In a common demonstration an instructor \"races\" various roundobjects by releasing them from rest at the top of an inclined planeand letting them roll down the plane. Before the objects arereleased the students guess which object will win.
A) Find the ratio of the final speeds for a solid sphere and asolid cylinder.
B) Assuming that the masses of the two cylinders are the same,what is the ratio of the rotational kinetic energy of the solidsphere to the solid cylinder at the bottom of the ramp?
SET UP We use conservation of energy, ignoringrolling friction and air drag. If the objects roll withoutslipping, then no work is done by friction and the total energy isconserved. Each object starts from rest at the top of an inclinewith height hh, so Ki=0Ki=0, Ui=mghUi=mgh, and Uf=0Uf=0 for each.The final kinetic energy is a combination of translational androtational energies:
Kf=12mvcm2+12Icmω2Kf=12mvcm2+12Icmω2
Both vcmvcm and ωω are unknown, but if we assume that theobjects roll without slipping, these two quantities areproportional. When an object with radius RR has rotatedthrough one complete revolution (2ππ radians), it has rolled adistance equal to its circumference (2πR)(2πR). Thus the distancetraveled during any time interval ΔtΔt is RR times theangular displacement during that interval, and it followsthat vcm=Rωvcm=Rω.
SOLVE For the cylindrical shell,Ishell=MR2Ishell=MR2. Conservation of energy then results in
0+Mghvcm====12Mvcm2+12Icmω212Mvcm2+12(MR2)(vcm/R)212Mvcm2+12Mvcm2=Mvcm2gh−−√0+Mgh=12Mvcm2+12Icmω2=12Mvcm2+12(MR2)(vcm/R)2=12Mvcm2+12Mvcm2=Mvcm2vcm=gh
For the solid cylinder, Isolid=12MR2Isolid=12MR2 and thecorresponding equations are:
0+Mghvcm====12Mvcm2+12Icmω212Mvcm2+12(12MR2)(vcm/R)212Mvcm2+14Mvcm2=34Mvcm243gh−−−√0+Mgh=12Mvcm2+12Icmω2=12Mvcm2+12(12MR2)(vcm/R)2=12Mvcm2+14Mvcm2=34Mvcm2vcm=43gh
We see that the solid cylinder's speed at the bottom of the hillis greater than that of the hollow cylinder by a factor of43−−√43.
We can generalize this result in an elegant way. We note thatthe moments of inertia of round objects about axes through theircenters of mass can be expressed as Icm=βMR2Icm=βMR2, where ββ is apure number between 0 and 1 that depends on the shape of the body.For a thin-walled hollow cylinder, β=1β=1; for a solid cylinder,β=12β=12; and so on. From conservation of energy,
0+Mghvcm====12Mvcm2+12Icmω212Mvcm2+12(βMR2)(vcm/R)212(1+β)Mvcm22gh1+β−−−√0+Mgh=12Mvcm2+12Icmω2=12Mvcm2+12(βMR2)(vcm/R)2=12(1+β)Mvcm2vcm=2gh1+β
REFLECT This is a fairly amazing result; thefinal speed of the center of mass doesn't depend on either the massMM of the body or its radius RR. All uniform solid cylinders havethe same speed at the bottom, even if their masses and radii aredifferent, because they have the same ββ. All solid spheres havethe same speed, and so on. The smaller the value of ββ, the fasterthe body is moving at the bottom (and at any point on the waydown). Small-ββ bodies always beat large-ββ bodies because theyhave less kinetic energy tied up in rotation and have moreavailable for translation. For the hollow cylinder (β=1β=1), thetranslational and rotational energies at any point are equal, butfor the solid cylinder (β=12)(β=12), the rotational energy at anypoint is half the translational energy. Considering the values ofββ for round objects for an axis through the center of mass, we seethat the order of finish is as follows: any solid sphere, any solidcylinder, any thin spherical shell, and any thin cylindricalshell.
