classical mechanics lecture 10 · pdf fileclassical mechanics lecture 10 ... forces come in...

Classical Mechanics Lecture 10

Today’sConcept: CenterofMass

! Findingit! Usingit

MechanicsLecture10,Slide1

Your Comments

Iwouldlikemoredayswherewecangooverdifficultproblemsinclass.ItwasreallyhelpfulhavingtheTA'stohelpusout,andIwouldlikeifwecoulddoitmoreoIen.Evenifwecouldspendmaybe30minutesofclassdoingproblems,itmadeitmucheasiertofinishthehomework,andhelpsuspreparefortheexam.! Realworldexample

Willweneedtoknowhowtouseintegralstofindthecenterofmass?! Orwillrcm=(m1r1...mnrn)/mtotalbeenough?

Idliketohaveadoughnutinmycenterofmass

whydoesn'tthemooncrashintotheearth,causingwidespreadpanicanddestrucVon?

goodlecturetheexamplesweregood.


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Center of Mass

This is it!

Ayummyglazeddoughnutisshownbelow.WhereisthecenterofmassofthisfantasVcculinarydelight?

A)Inthecenterofthehole.B)Anywherealongtheblue

dashedlinegoingthroughthesolidpartofthedough.

C)Thecenterofmassisnotdefinedincaseswherethereismissingmass.

x

TheyummydoughnutisroundandhasequalweightdistribuVonalloverthedoughnut,therefore,thecenterofmassmustbeinthemiddle

CheckPoint


x=Lx=0

Example

AuniformsVckoflengthLandmassM.


dm

=1L

12

x

2!

L

0


Cube Example


Cube Example: Using Volume Mass Density


Cube Example: CM Calculation


Center of Mass for System of Objects

ThediskshowninCase1clearlyhasitsCMatthecenter.SupposethediskiscutinhalfandthepiecesarrangedasshowninCase2

Inwhichcaseisthecenterofmasshighest?

Clicker Question


Case1 Case2

xCM

x

A)Case1B)Case2C)same

h =4r3�<

r2

Clicker Question

ThediskshowninCase1clearlyhasitsCMatthecenter.SupposethediskiscutinhalfandthepiecesarrangedasshowninCase2

Inwhichcaseisthecenterofmasshighest?

Case1 Case2

xCM

x

x


A)Case1B)Case2C)same

xx

h{

Top half hasmore mass

Bottom half hasless mass

xmiddle

centre of mass is here

Ifthetotalforceiszero,theCM won’taccelerate


Kinematic Quantities

Clicker Question

A)0 0 B)0 H/2C)0 H/3 D)H/4 H/4E)H/4 0

XCM YCM

x

y

H

ThreeVnyequalmassesareplacedatthecornersofanequilateraltriangle.Whenthemassesarereleased,theyaXractandquicklymovetoasinglepoint.Whatarethecoordinatesofthatpoint?


C)0 H/3

XCM YCM

(symmetry)

x

y

H

ThreeVnyequalmassesareplacedonahorizontalfricVonlesssurfaceatthecornersofanequilateraltriangle.Whenthemagnetsarereleased,theyaXractandquicklyslidetoasinglepoint.Whatarethecoordinatesofthatpoint?


= −1 m

Homework Problem


= 0.975 m

Homework Problem


m5 CMmovestowardm5

ABCD

cm


m5 CMmovestowardm5

cm

ABC


Clicker Question

Twopucksofequalmass,onafricVonlesstable,arebeingpulledatdifferentpointswithequalforces.Whichonegetstotheendofthetablefirst?

A)Puck1 B)Puck2 C)Same


2)

M

M

T

T

1)

a1

a2

Twopucksofequalmass,onafricVonlesstable,arebeingpulledatdifferentpointswithequalforces.Whichonegetstotheendofthetablefirst?

C)Same

SAME


2)

M

M

T

T

1)

Clicker Question

Pulling a disk

Acylinderrestsonasheetofpaperonatable.Youpullonthepapercausingthepapertoslidetotheright.ThisresultsinthecylinderrollingleIwardrelaVvetothepaper.

Howdoesthecenterofmassofthecylindermoverela&vetothetable?

A)LeIwards

B)No-wards

C)Rightwards

The Center of Mass S E C T I O N 5 - 5 | 155

in the system by some other particle in the system) and others are external forces(exerted on a particle in the system by something external to the system). Thus,

5-26

According to Newton’s third law, forces come in equal and opposite pairs.Therefore, for each internal force acting on a particle in the system there is an equaland opposite internal force acting on some other particle in the system. When wesum all the internal forces, each third-law force pair sums to zero, so Equation 5-22 then becomes

5-27

NEWTON’S SECOND LAW FOR A SYSTEM

That is, the net external force acting on the system equals the total mass M timesthe acceleration of the center of mass Thus,

The center of mass of a system moves like a particle of mass under the influence of the net external force acting on the system.

