Download October 24

Lesson 7
Physics 168
Monday, October 24, 16
Luis Anchordoqui
1
Eruption of a large volcano on Jupiter’s moon
When volcano erupts speed of effluence exceeds escape speed of Io
and so a stream of particles is projected into space
Material in stream can collide with and stick to
surface of asteroid passing through stream
We now consider effect of impact of this material on motion of asteroid
Monday, October 24, 16
2
Continuously varying mass
Overview
Consider continuous stream of matter moving at velocity
~u
which impacts object of mass M that is moving with velocity
This impacting particles stick to object
increasing its mass by M during time
v changes by
During time t velocity ~
~v
t
~v
Applying impulse momentum theorem to this system
F~net, ext
t =
C.
B.-Champagne
Luis
Anchordoqui
Monday, October 24, 16
P~ = P~f
~i = [(M +
P
M )(~v +
~v )]
[M~v +
M ~u]
2
3
Continuously varying mass (cont’d)
F~net, ext
Rearranging terms
Overview
t = M ~v +
M (~v
Dividing by
F~net, ext
Taking limit
t = M
~v
+
t
M
(~v
t
~u) +
M ~v
t
~u) +
t ! 0 that also means M ! 0 and
d~v
dM
~
Fnet, ext = M
+
(~v
~u)
dt
dt
M
t
~v
~v ! 0
Rearranging terms we obtain Newton’s second law
for a system that has a continuously changing mass
dM
d~v
~
Fnet, ext +
~vrel = M
dt
dt
~vrel = ~u ~v
C.
B.-Champagne
Luis
Anchordoqui
Monday, October 24, 16
2
4
Rocket Propulsion
Overview
Momentum conservation works for a rocket as long as we consider
rocket and its fuel to be one system and account for mass loss of rocket
C.
B.-Champagne
Luis
Anchordoqui
Monday, October 24, 16
2
5
Rocket Propulsion
Overview
Rocket propulsion is a striking example of conservation of momentum in action
Use Newton’s law in form
Fext = dP/dt
Consider a rocket moving with speed v relative to earth ☛
If fuel is burned at constant
rocket’s mass at time
t is ☛ m(t) = m0
Momentum of system at time
At a later time
t+
R = |dm/dt|
t
is ☛
Rt
Pi = mv
t rocket has expelled gas of mass R
If gas is exhausted at speed uex relative to rocket
velocity of gas relative to Earth is
Rocket then has a mass m
C.
B.-Champagne
Luis
Anchordoqui
Monday, October 24, 16
R
v
t
uex
t and is moving at speed v +
v
2
6
Rocket Propulsion
Momentum of system at
t + t is
Overview
Pf = (m
R
t)(v +
v) + R
= mv + m
v
vR t
⇡ mv + m
v
uex R
we dropped term R
R
t(v
t
uex )
v + vR
t
uex R
t
t
t v which is product of two very small quantities
Change in momentum is
P = Pf
Pi = m
v
uex R
t
and
P
= m
t
As
t approaches zero
C.
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Luis
Anchordoqui
Monday, October 24, 16
v
t
uex R
v/ t approaches derivate dv/dt ☛ acceleration
2
7
Rocket Propulsion
For a rocket moving upward near surface of earth
Setting
dP/dt = Fext =
mg
Fext =
mg
gives us rocket equation
dv
m
= Ruex + Fext = Ruex
dt
mg
rocket equation
or
dv
Ruex
=
dt
m
Quantity
Ruex
Ruex
g =
m0
Rt
g
( ✴)
is force exerted on rocket by exhausting fuel
This is called thrust
Fth = Ruex
Monday, October 24, 16
dm
=
uex
dt
8
Rocket Propulsion
(✴) is solved by integrating both sides with respect to time
For a rocket starting at rest at
t = 0 result is
⇣m
Rt ⌘
0
v = uex ln
gt
m0
as can be verified by taking time derivative of v
Payload of a rocket is final mass mf after all fuel has been burned
Rtb or
Burn time tb is given by mf = m0
tb =
m0
mf
R
A rocket starting at rest with mass m0 and payload of
attains a final speed
vf =
mf
uex ln
m0
gtb
mf
final speed of rocket
assuming acceleration of gravity to be constant
Monday, October 24, 16
9
Saturn V: America’s Moon Rocket
Overview
The Saturn V rocket used in Apollo moon-landing program had:
6
initial mass m0 = 2.85 ⇥ 10 kg 73% of which was fuel
a burn rate ↵ = 13.84 ⇥ 103 kg/s
and a thrust
Fth = 34 ⇥ 106 N
Find (a) the exhaust speed relative to the rocket; (b) the burn time; (c) the acceleration at liftoff
(d) the acceleration at just before burnout; (e) the final speed of the rocket
C.
B.-Champagne
Luis
Anchordoqui
Monday, October 24, 16
2
10
(a) Fth
Saturn V: America’s Moon Rocket
dm
=
uOverview
ex ) uex = 2.46 km/s
dt
mfuel = ↵ tb
(b) mb = 0.27m0 = 7.70 ⇥ 105 kg
mfuel
m0 mb
tb =
=
= 150 s
↵
↵
(c)
dvy
uex dm
=
dt
m0 dt
dvy
uex dm
(d)
=
dt
mb dt
(e) vy = uex ln
C.
