Transcript center of mass

Center of Mass
Linear Momentum
Chapter 9
1
Center of Mass
the point in which the mass of an object can be
considered to be concentrated
the objects motion acts as if it’s mass is
concentrated at that point
the translational acceleration calculated by Newton’s
second law Fnet = ma determines the motion of the
object at it’s center of mass
all gravitational forces exerted on the object is applied at that
point
2
Calculating Center of Mass
The center of mass of an object can be found in all dimensions.
(m1x1 + m2x2 + . . . + mnxn)
______________________
xcom =
(m1 + m2 + . . . + mn)
...
(m1y1 + m2y2 + + mnyn)
______________________
ycom =
(m1 + m2 + . . . + mn)
^
^
^
→
vectorcom = rcom = xcomi + ycomj + zcomk
→
r
com =
1
___
M
n
Σ
i=1
→
miri
(m1z1 + m2z2 + . . . + mnzn)
______________________
zcom =
(m1 + m2 + . . . + mn)
3
Newton’s Second Law for a System
of Particles
→
→
Fnet = Macom
→
Fnet : net force of all external forces (internal forces not included)
M = total mass of the system (closed system)
→
acom = acceration of the center of mass of the system
→
→
Fnet,x = Macom,x
→
→
Fnet,y = Macom,y
→
→
Fnet,z = Macom,z
4
Conversion of Center of Mass
Equation
→
rcom : gives position
→
→
→
→
Mrcom = m1r1 + m2r2 + . . . + mnrn
→
→
→
→
→
[Mrcom]` = Mvcom = m1v1 + m2v2 + … + mnvn
→
→
→
→
→
[Mvcom]` = Macom = m1a1 + m2a2 + … + mnan
→
dr
→ ___i
vi =
dt
→
dv
→ ___i
ai =
dt
→
→
→ …
→ → →
m1a1 + m2a2 + + mnan = F1 + F2 + … + Fn
→
→
Fi = miai
5
Linear Momentum
linear momentum of a particle = total mass of the
system * velocity of its center of mass
→ → →
→
P = p1 + p2 + … + pn
→
→
→
= m1v1 + m2v2 + … + mnvn
→
→
P = Mvcom
→
→
p = mv
the momentum of an particle acts in the same
direction as the velocity of that particle because
mass is a positive scaler quantity
time rate of change of the momentum of a particle = net force acting
on that particle and in the same direction
→
→
dP
Fnet = ___
dt
→
→
→
dp
d
→
dv
com
→
Fnet = ___ = ___ (Mvcom) = M _____
=
Ma
com
dt
dt
dt
therefore net external force changes the momentum of an object,
thus momentum can only change due to an external force
6
Impulse
impulse is defined as:
∫
tf
→
dp =
ti
∫
tf
→
F(t) dt
ti
→ →
→
pf – pi = ∆p
→
J=
∫
tf
→
F(t)dt
ti
→ →
∆p = J linear momentum – impulse theorem
→ →
J = Favg∆t
→
→
J = -n∆p
n = # of projectiles
series of collisions :
thus
→
→
→ __
→
→
__
J
__ ∆p
-n
-n m∆v
___ ∆v
∆m
–
Favg =
=
=
=
∆t
∆t
∆t
∆t
7
Conservation of Linear Momentum
no external forces :
→
Fnet = 0
momentum stays constant in a closed system
→ →
Pi = Pf
law of conservation of linear momentum
if a component of net external force in a closed system = 0 along an
axis, then the component of the linear momentum of the system along
that system can not change
8
Collisions
elastic collision: momentum AND kinetic energy conserved
→
→
→
→
P1,i + P2,i = P1,f + P2,f
→
→
→
→
m1v1,i + m2v2,i = m1v1,f + m2v2,f
conservation of momentum
K1,i + K2,i = K1,f + K2,f
conservation of kinetic energy
1/ m →
2 + 1/ m →
2 = 1/ m →
2 + 1/ m →
2
v
v
v
2 1 1,i
2 2 2,1
2 1 1,f
2 2v2,f
inelastic collision: momentum is conserved, kinetic energy IS NOT conserved
→
→
→
→
P1,i + P2,i = P1,f + P2,f
→
→
→
→
m1v1,i + m2v2,i = m1v1,f + m2v2,f
conservation of momentum
completely inelastic: bodies stick together
9
Systems With Varying Mass: A
Rocket
→ →
Pi = Pf
m
velocity of rocket
relative to frame =
M + dM
-dM
→
v
vR-F
=
velocity of rocket + velocity of products
relative to frame
relative to products
→
+
vR-P
→
vP-F
→
vrel
v + dv
U
→
→
→ →
Mv = -dMU + (M +dM)(v + dv)
→
P of gas
∫
→
P of rocket afterwards
→
→ dM
dv = – vrel ___
M
vf
→
→
dv = – vrel
∫
vi
Mf
Mi
M
→
→ →
v - v = v ln __i
f
i
rel
||
Mf
→
→ → →
(v + dv) = vrel + U
→ → → →
U = v + dv - vrel
→
→
-dMvrel = Mdv
→
__
dv
→
____
-dM V = M
rel
dt
dt
___
dM
M
second rocket equation
→ = Ma
→
Rv
rel
first rocket equation
10
What is Newton’s Second Law for a
system of particles?
