Download Chapter 09 Center of Mass and Linear Momentum

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
Consider two masses m1 at x = x1 and m2 at x2.
x c. m .
m1x1 + m 2 x 2
≡
m1 + m 2
• Center of mass for a system of n particles:
x c. m .
n
1
= ∑ mi x i
M i =1
y c. m .
• M is just the total mass
of the system
1 n
= ∑ mi yi
M i =1
z c.m .
1 n
= ∑ mi zi
M i =1
n
M = ∑ mi
i =1
v
• Using vectors, we have: rc.m. = x c.m. î + y c.m.ˆj + z c.m. k̂
• Therefore:
1 n
v
v
rc.m. = ∑ m i ri
M i =1
• For a solid body, we can treat it as a continuous
distribution of matter dm
x c.m .
1
= ∫ x dm
M
y c.m. =
1
y dm
∫
M
z c. m .
1
= ∫ z dm
M
• If the object has uniform density,
dm = (M/V)dV
x c. m .
1
= ∫ x dV
V
y c.m .
1
= ∫ y dV
V
z c.m .
1
= ∫ z dV
V
• 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.
• 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.
Technique: Use symmetry as
much as possible.
C
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.
The hole is as if the smaller disk
had “negative” mass to
counteract the solid larger mass
disk.
S
Superimpose the two disks. The
overlap of the “positive” mass
with the “negative” mass will
result mathematically in a hole.
S
C
The center of mass will be
somewhere over here.
center of mass (c.m. or com) of Plate C: xC = 0
center of mass (c.m. or com) of Disk S: xS = −R
x c. m .
m C x C + mS x S 0 + mS (−R )  − mS 
=
=
=
R
m C + mS
m C + mS
 m C + mS 
Superimpose the two disks. The
overlap of the “positive” mass
with the “negative” mass will
result mathematically in a hole.
S
Now we need mC and mS. Both
disks have the same uniform mass
density ρ (but with different
“signs”). Thus,
(
) (
m C = ρC VC = ρ π(2R ) 2 = ρ 4πR 2
Likewise,
(
)
(
m S = ρS VS = −ρ π(R ) = −ρ πR
2
2
)
)
C
Superimpose the two disks. The
overlap of the “positive” mass
with the “negative” mass will
result mathematically in a hole.
S
(
) ( )
= −ρ(π(R ) ) = −ρ(πR )
m C = ρC VC = ρ π(2R ) = ρ 4πR
2
m S = ρS VS
x c.m .
C
2
2
2
m C x C + mS x S 0 + mS (−R )  − mS 
=
=
=
R
m C + mS
m C + mS
 m C + mS 
2


