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Physics 111: Lecture 13, Pg 1

Physics 111: Lecture 13Physics 111: Lecture 13

Today’s AgendaToday’s Agenda Potential Energy & Force

Systems of Particles

Center of mass

Velocity and acceleration of the center of mass

Dynamics of the center of mass Linear Momentum

Example problems


Potential Energy & ForcePotential Energy & Force For a conservative force we define the potential energy

function:

Therefore:

Consider some potential energy functions we know, and find the forces:

Spring:

Gravity near earth:

Newton’s Gravity:

CdxFU

dxFWU2

1

x

x

U kx Cx 12

2

U mgy Cy

F dUdx

kxx

F dUdy

mgy

U GMmR

CR F dUdR

GMmRR 2

It’s true!!It’s true!!

dxdUF


Potential Energy DiagramsPotential Energy Diagrams

Consider a block sliding on a frictionless surface, attached to an ideal spring.

0 x

U

mx

U kxs 12

2



Consider a block sliding on a frictionless surface, attached to an ideal spring.

F = -dU/dx = -slope

mx

x 0 x

U

FF

FF

U kxs 12

2



The potential energy of the block is the same as that of an object sliding in a frictionless “bowl”:

Ug = mgy = 1/2 kx2 = Us

0 x

U

m

2xgm2ky

is the height of an object in the bowl at position x


EquilibriumEquilibrium

F = -dU/dx = -slope So F = 0 if slope = 0. This is the case at the

minimum or maximum of U(x).

This is called an equilibrium position.If we place the block at

rest at x = 0, it won’t move.

0 x

U

mx



If small displacements from the equilibrium position result in a force that tends to move the system back to its equilibrium position, the equilibrium is said to be stable.

This is the case if U is a minimum at the equilibrium position.

In calculus language, the equilibrium is stable if the curvature (second derivative) is positive.

0 x

U

mx

FF

FF



Suppose U(x) looked like this:

This has two equilibrium positions, one is stable (+ curvature) and one is unstable (- curvature).

Think of a small object sliding on the U(x) surface:If it wants to keep sliding when you give it a little push, the

equilibrium is unstable.If it returns to the equilibrium position when you give it a little

push, the equilibrium is stable.If the curvature is zero (flat line) the equilibrium is neutral.

U

x0

stable

unstable

neutral

Balancecone

Birds


System of ParticlesSystem of Particles

Until now, we have considered the behavior of very simple systems (one or two masses).

But real life is usually much more interesting! For example, consider a simple rotating disk.

An extended solid object (like a disk) can be thought of as a collection of parts. The motion of each little part depends on where it is in the object!


System of Particles: Center of MassSystem of Particles: Center of Mass

How do we describe the “position” of a system made up of many parts?

Define the Center of MassCenter of Mass (average position):For a collection of N individual pointlike particles whose

masses and positions we know:

Rr

CM

i ii

N

ii

N

m

m

1

1

y

x

rr1

m1

rr3

rr2

rr4m4

m2

m3

RRCM

(In this case, N = 4)

Icetable



If the system is made up of only two particles:

Rr r r

CM

i ii

N

ii

N

m

m

m mm m

1

1

1 1 2 2

1 2

y

x

rr2rr1

m1

m2

RRCM

rr2 - r r1

m m m

m m1 2 1 2 2 1

1 2

r r r

R r r rCMmM

12

2 1

where M = m1 + m2

So:




y

x

rr2rr1

m1

m2

RRCM

rr2 - r r1

where M = m1 + m2

+

R r r rCMmM

12

2 1

If m1 = m2

the CM is halfway betweenthe masses.

R r r rCM 1 2 112




y

x

rr2rr1

m1

m2

RRCM

rr2 - r r1

where M = m1 + m2

+

If m1 = 3m2

the CM is now closer tothe heavy mass.

