Chapter 09 Center of Mass and Linear Momentumastro1.panet.utoledo.edu/~khare/teaching/phys2130h...Chapter 09 Center of Mass and Linear Momentum • Center of mass: The center of mass

Chapter 09 Center of Mass and Linear Momentum

• Center of mass: The center of mass of a body or a system of bodies is the point that moves as if all of the mass are concentrated there and all external forces are applied there.

• Note that HRW uses “com” but I will use “c.m.” because “c.m.” is more standard notation for the “center of mass”.

System of Particles

21

2211.m.c mm

xmxmx ++≡

Consider two masses m1 at x = x1 and m2 at x2.

• Center of mass for a system of n particles:

∑=

=n

1iii.m.c xm

M1x ∑

=

=n

1iii.m.c zm

M1z∑

=

=n

1iii.m.c ym

M1y

∑=

=n

1iimM

kzjyixr .m.c.m.c.m.c.m.c ++=v

∑=

=n

1iii.m.c rm

M1r vv

• M is just the total mass of the system

• Therefore:

• Using vectors, we have:

• For a solid body, we can treat it as a continuous distribution of matter dm

• If the object has uniform density, dm = (M/V)dV

• If an object has a point, a line or a plane of symmetry, the center of mass of such an object then lies at that point, on that line or in that plane.

∫= dVyV1y .m.c ∫= dVz

V1z .m.c∫= dVx

V1x .m.c

∫= dmzM1z .m.c∫= dmx

M1x .m.c ∫= dmy

M1y .m.c

• Sample 9-2: the figure shows a uniform metal plate P of radius 2R from which a disk of radius R has been stamped out (removed). Using the x-y coordinate system shown, locate the center of mass of the plate.

Notice that the object is symmetric about the x-axis, so yc.m. = 0. We just need to calculate xc.m. = 0.

The center of mass of the large disk must be at the center of the disk by symmetry. The same holds true for the small disk.

Technique: Use symmetry as much as possible.

The hole is as if the smaller disk had “negative” mass to counteract the solid larger mass disk.

C

S

Superimpose the two disks. The overlap of the “positive” mass with the “negative” mass will result mathematically in a hole.

The center of mass will be somewhere over here.center of mass (c.m. or com) of Plate C: xC = 0

CS

center of mass (c.m. or com) of Disk S: xS = −R

Rmmm

mm)R(m0

mmxmxmx

SC

S

SC

S

SC

SSCC.m.c

+−=+

−+=++=


CS

( ) ( )22CCC R4)R2(Vm πρ=πρ=ρ=

Now we need mC and mS. Both disks have the same uniform mass density ρ (but with different “signs”). Thus,

Likewise,

( ) ( )22SSS R)R(Vm πρ−=πρ−=ρ=


CS

( ) ( )22CCC R4)R2(Vm πρ=πρ=ρ=

( ) ( )22SSS R)R(Vm πρ−=πρ−=ρ=

R31R

)R()R4()R(

Rmmm

mm)R(m0

mmxmxmx

22

2

SC

S

SC

S

SC

SSCC.m.c

=

πρ−πρ

πρ−=

+−=+

−+=++=

• Newton’s 2nd law for a system of particlesWe know

take derivative with respect to time

take derivative with respect to time again

ii

n

1i.m.c rmrM rr=Σ=

netn21.m.c FF...FFaMrrrrr

=+++=

nn2211.m.c vm...vmvmvM rrrr +++=

nn2211.m.c am...amamaM rrrr+++=

∑=

=n

1iii.m.c rm

M1r rr

.m.cnet

z.,m.cz,nety.,m.cy,netx.,m.cx,net

aMF

aMFaMFaMFvv

=⇒

===

Newton’s second law for a system of particles

Fnet is the net force of all external forces that act on the system.M is the total mass of the system.ac.m. is the acceleration of the center of the mass

Linear Momentum

• The linear momentum of a particle is a vector defined as

• Newton’s second law in terms of momentum

dtPdFnet

vv

=

netFamdtdmv

dtvdm)vm(

dtd

dtpd vvv

vv

v==+==

vmp vv =

Most of the time the mass doesn’t change, so this term is zero. Exceptions are rockets (Monday)

The figure gives the linear momentum versus time for a particle moving along an axis. A force directed along the axis acts on the particle. (a) Rank the four regions indicated according to the magnitude of the force, greatest first

(b) In which region is the particle slowing?

