonstration that uses these shapes as well:

In the photograph above is an incline, a sphere, disc and hoop set having a radius of 50 mm, and a second similar set with a radius of 25 mm. At the right is a stop to prevent the objects from rolling off the table, held with two C-clamps. At center is a ruler, with which you can line up the sphere, disc and hoop of one set, and then release them. The demonstration works on a five-foot long table, as shown, but it is better to use a longer table to allow for greater separation of the objects when they arrive at the stop. If you wished, you could probably carry this demonstration to a remote location, as long as a table was available there, whose length was at least five feet.

The analyses for this system are identical to those above, except that one of the moments of inertia, that of the sphere, is different. While the mass distribution in the sphere is uniform, it is not uniform with respect to a rotational axis through its center, and the resulting moment of inertia is 2

MR^{2}/5. Putting this into the equation above for final velocity givesv= √(2gh/(1 + (2/5))), orv= √(2gh/(7/5)) =v= √(10gh/7) =v= √(1.43gh) = 1.20√(gh). This is close to that for the disc, which is 1.15√(2gh) – about a 4% difference. Though this difference is small, it is enough that by the time the sphere has traveled the length of the ramp, its separation from the disc is noticeable. As mentioned earlier, it is more easily noticeable if you set the demonstration up on a long table, where the objects can travel some distance after they have reached the bottom of the ramp.Also as noted above, what determines an object’s rate of acceleration down the ramp (and thus its final velocity) is not its mass or its size, but only the way its mass is distributed. This is perhaps well illustrated by the fact that while the masses of the cylinders in the large demonstration are all the same, the masses of the objects in this smaller demonstration are all different. For the large set, the sphere is 0.738 kg, the disc is 0.530 kg, and the hoop is 0.211 kg. For the small set, the sphere is 0.116 kg, the disc is 0.108 kg, and the hoop is 0.091 kg. Still, for each set, the orders in which the objects arrive at the end of the ramp are similar to those for the large set of cylinders, and, as indicated by our calculations, the lag between the disc and the cylinder with the copper core is greater than that between the disc and the sphere for either set. The lags between the smaller discs and hoops are slightly different, because the ~6-mm wall thickness represents 0.12 times the 50-mm radius of the large hoop, but 0.24 times that of the smaller hoop. This results in about a 3 percent difference in their final velocities.

References:1) Resnick, Robert and Halliday, David.

Physics, Part One, Third Edition(New York: John Wiley and Sons, 1977), pp. 241, 250-252.