R r r rCMmM

12

2 1

R r r rCM 1 2 114



The center of mass is where the system is balanced!Building a mobile is an exercise in finding centers of mass.

m1m2

+m1 m2

+

Baton



We can consider the components of RRCM separately:

( , , ) , ,X Y Zm x

M

m y

M

m z

MCM CM CMi ii i ii i ii

y

x

rr1

m1

rr3

rr2

rr4m4

m2

m3

RRCM

(In this case, N = 4)


Example Calculation:Example Calculation: Consider the following mass distribution:

(24,0)

12m4

24m12)m2(0mM

xmX i iiCM

6m4

0m12)m2(0mM

ymY i iiCM

(0,0)

(12,12)

m

2m

m

RCM = (12,6)


Rr r

CMdmdm

dmM


For a continuous solid, we have to do an integral.

y

x

dm

rrwhere dm is an infinitesimal mass element.



We find that the Center of Mass is at the “center” of the object.

y

x

RRCM



The location of the center The location of the center of mass is an intrinsic of mass is an intrinsic property of the object!!property of the object!!(it does not depend on whereyou choose the origin or coordinates whencalculating it).

y

xRRCM

We find that the Center of Mass is at the “center” of the object.


System of Particles: Center of MassSystem of Particles: Center of Mass We can use intuition to find the location of the center of mass for symmetric objects that have uniform density: It will simply be at the geometrical center !

+CM

+ +

+ + +



The center of mass for a combination of objects is the average center of mass location of the objects:

+

+y

x

RR2

RR1

RRCM

m1

m2

+

21

2211CM

N

1ii

N

1iii

CM

mmmm

m

m

RRR

RR

RR2 - R R1

so if we have two objects:

122

1 Mm RRR

Pisa

Bottle


Lecture 13, Lecture 13, Act 1Act 1Center of MassCenter of Mass

The disk shown below (1) clearly has its CM at the center. Suppose the disk is cut in half and the pieces arranged as shown in (2):

Where is the CM of (2) as compared to (1)?

(a)(a) higher (b)(b) lower (c)(c) same

(1) (2)

XCM


Lecture 13, Lecture 13, Act 1Act 1SolutionSolution

The CM of each half-disk will be closer to the fat end than to the thin end (think of where it would balance).

(1) (2)

X

X

The CM of the compound object will be halfway between theCMs of the two halves.

XXCM

This is higher than the CM of the disk



The center of mass (CM) of an object is where we can freely pivot that object.

Gravity acts on the CM of an object (show later)

If we pivot the objectsomewhere else, it willorient itself so that theCM is directly below the pivot.

This fact can be used to findthe CM of odd-shaped objects.

+CM

pivot

+ CM

pivot

+

pivot

CM

mg

Doublecone



Hang the object from several pivots and see where the vertical lines through each pivot intersect!

pivot

pivotpivot

+CM

The intersection point must be at the CM.

Oddshapes


Lecture 13, Lecture 13, Act 2Act 2Center of MassCenter of Mass

An object with three prongs of equal mass is balanced on a wire (equal angles between prongs). What kind of equilibrium is this position?

a) stable

b) neutral

c) unstable

3 prongedobject

Fork, spoon,and match



The center of mass of the object is at its center and is initially directly over the wire

If the object is pushed slightly to the left or right, its center of mass will not be above the wire and gravity will make the object fall off

++ CM ++ CM

mgmg

(front view)


Consider also the case in which the two lower prongs have balls of equal mass attached to them:

++ CM

mg


In this case, the center of mass of the object is below the wire

mg

++ CM

When the object is pushed slightly, gravity provides a restoring force, creating a stable equilibrium


Velocity and AccelerationVelocity and Accelerationof the Center of Massof the Center of Mass

If its particles are moving, the CM of a system can also move. Suppose we know the position rri of every particle in the system

as a function of time.

R rCM i ii

N

Mm

11

V R r vCMCM

ii

i

N

i ii

Nddt M

m ddt M

m

1 11 1

So:

A V v aCMCM

ii

i

N

i ii

Nddt M

m ddt M

m

1 11 1

And:

The velocity and acceleration of the CM is just the weighted average velocity and acceleration of all the particles.