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Luis
Anchordoqui
Monday, October 24, 16
✓
g = 2.14 m/s
g = 34.3 m/s
m0
m0
↵t
◆
2
2
gt = 1.75 km/s
2
11
Rotational Dynamics
Monday, October 24, 16
12
Angular Quantities
Overview
In purely rotational motion
all points on object move in circles around axis of rotation ☛ O
Radius of circle is
r
All points on straight line drawn through axis
move through same angle in same time
Angle
✓
in radians is defined ☛
C.
B.-Champagne
Luis
Anchordoqui
Monday, October 24, 16
l
✓ =
r
arc length
2
13
titulo
A laser jet is directedOverview
at Moon, 380.000 km from Earth
Beam diverges at an angle ✓ = 1.4 ⇥ 10 5
What diameter spot will it make on Moon?
C.
B.-Champagne
Luis
Anchordoqui
Monday, October 24, 16
2
14
titulo
A laser jet is directedOverview
at Moon, 380.000 km from Earth
Beam diverges at an angle ✓ = 1.4 ⇥ 10 5
What diameter spot will it make on Moon?
diameter
✓=
) diameter = ✓rEM = 5.3 ⇥ 103 m
rEM
C.
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Luis
Anchordoqui
Monday, October 24, 16
2
15
Angular Quantities (cont’d)
Angular displacement
Overview
✓ = ✓2
✓1
Average angular velocity is defined as total
angular displacement divided by time
!
¯ =
✓
t
Instantaneous angular velocity:
! = lim
t!0
C.
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Luis
Anchordoqui
Monday, October 24, 16
✓
t
2
16
Angular Quantities (cont’d)
Overview
Angular acceleration is rate at which angular velocity changes with time
↵
¯=
!2
!1
t
!
t
=
Instantaneous acceleration
↵ = lim
a
t!0
C.
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Luis
Anchordoqui
Monday, October 24, 16
!
t
2
17
Angular Quantities
Overview
Every point on a rotating body has an angular velocity !
and a linear velocity v
They are related ☛
C.
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Luis
Anchordoqui
Monday, October 24, 16
v = r!
2
18
Angular Quantities (cont’d)
Overview
Therefore objects farther from axis of rotation will move faster
C.
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Luis
Anchordoqui
Monday, October 24, 16
2
19
Star tracks in a time exposure of night sky
Monday, October 24, 16
20
Angular Quantities (cont’d)
Overview
If angular velocity of a rotating object changes, it has a tangential acceleration
atan = r ↵
Even if angular velocity is constant each point on object
has centripetal acceleration
aR
C.
B.-Champagne
Luis
Anchordoqui
Monday, October 24, 16
v2
(r!)2
=
=
= !2 r
r
r
2
21
Angular Quantities
Overview
Here is correspondence between linear and rotational quantities
LINEAR
TYPE
ROTATIONAL
x
DISPLACEMENT
✓
x = r✓
v
VELOCITY
!
v = r!
atan
ACCELERATION
↵
atan = r ↵
C.
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Luis
Anchordoqui
Monday, October 24, 16
RELATION
2
22
Angular Quantities (cont’d)
Overview
✓ Frequency is number of complete revolutions per second
!
f =
2⇡
✓ Frequencies are measured in hertz
1 Hz = 1 s
1
✓ Period is time one revolution takes
1
T =
f
C.
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Luis
Anchordoqui
Monday, October 24, 16
2
23
Rotational Kinetic Energy
Overview
Kinetic energy of rigid object rotating about fixed axis
is sum of kinetic energy of individual particles that collectively make object
1
K = mi vi2
2
vi = ri ! gives rotational kinetic energy
Kinetic energy of the i-th particle ☛
Summing over all particles using
K=
X1
i
I=
X
2
mi vi2
1 2X
1 2
1X
2 2
2
mi ri ! = !
mi ri = I!
=
2 i
2
2
i
mi ri2 ☛
moment of inertia for axis of rotation
i
Object that has both translational and rotational motion
also has both translational and rotational kinetic energy
1
1
2
KE = M vCM + ICM ! 2
2
2
C.
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Luis
Anchordoqui
Monday, October 24, 16
2
24
Quantity
I =
X
Moment of Inertia
is called rotational inertia of an object
Overview
mi ri2
Distribution of mass matters here
these two objects have same mass
but one on left has a greater rotational inertia
as so much of its mass is far from axis of rotation
Integral form ☛
C.
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Luis
Anchordoqui
Monday, October 24, 16
I =
Z
r2 dm
2
25
Estimating moment of inertia
Overview
Estimate moment of inertia of a thin uniform rod of length L and mass M
about an axis perpendicular to rod and through one end
Execute this estimation by modeling rod as three point masses
each point mass representing 1/3 of rod
C.
