In a common lecture demonstration, an instructor \"races\" various round objects
ID: 1352690 • Letter: I
Question
In a common lecture 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 (Figure 1) . Before the objects are released, the students guess which object will win. How much faster is the solid cylinder moving at the bottom of the ramp than the hollow cylinder?
SOLUTION
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 h, so Ki=0, Ui=mgh, and Uf=0 for each. The final kinetic energy is a combination of translational and rotational energies:
Kf=12mvcm2+12Icm2
Both vcm and are unknown, but if we assume that the objects roll without slipping, these two quantities are proportional. When an object with radius R has rotated through one complete revolution (2 radians), it has rolled a distance equal to its circumference (2R). Thus the distance traveled during any time interval t is R times the angular displacement during that interval, and it follows that vcm=R.
SOLVE For the cylindrical shell, Ishell=MR2. Conservation of energy then results in
0+Mghvcm====12Mvcm2+12Icm212Mvcm2+12(MR2)(vcm/R)212Mvcm2+12Mvcm2=Mvcm2gh
For the solid cylinder, Isolid=12MR2 and the corresponding equations are:
0+Mghvcm====12Mvcm2+12Icm212Mvcm2+12(12MR2)(vcm/R)212Mvcm2+14Mvcm2=34Mvcm243gh
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.
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=MR2, where is a pure number between 0 and 1 that depends on the shape of the body. For a thin-walled hollow cylinder, =1; for a solid cylinder, =12; and so on. From conservation of energy,
0+Mghvcm====12Mvcm2+12Icm212Mvcm2+12(MR2)(vcm/R)212(1+)Mvcm22gh1+
REFLECT This is a fairly amazing result; the final speed of the center of mass doesn't depend on either the mass M of the body or its radius R. All uniform solid cylinders have the same speed at the bottom, even if their masses and radii are different, because they have the same . All solid spheres have the same speed, and so on. The smaller the value of , the faster the body is moving at the bottom (and at any point on the way down). Small- bodies always beat large- bodies because they have less kinetic energy tied up in rotation and have more available for translation. For the hollow cylinder (=1), the translational and rotational energies at any point are equal, but for the solid cylinder (=12), the rotational energy at any point is half the translational energy. Considering the values of 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.
Part A - Practice Problem:
Find the ratio of the final speeds for a solid sphere and a solid cylinder.
Express your answer to four significant figures.
Part B - Practice Problem:
What is the ratio of the rotational kinetic energy of the solid sphere to the solid cylinder at the bottom of the ramp?
Express your answer to three significant figures.
Explanation / Answer
Given data:
solid cylinder
moment of inertia=(1/2)(Mr2)
hallow cylinder
moment of inertia=mr2
Part A) from the conservation of energy
(kinetic energy+potentiial energy)initial=(kinetic energy+potentiial energy)final
released from rest ,so initial K.E=0
finally the object reaches the bottom,so final potential energy=0
K.Efinal=K.Erotational+K.Etranslational
K.Efinal=(1/2)I(omega)2+(1/2)mv2
as we know that v=r*omega
omega=v/r
for solid cylinder:
K.Efinal=(1/2)(1/2*mr2)(v/r)2+(1/2)mv2
K.Efinal=(3/4)mv2
for hallow hallow cylinder
K.Efinal=(1/2)(mr2)(v/r)2+(1/2)mv2=mv2
substitute in conservation of energy
mgh=(3/4)mvsolid2 ------------(eq.1)
mgh=mvhallow2----------------(eq2)
eq1/eq2--------vsolid/vhallow=4/3
PartB)
K.Esphere/K.Ecylinder=(1/2)I(omega)2/(1/2)I(omega)2
K.Esphere/K.Ecylinder=Isphere/Icylinder
K.Esphere/K.Ecylinder=(2/5)mr2/(1/2)mr2
K.Esphere/K.Ecylinder=4/5