• Linear momentum of a system of particles

• Newton’s 2nd law for a system of particles

.m.c

n

1iii

n

1ii vMvmpP vvvv

=== ∑∑==

net.m.c.m.c FaM

dtvdM

dtPd vv

vv

===

.m.cvMP vv=

dtPdFnet

vv

=

A helium atom and a hydrogen atom can bind to form the metastable molecule HeH (lifetime of about 1µs). Consider one such molecule at rest in the lab frame at the origin. This molecule then dissociates with the hydrogen atom having momentum mpvalong the +x axis. What happens to the helium atom?

Daily Quiz, February 16, 2005

A helium atom and a hydrogen atom can bind to form the metastable molecule HeH (lifetime of about 1µs). Consider one such molecule at rest in the lab frame at the origin. This molecule then dissociates with the hydrogen atom having momentum mpvalong the +x axis. What happens to the helium atom?


A helium atom and a hydrogen atom can bind to form the metastable molecule HeH (lifetime of about 1µs). Consider one such molecule at rest in the lab frame at the origin. This molecule then dissociates with the hydrogen atom having momentum mpvalong the +x axis. What happens to the helium (mHe = 4mp) atom?


1) stays at x=0 2) goes along +x at speed v3) goes along −x at speed v 4) goes along −x at speed v/4

5) none of the above

A helium atom and a hydrogen atom can bind to form the metastable molecule HeH (lifetime of about 1µs). Consider one such molecule at rest in the lab frame at the origin. This molecule then dissociates with the hydrogen atom having momentum mpvalong the +x axis. What happens to the helium (mHe = 4mp) atom?


pinitial = 0 means pfinal = 04) goes along −x at speed v/4mpv + mHevHe = 0 =

mpv + 4mpvHe = 0 => vHe = −mpv/4

Collisions take time!

Even something that seems instantaneous to us takes a finite amount of time to happen.

Impulse and Change in Momentum

Call this change in momentum the “Impulse” and give it the symbol J.

∫==∆=− f

i

t

tif dtFJpppvvvvv

∫=∆=−⇒=⇒= f

i

t

tifnet dtFpppdtFpddtpdF

vvvvvvvv

Actual force function versus time.

Average force versus collision time.

Impulse and Change in MomentumCall this change in momentum the “Impulse” and give it the symbol J.

∫==∆=− f

i

t

tif dtFJpppvvvvv

The instantaneous force is hard (or very difficult) to know in a real collision, but we can use the average force Favg. Thus,

J = Favg∆t.

An egg was dropped on the table broke, but the egg dropped on the foam pad didn’t break. Why didn’t this egg break?


Foam Table



Foam Table

1) Foam egg’s speed was less.2) Foam egg’s change in momentum was less.3) Foam egg’s collision time was greater.4) What do you mean, it did break! 0) none of the above



Foam Table

1) Foam egg’s speed was less.

The eggs were dropped from same height, so their speeds were the same. mgh = 1/2mv2



Foam Table

The table egg’s momentum stopped at the table, but the foam egg bounced up making its change in momentum greater than the table egg!

2) Foam egg’s change in momentum was less.



Foam Table

3) Foam egg’s collision time was greater.

J = Favg∆t.Since J constant, if ∆t is large then Favg will be small.

Conservation of Linear Momentum

• For a system of particles, if it is both isolated (the net external force acting on the system is zero) and closed ( no particles leave or enter the system )….If then

Therefore or

then the total linear momentum of the system cannot change.Law of conservation of linear momentum

0F =Σr

0dtPd =r

ttanconsP =r

fi PPrr

=

• Conservation of linear momentum along a specific direction: If Σ Fx = 0 Then Pi, x = Pf, x

If Σ Fy = 0 Then Pi, y = Pf, y

If the component of the net external force on a closed system is zero along an axis, then the component of the linear momentum of the system along that axis cannot change.