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Luis
Anchordoqui
Monday, October 24, 16
2
26
Estimating moment of inertia
Overview
I=
X
mi ri2 =
C.
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Luis
Anchordoqui
Monday, October 24, 16
✓
1
M
3
◆✓
1
L
6
◆2
+
✓
1
M
3
◆✓
3
L
6
◆2
+
✓
1
M
3
◆✓
5
L
6
◆2
=
35
M L2
108
2
27
Moment of inertia of a thin uniform road
Overview
Find moment of inertia of a thin uniform rod of length L and mass M
about an axis perpendicular to rod and through one end
C.
B.-Champagne
Luis
Anchordoqui
Monday, October 24, 16
2
28
Moment of inertia of a thin uniform road
Overview
I=
Z
r2 dm =
Z
x2 dm
M
dm = dx =
dx
L
I=
C.
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Luis
Anchordoqui
Monday, October 24, 16
Z
2
x dm =
Z
L
0
2M
1
x
dx = M L2
L
3
2
29
Steiner’s Theorem
Overview
Parallel-axis theorem relates moment of inertia about an axis through CM
to moment of inertia about a second parallel axis
Let
I
be moment of inertia
Let Icm be moment of inertia about a parallel-axis through center of mass
Let M be total mass of object and
h distance between two axes
Parallel axis theorem states that ☛
I = Icm + M h2
C.
B.-Champagne
Luis
Anchordoqui
Monday, October 24, 16
2
30
Steiner’s Theorem (cont’d)
Overview
Consider object rotating about fixed axis that does not pass through CM
Kinetic energy of such a system is
1 2
K = I!
2
Moment of inertia about fixed axis
Kinetic energy of a system can be written as
sum of its translational kinetic energy and kinetic energy relative to its CM
For an object that is rotating kinetic energy relative to its CM
1
Icm ! 2
2
Moment of inertia about axis through cm
Total kinetic energy of object is ☛
C.
B.-Champagne
Luis
Anchordoqui
Monday, October 24, 16
1
1
2
K = M vcm + Icm ! 2
2
2
2
31
Steiner’s Theorem (cont’d)
Overview
cm moves along a circular path of radius
Substituting
h ! vcm = h!
1 2
1
1
2 2
I! = M h ! + Icm ! 2
2
2
2
Multiplying through this equation by
2
2/! 2
leads to
I = M h + Icm
C.
B.-Champagne
Luis
Anchordoqui
Monday, October 24, 16
2
32
Application of Steiner’s theorem
A thin uniform rod of mass
M and length L on x axis has one end at origin
Overview
Using parallel-axis theorem, find moment of inertia about
parallel to
y
C.
B.-Champagne
Luis
Anchordoqui
Monday, October 24, 16
y 0 axis, which is
axis, and through center of rod
2
33
Application of Steiner’s theorem
Overview
Iy = ICM + M h2 ) ICM = Iy
C.
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Luis
Anchordoqui
Monday, October 24, 16
M h2 =
1
M L2
3
1
1
M L2 =
M L2
4
12
2
34
Torque
Overview
To make an object start rotating a force is needed
Position and direction of force matter as well
Perpendicular distance from axis of rotation
to line along which force acts is called lever arm
C.
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Luis
Anchordoqui
Monday, October 24, 16
2
35
Torque (Cont’d)
Overview
Torque is defined as ☛
C.
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Luis
Anchordoqui
Monday, October 24, 16
⌧ = r? F
2
36
Torque (Cont’d)
Lever arm for
Overview
FA is distance
from knob to hinge
Lever arm for
FD
is zero
Lever arm for
FC
is as shown
C.
B.-Champagne
Luis
Anchordoqui
Monday, October 24, 16
2
37
Rotational Dynamics ☛ Torque and Rotational Inertia
Knowing that
Overview
F = ma ☛ ⌧ = mr2 ↵
This is for a single point mass
What about an extended object?
F~
r
m
As angular acceleration is same for whole object we can write
C.
B.-Champagne
Luis
Anchordoqui
Monday, October 24, 16
X
⌧i, net = (
X
mi ri2 ) ↵
2
38
titulo
Overview
Rotational inertia of an object depends
not only on its mass distribution but
also location of axis of rotation
– compare (f) and (g), for example -
C.
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Luis
Anchordoqui
Monday, October 24, 16
2
39
titulo
Overview
To get a flat, uniform cylindrical satellite spinning at correct rate,
engineers fire four tangential rockets as shown in figure
If satellite has a mass of 3600 kg and a radius of 4 m, what is required
steady force of each rocket if satellite is to reach 32 rpm in 5 min?
End view of cylindrical satellite
C.
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Luis
Anchordoqui
Monday, October 24, 16
2
40
The ring of rockets will create a torque with zero net force
Overview
⌧net = 4F R
⌧net = I↵
1
I = M R2
2
↵=
1
4F R = I↵ = M R2
2
C.
B.-Champagne
Luis
Anchordoqui
Monday, October 24, 16
!
t
!
1
!
) F = MR
⇡ 20 N
t
8
t
2
41