→
→
Fnet = Macom
11
How does
→ Mrcom become
→
Mvcom?
→
dr
→
com
____
Mvcom = M dt
12
Why is the direction of a particle’s
momentum the same as the
direction of its velocity?
mass is a scaler
quantity in the equation:
→
→
p = mv
13
If there is no external force
(Fnet = 0), P is constant in a
_________
system.
closed
14
The equation for the linear
momentum of a system is?
→
→
p = mv
15
Impulse is defined as?
→
J=
∫
tf→
F(t)dt
ti
16
What is the impulse – momentum
theorem?
→ → → →
J = ∆P = Pf - Pi
17
If there is no net external force
momentum ( does, does not )
change.
18
In an elastic collision both
kinetic energy
_______________
and
momentum
_______________
are conserved.
19
In an inelastic collision
momentum
_____________
is conserved and
kinetic energy is not conserved.
_____________
20
In a completely inelastic collision
what occurs?
the bodies stick
together
21
What is the Law of Conservation of
Momentum?
Pi = P f
22
vR-P
vP-F
vR-F = _______
+ ________
where P = products
R = rocket
F = frame of reference
23
vR-G
Vrel = ________
24
The vcom ( is, isn’t ) changed by a
collision.
25
In the absence of external forces a
rocket accelerates at an
instonaneous rate given by: (first
rocket equation)
Rvrel = Ma
26
The point in which the mass of an
object could be concentrated is?
center of mass
27
What is the equation for the center
of mass of an object on the x axis?
... + m x )
(m
x
+
m
x
+
____________________
1 1
2 2
n n
xcom = (m + m + . . . + m )
1
2
n
28
The thrust of an object is given by:
Rvrel
29
All gravitation forces are exterted
on what point?
center of mass
30
The path in which a baseball bat’s
center of mass is thrown into the air
is in the shape of a
parabola
______________.
31
Find the center of mass of these 3
objects:
m
2m
2m
d
3d
d
32
Find the impulse from 1 second to
5 seconds if Fnet(t) = t.
12
33
In a completely elastic collision:
v
2m
m
Find the final velocity of the
second particle.
→
→
V2 = 2v
34
In a completely inelastic collision:
v
2m
m
Find the final velocity.
vf = 2/3v
35
If a boat is moving at 10 mph
against a river current of 2 mph,
how fast is the boat going past the
shoreline?
6 mph
36
If a 2000 kg (including fuel) rocket
burns 10 kg of fuel per second
going at 10000 m/s past earth
ejecting fuel at 500 m/s past earth.
What is the rocket’s acceleration?
4.75 m/s
2
37
A 2 kg mass hits a 4 kg mass at 5
m/ in an elastic collision. Find the
s
final velocity of the 2 kg mass.
– 1.67 m/s
38
In a completely in elastic collision:
2m
2v
v
4m
What is the final velocity?
0 m/s
39
Find the center of mass:
m
2
2m
2
-2
4m
3m
-2
(0.1,- 0.8)
40