1
− ρ(πR )
R = R
= 
2
2 
3
 ρ(4πR ) − ρ(πR ) 
• Newton’s 2nd law for a system of particles
We know
r
r
1 n
rc.m. = ∑ m i ri
M i =1
n
r
r
M rc.m. = Σ m i ri
i =1
take derivative with respect to time
r
r
r
r
M v c.m. = m1v1 + m 2 v 2 + ... + m n v n
take derivative with respect to time again
r
r
r
r
M a c.m. = m1a 1 + m 2 a 2 + ... + m n a n
r r
r r
r
M a c.m. = F1 + F2 + ... + Fn = Fnet
Newton’s second law for a system of particles
Fnet , x = M a c.m., x
Fnet , y = M a c.m., y
v
v
Fnet = M a c.m.
⇒
Fnet ,z = M a c.m.,z
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
v
defined as
v
p = mv
• Newton’s second law in terms of momentum
v
v
dp d
dv v dm
v v
v
= (mv) = m + v
= ma = Fnet
dt dt
dt
dt
v
v
dP
Fnet =
dt
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
n
n
v
v
v
v
P = ∑ p i = ∑ m i v i = M v c .m .
i =1
i =1
v
v
P = Mv c . m .
• Newton’s 2nd law for a system of particles
v
v
v
dv c . m .
dP
v
=M
= M a c.m. = Fnet
dt
dt
v
v
dP
Fnet =
dt
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 mpv
along 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 mpv
along 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 mpv
along the +x axis. What happens to the helium (mHe = 4mp)
atom?
1) stays at x=0
2) goes along +x at speed v
3) goes along −x at speed v
4) goes along −x at speed v/4
5) none of the above
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 mpv
along the +x axis. What happens to the helium (mHe = 4mp)
atom?
pinitial = 0 means pfinal = 0
mpv + mHevHe = 0 =
mpv + 4mpvHe = 0
4) goes along −x at speed v/4
=> 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
v
v
tf v
dp
v v
v v
v
Fnet =
⇒ dp = Fdt ⇒ p f − p i = ∆p = ∫ Fdt
ti
dt
Call this change in momentum the “Impulse” and
give it the symbol J.
tf v
v v
v v
p f − p i = ∆p = J = ∫ Fdt
ti
Actual force function
versus time.
Average force versus
collision time.
Impulse and Change in Momentum
Call this change in momentum the “Impulse” and
give it the symbol J.
tf v
v v
v v
p f − p i = ∆p = J = ∫ Fdt
ti
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.
Daily Quiz, February 18, 2005
Table
Foam
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?
Daily Quiz, February 18, 2005
Foam
Table
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?
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
Daily Quiz, February 18, 2005
Foam
Table
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?
1) Foam egg’s speed was less.
The eggs were dropped from same height, so their
speeds were the same. mgh = 1/2mv2
Daily Quiz, February 18, 2005
Foam
Table
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?
2) Foam egg’s change in momentum was less.
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!
Daily Quiz, February 18, 2005
Foam
Table
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?
3) Foam egg’s collision time was greater.
Since J constant, if ∆t is large then Favg will be small.
J = Favg∆t.
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 ror enter the system )….
r
dP
If
then
=0
ΣF = 0
dt
Therefore
r
P = cons tan t
or
r r
Pi = P f
then the total linear momentum of the system
cannot change.
Law of conservation of linear momentum
• 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
v
defined as
v
p = mv
• Newton’s second law in terms of momentum
v
v
dp d
dv v dm
v v
v
= (mv) = m + v
= ma = Fnet
dt dt
dt
dt
v
v
dP
Fnet =
dt
Most of the time the mass
doesn’t change, so this term
is zero. Exceptions are
rockets (Tuesday)
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 ror enter the system )….
r
dP
If
then
=0
ΣF = 0
dt
Therefore
r
P = cons tan t
or
r r
Pi = P f
then the total linear momentum of the system
cannot change.
Law of conservation of linear momentum
• 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
In absence of external forces,
Linear momentum is conserved.
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.
Conservation of Linear Momentum
v
v
pi = pf ⇒
v
v
p 1i + p 2 i =
v
v
m 1 v 1i + m 2 v 2 i
=
v
v
p 1f + p 2 f
v
v
m 1 v 1f + m 2 v 2 f
Center of Mass motion is constant
Center of Mass Motion
v
v
v
v
v
P = p 1i + p 2 i = M v c .m . = ( m 1 + m 2 ) v c .m .
v
P
v
⇒ v c .m . =
m1 + m 2
Inelastic Collisions
Perfectly inelastic collision: The two masses stick together
v
v
v
v
v
m 1 v 1 i + m 2 v 2 i = m 1 v 1 f + m 2 v 2 f = (m 1 + m 2 )V f
v
Vf
0
=
m1
v
v 1i
m1 + m 2
v
( = v c .m . )
Inelastic Collisions
Was the mechanical energy:
conserved (Ei = Ef);
lost
(Ei > Ef); or
gained
(Ei < Ef);
in the collision?
Inelastic Collisions
How much mechanical energy was lost in the collision?
v2

1
1
(m 1 + m 2 )V f = (m 1 + m 2 ) m 1
2
2
 m1 + m 2
E lost = E i − E f =
E lost
1
1
m 1 v 12i − 
2
2
1
=
2
2
 2
 2
1
m 12
 v 1 i = 
 v 1 i
2  m1 + m 2 

 2
m 12
1  m 1m 2  2
 v 1 i = 
 v 1 i
m1 + m 2 
2  m1 + m 2 
 m 1m 2

 m1 + m 2
 2
 v 1 i

⇒
Elastic Collisions
Perfectly elastic collision: Mechanical energy is conserved
v
v
v
v
v
m 1 v 1 i + m 2 v 2 i = m 1 v 1 f + m 2 v 2 f (= M v c . m . )
1
1
v2
v2
E i = m 1 v 1i + m 2 v 2 i
2
2
=
1
1
v2
v2
m 1 v 1f + m 2 v 2 f = E f
2
2
Elastic Collisions
Perfectly elastic collision: Mechanical energy is conserved
v
v
v
v
v
m 1 v 1 i + m 2 v 2 i = m 1 v 1 f + m 2 v 2 f (= M v c . m . )
v
v 1f
v
v 2f
=
m1 − m 2 v
v 1i +
m1 + m 2
2m 2 v
v 2i
m1 + m 2
=
2m1 v
v 1i +
m1 + m 2
m 2 − m1 v
v 2i
m1 + m 2
What about two (or more) dimensions?
Simply break the momenta and velocities into
their x-, y-, and z-components.
• Inelastic collisions in two dimensions
r r
Pi = Pf
Pi x = Pf x
Pi y = Pf y
For the case shown here:
m1v1i = m1v1 f cos θ1 + m2 v2 f cos θ2
0 = m1v1 f sin θ1 + m2 v2 f sin θ2
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 ror enter the system )….
r
dP
If
then
=0
ΣF = 0
dt
Therefore
r
P = cons tan t
or
r r
Pi = P f
then the total linear momentum of the system
cannot change.
Law of conservation of linear momentum
Linear Momentum is conserved
v v
Pf = Pi
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.
Mv = −dM U + (M + dM )( v + dv)
 Velocity of rocket   Velocity of rocket   Velocity of exhaust 

