Question

In a common demonstration an instructor "races" various round objects by releasing them from rest at the top of an inclined plane and letting them roll down the plane. Before the objects are released the students guess which object will win.

A) Find the ratio of the final speeds for a solid sphere and a solid cylinder.

B) Assuming that the masses of the two cylinders are the same, what is the ratio of the rotational kinetic energy of the solid sphere to the solid cylinder at the bottom of the ramp?

SET UP We use conservation of energy, ignoring rolling friction and air drag. If the objects roll without slipping, then no work is done by friction and the total energy is conserved. Each object starts from rest at the top of an incline with height hh, so Ki=0Ki=0, Ui=mghUi=mgh, and Uf=0Uf=0 for each. The final kinetic energy is a combination of translational and rotational energies:

Kf=12mvcm2+12Icmω2Kf=12mvcm2+12Icmω2

Both vcmvcm and ωω are unknown, but if we assume that the objects roll without slipping, these two quantities are proportional. When an object with radius RR has rotated through one complete revolution (2ππ radians), it has rolled a distance equal to its circumference (2πR)(2πR). Thus the distance traveled during any time interval ΔtΔt is RR times the angular displacement during that interval, and it follows that 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 the corresponding 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 hill is greater than that of the hollow cylinder by a factor of 43−−√43.

We can generalize this result in an elegant way. We note that the moments of inertia of round objects about axes through their centers of mass can be expressed as Icm=βMR2Icm=βMR2, where ββ is a pure 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; the final speed of the center of mass doesn't depend on either the mass MM of the body or its radius RR. All uniform solid cylinders have the same speed at the bottom, even if their masses and radii are different, because they have the same ββ. All solid spheres have the same speed, and so on. The smaller the value of ββ, the faster the body is moving at the bottom (and at any point on the way down). Small-ββ bodies always beat large-ββ bodies because they have less kinetic energy tied up in rotation and have more available for translation. For the hollow cylinder (β=1β=1), the translational and rotational energies at any point are equal, but for the solid cylinder (β=12)(β=12), the rotational energy at any point is half the translational energy. Considering the values of ββ for round objects for an axis through the center of mass, we see that the order of finish is as follows: any solid sphere, any solid cylinder, any thin spherical shell, and any thin cylindrical shell.

Answer 1

A) Find the ratio of the final speeds for a solid sphere and a solid cylinder.

We know that P.E at the top gets converted into K.E at the bottom, and this K.E has a translational as well as a rotational component.

For solid sphere,

using conservation of energy

mgh = 1/2mv² + 1/2Iw²

mgh = 1/2mv² + 1/2 * 2/5 * m * r² * v² / r²

gh = 1/2v² + 1/5v²

gh = 7/10v²

v = sqrt ( 10gh / 7)

Now,

For solid cylinder

mgh = 1/2mv² + 1/2 * 1/2 * m * r² * v² / r²

gh = 1/2v² + 1/4v²

gh = 3/4 v²

so,

v = sqrt ( 4gh / 3)

so,

ratio is

v_{solid sphere} / v_{solid cylinder} = sqrt ( 10gh / 7) / sqrt ( 4gh / 3)

v_{solid sphere} / v_{solid cylinder} = 1.035

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B) Assuming that the masses of the two cylinders are the same, what is the ratio of the rotational kinetic energy of the solid sphere to the solid cylinder at the bottom of the ramp

For solid sphere, we have

1/2 * m * 10gh / 7 + K.E = mgh

5mgh / 7 + K.E = mgh

so,

K.E = 0.2857mgh

and

for solid cylinder, we have

1/2 * m * 4gh / 3 + K.E = mgh

2mgh / 3 + K.E = mgh

K.E = 0.333 mgh

so,

ratio = 0.857