Linear Momentum

• The linear momentum of a particle is a vector defined as

• Newton’s second law in terms of momentum

dtPdFnet

vv

=

netFamdtdmv

dtvdm)vm(

dtd

dtpd vvv

vv

v==+==

vmp vv =

Most of the time the mass doesn’t change, so this term is zero. Exceptions are rockets (Tuesday)



Therefore or


0F =Σr

0dtPd =r

ttanconsP =r

fi PPrr

=

• Conservation of linear momentum along a specific direction: If Σ Fx = 0 Then Pi, x = Pf, x

If Σ Fy = 0 Then Pi, y = Pf, y

If the component of the net external force on a closed system is zero along an axis, then the component of the linear momentum of the system along that axis cannot change.

Collisions

Linear momentum is conserved.

In absence of external forces,

Mechanical energy may or may not be conserved.

Elastic collisions: Mechanical energy is conserved.

Inelastic collisions: Mechanical energy is NOT conserved.

But, Linear momentum is always conserved.


f2f1i2i1

fi

pppppp

vvvv

vv

+=+⇒=

f22f11i22i11 vmvmvmvm vvvv +=+

Center of Mass motion is constant

Center of Mass Motion

21.m.c

.m.c21.m.ci2i1

mmPv

v)mm(vMppP

+=⇒

+==+=v

v

vvvvv

Inelastic Collisions

( ) f21f22f11i22i11 Vmmvmvmvmvmvvvvv +=+=+

Perfectly inelastic collision: The two masses stick together

0

)v(vmm

mV .m.ci121

1f

vvv=

+=


Was the mechanical energy:conserved (Ei = Ef);lost (Ei > Ef); or gained (Ei < Ef);

in the collision?


( ) ( )

2i1

21

212i1

21

212

i11filost

2i1

21

212

i1

2

21

121

2f21

vmm

mm21v

mmm

21vm

21EEE

vmm

m21v

mmmmm

21Vmm

21

+

=

+

−=−=

⇒

+

=

+

+=+v

How much mechanical energy was lost in the collision?

2i1

21

21lost v

mmmm

21E

+

=

Elastic Collisions

( ).m.cf22f11i22i11 vMvmvmvmvm vvvvv =+=+

Perfectly elastic collision: Mechanical energy is conserved

f2

f222f11

2i22

2i11i Evm

21vm

21vm

21vm

21E =+=+= vvvv

Elastic Collisions

( ).m.cf22f11i22i11 vMvmvmvmvm vvvvv =+=+

Perfectly elastic collision: Mechanical energy is conserved

i221

12i1

21

1f2

i221

2i1

21

21f1

vmmmmv

mmm2v

vmm

m2vmmmmv

vvv

vvv

+−+

+=

++

+−=

What about two (or more) dimensions?

Simply break the momenta and velocities into their x-, y-, and z-components.

• Inelastic collisions in two dimensions

Pi x = Pf x Pi y = Pf y

For the case shown here:

fi PPrr

=

22211111 coscos θvmθvmvm ffi +=

222111 sinsin0 θvmθvm ff +=



Therefore or


0F =Σr

0dtPd =r

ttanconsP =r

fi PPrr

=

Linear Momentum is conserved

if PPvv

=Let the mass M change (M + dM), which in turn makes the velocity change. Rocket exhausts −dM in time dt at a velocity U relative to our inertial reference frame.

)dvv)(dMM(UdMMv +++−=

+

=

frame.reftorelative

exhaustofVelocityexhausttorelative

rocketofVelocityframe.reftorelative

rocketofVelocity

Uv)dvv( rel +=+

vrel v + dv

U USA



relv)dvv(U −+=

vrel

Substitute and divide by dt

dtdvMv

dtdM

rel =−

v + dv

USA

if PPvv

=

U



relv)dvv(U −+=

vrel

Substitute and divide by dt

=−⇒

−=⇒−= ∫∫

f

irelif

M

Mrel

v

vrel

MMlnvvv

MdMvdvv

MdMdv f

i

f

i

if PPvv

=

v + dv

USA

Problem 09-05

Find the center of mass of the ammonia molecule.