N

1iimM


Linear Momentum:Linear Momentum:

Definition: Definition: For a single particle, the momentum pp is defined as:

Newton’s 2nd Law:

FF = maa

dtm

pp = mvv(pp is a vector since vv is a

vector).

So px = mvx etc.

dv )m(dtd v

dtdF p

Units of linear momentum are kg m/s.



P p v ii

N

i ii

Nm

1 1

For a system of particles the total momentum P is the vector sum of the individual particle momenta:

m Mi ii

N

CMv V

1But we just showed that

P VM CMSo

N

1iiiCM m

M1 vV



So the total momentum of a system of particles is just the total mass times the velocity of the center of mass.

Observe:

We are interested in so we need to figure out

ddt

M ddt

M mCMCM i i

ii net

i

P V A a F ,

ddtP Fi net

i,

P VM CM



Suppose we have a system of three particles as shown. Each particle interacts with every other, and in addition there is an external force pushing on particle 1.

m1

m3

m2

FF13

FF31 FF32

FF23

FF21FF12

FF1,EXT

EXT1

3231

2321

EXT11213i NETi

,

,,

FFFFF

FFFF

(since the other forces cancel in pairs...Newton’s 3rd Law)

All of the “internal” forces cancel !!Only the “external” force matters !!



Only the total external force matters!

m1

m3

m2

FF1,EXT

EXTNETi EXTidtd

,, FFP

CMEXTNET Mdtd APF ,

Which is the same as:

Newton’s 2nd law applied to systems!


Center of Mass Motion: RecapCenter of Mass Motion: Recap

We have the following law for CM motion:

This has several interesting implications:

It tells us that the CM of an extended object behaves like a simple point mass under the influence of external forces: We can use it to relate FF and AA like we are used to doing.

It tells us that if FFEXT = 0, the total momentum of the system can not change.The total momentum of a system is conserved if there are no

external forces acting.

F P AEXT CMddt

M

Pork chop

Pendulum


Example: Astronauts & RopeExample: Astronauts & Rope Two astronauts at rest in outer space are connected by a light rope. They begin to pull towards each other. Where

do they meet?

M = 1.5m m


Example: Astronauts & Rope...Example: Astronauts & Rope...

They start at rest, so VCM = 0. VCM remains zero because

there are no external forces. So, the CM does not move! They will meet at the CM.

Finding the CM:

M = 1.5m m

CM

L

If we take the astronaut on the left to be at x = 0:

x M m(LM m

m(Lm

Lcm

( ) ) )

.0

2 525

x=0 x=L


Lecture 13, Lecture 13, Act 3Act 3Center of Mass MotionCenter of Mass Motion

A man weighs exactly as much as his 20 foot long canoe. Initially he stands in the center of the motionless canoe, a distance of 20 feet from shore. Next he walks toward the shore until he gets to the end of the canoe.

What is his new distance from the shore. (There no horizontal force on the canoe by the water).

20 ft

? ft

20 ft

(a)(a) 10 ft (b)(b) 15 ft

(c)(c) 16.7 ft

before

after



XX

Since the man and the canoe have the same mass, the CM of the man-canoe system will be halfway between the CM of the man and the CM of the canoe.

x 20 ft

Initially the CM of the system is 20 ft from shore.

CM of system



Since there is no force acting on the canoe in the x-direction, thelocation of the CM of the system can’t change!

20 ftx

XX

CM of system

Therefore, the man ends up 5 ft to the left of the system CM, and the center of the canoe ends up 5 ft to the right.

10 ft

5 ft

15 ft

He ends up moving 5 ft toward the shore (15 ft away).


Recap of today’s lectureRecap of today’s lecture

Systems of particles (Text: 8-1)

Center of mass (Text: 8-1 & 12-6)

Velocity and acceleration of the center of mass (Text:8-3)

Dynamics of the center of mass (Text: 8-3 to 8-4)Linear Momentum

Example problems

Look at textbook problems Look at textbook problems Chapter 8: # 3, 7, 17, 29, 35, 77, 111

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