This theorem is important because it describes the motion of the center of massfor any system of particles: The center of mass moves exactly like a single point par-ticle of mass M acted on by only the external forces. The individual motion of a par-ticle in the system is typically much more complex and is not described by Equation5-27. The hammer thrown into the air in Figure 5-48 is an example. The only exter-nal force acting is gravity, so the center of mass of the hammer moves in a simpleparabolic path, as would a point particle. However, Equation 5-27 does not describethe rotational motion of the head of the hammer about the center of mass.

If a system has a zero net external force acting on it, then . In this casethe center of mass either remains at rest or moves with constant velocity. The in-ternal forces and motion may be complex, but the motion of the center of mass issimple. Further, if the component of the net next force in a given direction, say thex direction, remains zero, then remains zero and remains constant. An ex-ample of this is a projectile in the absence of air drag. The net external force on theprojectile is the gravitational force. This force acts straight downward, so its com-ponent in any horizontal direction remains zero. It follows that the horizontalcomponent of the velocity of the center of mass remains constant.

vcmxacmx

aScm ! 0

M ! gmiaScm.

FS

net ext ! ai

FS

iext !MaScm

g FS

i int ! 0.

MaScm ! ai

FS

i int " ai

FS

iext

CONCEPT CHECK 5-2

A cylinder rests on a sheet ofpaper on a table (Figure 5-49). Youpull on the paper causing thepaper to slide to the right. Thisresults in the cylinder rolling left-ward relative to the paper. Howdoes the center of mass of thecylinder move relative to the table?

✓

PapercmM

F I G U R E 5 - 4 9

Example 5-16 An Exploding Projectile

A projectile is fired into the air over level ground on a trajectory thatwould result in it landing 55 m away. However, at its highest point itexplodes into two fragments of equal mass. Immediately following theexplosion one fragment has a momentary speed of zero and then fallsstraight down to the ground. Where does the other fragment land?Neglect air resistance.

PICTURE Let the projectile be the system. Then, the forces of the ex-plosion are all internal forces. Because the only external force acting onthe system is that due to gravity, the center of mass, which is midwaybetween the two fragments, continues on its parabolic path as if therehad been no explosion (Figure 5-50).

2m

m m

m

cm

cmm

F I G U R E 5 - 5 0

Rai Stones

MoneyinYap

CheckPoint

Twoobjects,onehavingtwicethemassoftheother,areiniVallyatrest.Twoforces,onetwiceasbigastheother,actontheobjectsinoppositedirecVonsasshown.

A)aCM = F/M totherightB)aCM = F/(3M) totherightC)aCM = 0D)aCM = F/(3M) totheleIE)aCM = F/M totheleI

WhichofthefollowingstatementsabouttheacceleraVonofthecenterofmassofthesystemistrue:


F2F M2M

WhichofthefollowingstatementsabouttheacceleraVonofthecenterofmassofthesystemistrue:

C)F=Ma.The2Mandthe2F wouldcanceleachotherout,makingthebiggerobjectaccelerateatthesamerateasthesmallerball.Therefore,theacceleraVonwouldbezero.

D)AcceleraVonisequaltothenetforcedividedbythetotalmass.Thenetforceis FtotheleIandthetotalmassis3M

E)ACM = F total/M total = 3F/3M = F/M

C)aCM = 0D)aCM = F/(3M) totheleIE)aCM = F/M totheleI


F2F M2M

CheckPoint

CheckPoint

TwoguyswhoweightthesameareholdingontoamasslesspolewhilestandingonhorizontalfricVonlessice.IftheguyontheleIstartstopullonthepole,wheredotheymeet?

A)−3 m B)0 m C)3 m

−3 m 0 m 3 m


Clicker Question

A)−3 mB)−1 mC)0 mD)1 mE)3 m

Alargeskinnyguywithmass2MandasmallerguywithmassMareholdingontoamasslesspolewhilestandingonfricVonlessice,asshownbelow.IftheliXleguypullshimselftowardthebigguy,wherewouldtheymeet?


−3 m 0 m 3 m

2M M

A)−3 mB)−1 mC)0 mD)1 mE)3 m

Alargeskinnyguywithmass2MandasmallerguywithmassMareholdingontoamasslesspolewhilestandingonfricVonlessice,asshownbelow.IftheliXleguypullshimselftowardthebigguy,wherewouldtheymeet?


−3 m 0 m 3 m

2M M

m m = −1 m

classical mechanics lecture 10 · pdf fileclassical mechanics lecture 10 ... forces come in...

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