Mass ratio: N/H = 13.9H to triangle center: d = 9.40x10-11mN to hydrogen: L = 10.14x10-11m

Problem 09-05Find the center of mass of the ammonia molecule.

Problem 09-05Find the center of mass of the ammonia molecule.

Problem 09-54Find the distance the spring is compressed. m1=2.0kg, m2=1.0kg.

Problem 09-54Find the distance the spring is compressed. m1=2.0kg, m2=1.0kg.

Problem 09-56Find the final velocity of A and is the collision elastic?Mass A: mA=1.6kg, vAi=5.5 m/sMass B: mB=2.4kg, vBi=2.5 m/s, vBf=4.9 m/s



Problem 09-60Find the final speeds of the ball and block.Mass 1 (ball): m1=0.5kg, h=0.70mMass 2 (block): m2=2.5kg, v2i=0.0 m/s

Problem 09-60Find the final speeds of the ball and block.Mass 1 (ball): m1=0.5kg, h=0.70mMass 2 (block): m2=2.5kg, v2i=0.0 m/s

Problem 09-63Find the mass m to stop M and the final height of m.Mass 1 (baseball): m=?, hinitial =1.8mMass 2 (basketball): M=0.63kg, vMf=0.0 m/s

Elastic collisions: Mechanical energy is conserved.

gh2vMv21Mgh 2 =⇒=Basketball:

rebounds upward with same speed -- only reversed direction

Problem 09-63Find the mass m to stop M and the final height of m.Mass 1 (baseball): m=?, hinitial =1.8mMass 2 (basketball): M=0.63kg, vMf=0.0 m/s

kg21.03/Mm

0gh2mMm3M

gh2mM

m2gh2mMmMv

mMm2v

mMmMv miMiMf

==⇒

=+−=

+−

+−=

++

+−=

gh2vmv21mgh 2 =⇒=Baseball:

Problem 09-63Find the mass m to stop M and final height of m.Mass 1 (ball): m1=0.5kg, h=0.70mMass 2 (block): m2=2.5kg, v2i=0.0 m/s

Problem 09-97Find the final speeds of the sleds.Mass 1(sleds): M=22.7kg,Mass 2 (cat): m=3.63kg, vi=3.05 m/s

Problem 09-97

Problem 09-97

Problem 09-130

Ball 1 vo=10.0m/s at contact point of balls 2 and 3. All three balls have mass m. Find the final velocities of all three balls.

Problem 09-130


1

2

3

θ = 30o since all three balls are identical

Pi = mvo

Pf = mV + 2mv cosθball1 balls 2 & 3

Problem 09-130


Problem 09-130


Problem 09-130


A Quiz

CERN

Consider a proton (mass mp and kinetic energy Ep) colliding head on with an electron (mass me) initially at rest. What is the maximum kinetic energy (Ee) that can be delivered to the electron?

x

y

pe−

A Quiz

CERNConsider a proton (mass mp = 1836me and kinetic energy Ep) colliding head on with an electron (mass me) initially at rest. What is the maximum kinetic energy (Ee) that can be delivered to the electron?

x

y

pe−

1) Ee = Ep 2) Ee < Ep 3) Ee > Ep4) not enough information

A Quiz

CERNConsider a proton (mass mp = 1836me and kinetic energy Ep) colliding head on with an electron (mass me) initially at rest. What is the maximum kinetic energy (Ee) that can be delivered to the electron?

x

y

pe−

1) Ee = Ep 2) Ee < Ep 3) Ee > Ep4) not enough information

fpfpip vmVmVm +=Conservation of Momentum

Conservation of Energy 2fe

2fp

2ip vm

21Vm

21Vm

21 +=

2fee

2ipp vm

21EandVm

21E =≡

( ) pp2pe

pee EE

m/m1m/m4

E <+

=

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Chapter 09 Center of Mass and Linear Momentumastro1.panet.utoledo.edu/~khare/teaching/phys2130h...Chapter 09 Center of Mass and Linear Momentum • Center of mass: The center of mass