 = 
 + 

 relative to ref . frame   relative to exhaust   relative to ref . frame 
( v + dv) = v rel
U
+ U
U
S
A
vrel
v + dv
Linear Momentum is conserved
v v
Pf = Pi
U = ( v + dv) − v rel
Substitute and divide by dt
Mv = −dM U + (M + dM )( v + dv)
dv
dM
v rel = M
−
dt
dt
U
U
S
A
vrel
v + dv
Linear Momentum is conserved
v v
Pf = Pi
U = ( v + dv) − v rel
Substitute and divide by dt
Mv = −dM U + (M + dM )( v + dv)
dM
dv = −
v rel ⇒
M
∫
vf
vi
dv = − v rel ∫
 Mi
⇒ v f − v i = v rel ln
 Mf
Mf
Mi
dM
M



U
S
A
vrel
v + dv
Problem 09-05
Find the center of mass of the ammonia molecule.
Mass ratio: N/H = 13.9
H to triangle center: d = 9.40x10-11m
N to hydrogen: L = 10.14x10-11m
Problem 09-05
Find the center of mass of the
ammonia molecule.
Problem 09-05
Find the center of mass of the
ammonia molecule.
Problem 09-54
Find the distance the spring is
compressed. m1=2.0kg, m2=1.0kg.
Problem 09-54
Find the distance the spring is
compressed. m1=2.0kg, m2=1.0kg.
Problem 09-56
Find the final velocity of A and is the collision
elastic?
Mass A: mA=1.6kg, vAi=5.5 m/s
Mass B: mB=2.4kg, vBi=2.5 m/s, vBf=4.9 m/s
Problem 09-56
Find the final velocity of A and is the collision
elastic?
Mass A: mA=1.6kg, vAi=5.5 m/s
Mass B: mB=2.4kg, vBi=2.5 m/s, vBf=4.9 m/s
Problem 09-56
Find the final velocity of A and is the collision
elastic?
Mass A: mA=1.6kg, vAi=5.5 m/s
Mass B: mB=2.4kg, vBi=2.5 m/s, vBf=4.9 m/s
Problem 09-60
Find the final speeds of the ball and block.
Mass 1 (ball): m1=0.5kg, h=0.70m
Mass 2 (block): m2=2.5kg, v2i=0.0 m/s
Problem 09-60
Find the final speeds of the ball and block.
Mass 1 (ball): m1=0.5kg, h=0.70m
Mass 2 (block): m2=2.5kg, v2i=0.0 m/s
Problem 09-63
Find the mass m to stop M and the final height of m.
Mass 1 (baseball): m=?, hinitial =1.8m
Mass 2 (basketball): M=0.63kg, vMf=0.0 m/s
Elastic collisions: Mechanical energy is conserved.
Basketball:
1
Mgh = Mv 2
2
⇒ v = 2gh
rebounds upward with same speed -- only reversed direction
Problem 09-63
Find the mass m to stop M and the final height of m.
Mass 1 (baseball): m=?, hinitial =1.8m
Mass 2 (basketball): M=0.63kg, vMf=0.0 m/s
Baseball:
1
mgh = mv 2
2
⇒ v = 2gh
M−m
2m
M−m
2m
v Mf =
v Mi +
v mi =
2gh −
2gh
M+m
M+m
M+m
M+m
M − 3m
=
2gh = 0
M+m
⇒ m = M / 3 = 0.21kg
Problem 09-63
Find the mass m to stop M and final height of m.
Mass 1 (ball): m1=0.5kg, h=0.70m
Mass 2 (block): m2=2.5kg, v2i=0.0 m/s
Problem 09-97
Find 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
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.
Pi = mvo
2
ball1 balls 2 & 3
Pf = mV + 2mv cosθ
1
3
θ = 30o since all three balls are identical
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
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
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.
A Quiz
y
p
e−
x
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?
A Quiz
Consider 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?
y
CERN
p
e−
x
1) Ee = Ep 2) Ee < Ep 3) Ee > Ep
4) not enough information
A Quiz
Conservation of Momentum
m p Vi = m p Vf + m p v f
1
1
1
2
2
2
m
V
m
V
m
v
=
+
Conservation of Energy
p i
p f
e f
2
2
2
1
1
2
E p ≡ m p Vi and E e = m e v f2
2
2
4m e /m p
Ee =
Ep < Ep
2
(1 + me /m p )
y
Consider a proton (mass mp = 1836me and
CERN
kinetic energy Ep) colliding head on with an
p
electron (mass me) initially at rest. What is
−
e
the maximum kinetic energy (Ee) that can be
delivered to the electron?
x
1) Ee = Ep 2) Ee < Ep 3) Ee > Ep
4) not enough information