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Rockets (Development)
Next midterm a week from
Thursday (3/15). Chapters
69 will be covered
Introduce change of
momentum due to varyingmass.
Material from Chapter 9.12
Webpage:http://www.colorado.edu/physics/phys1110/phys1110_sp12/
Announcements:
7/26/2019 Momentum for Changing Mass
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Clickers Not Synched with Students
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I have scores for these
clickers, but they are not
registered. Check D2L
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A rocket is launched straight up, and the net force
on the rocket as a function of time is shown in this
graph. The rocket travels straight up (defined to be
positive here.)
Between t=1 hr and 3 hrs, what happens to the speed of the rocket?
A). The rocket is speeding up the whole time
B) The rocket is slowing down the whole time.
C) The rocket is moving with a constant speed the whole time.
D) The rocket is speeding up for awhile, and then it slows down.
E)There is not enough information given to decide.
Clicker question 1 Set frequency to BA
7/26/2019 Momentum for Changing Mass
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A rocket is launched straight up, and the net force
on the rocket as a function of time is shown in this
graph. The rocket travels straight up (defined to be
positive here.)
Between t=1 hr and 3 hrs, what happens to the speed of the rocket?
A). The rocket is speeding up the whole time
B) The rocket is slowing down the whole time.
C) The rocket is moving with a constant speed the whole time.
D) The rocket is speeding up for awhile, and then it slows down.
E)There is not enough information given to decide.
The force is positive the whole time, which means a is positive the whole time, and
positive acceleration, starting from rest, means you speed up the WHOLE time!
Clicker question 1 Set frequency to BA
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Set frequency to BAClicker question 2
Inside the rocket, there is some payload, asshown. On the left is Rover I (RI). On the
right is Rover II (RII), which is identical to
RI. Sitting on top of RII is a sensor package
(S) which weighs much less than the Rovers
do.
At t=1 hr, how does the acceleration of sensor (S) compare to the acceleration
of Rover II?
A)a(S) = a(RII) B)a(S) < a(RII) C)a(S) > a(RII)
D)Without knowing more numbers, we cannot decide how the accelerations of
S and RII compare.
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Set frequency to BAClicker question 2
Inside the rocket, there is some payload, asshown. On the left is Rover I (RI). On the
right is Rover II (RII), which is identical to
RI. Sitting on top of RII is a sensor package
(S) which weighs much less than the Rovers
do.
At t=1 hr, how does the acceleration of sensor (S) compare to the acceleration
of Rover II?
A)a(S) = a(RII) B)a(S) < a(RII) C)a(S) > a(RII)
D)Without knowing more numbers, we cannot decide how the accelerations of
S and RII compare.
The two packages touch each other, they have the same motion,
the same acceleration.
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Set frequency to BAClicker question 3
Consider the following two statements and
decide if they are true or false.
i) At any given instant in time between t=1 and
t=4 hrs, the magnitude of the normal force of
the floor on RI is equal to the magnitude of the
normal force of the floor on RII
ii) Between t=0 and 5 hrs, the absolute value of the work
done by the normal force of the floor on RI is equal to the
absolute value of the work done by the normal force of the
floor on RII
A) Without knowing more numbers, we cannot decide whether thesestatements are true or false.
B) Both statements are true
C) i is true, but ii is false.
D) i is false, but ii is true
E) Both statements are false.
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Set frequency to BAClicker question 3
Consider the following two statements and
decide if they are true or false.
i) At any given instant in time between t=1 and
t=4 hrs, the magnitude of the normal force of
the floor on RI is equal to the magnitude of the
normal force of the floor on RII
ii) Between t=0 and 5 hrs, the absolute value of the work
done by the normal force of the floor on RI is equal to the
absolute value of the work done by the normal force of the
floor on RII
A) Without knowing more numbers, we cannot decide whether thesestatements are true or false.
B) Both statements are true
C) i is true, but ii is false.
D) i is false, but ii is true
E) Both statements are false.
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Rover Question Response
i) This statement is false. Just think about it: the floor has to "holdup" (even accelerate!) more on the right than it does on the left.
The normal force of the floor on rover I is given by F_net = ma, which
meansN-mg = ma, or N (on I) =m(g+a), where m is the mass of
Rover I. For Rover II, consider the SYSTEM RoverII+sensor: thenormal force on this is given by N (on II) = (m+msensor)*(g+a),
which is manifestly bigger than N(on I)
ii) This is false. Since the normal force is bigger on the RII, the work
will also be larger. (Distance is same)
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A student is asked to sketch a force diagram for Rover II only, as
viewed from the ground just after launch ("N" represents "normal",
W represents "Weight") Note that none of the diagrams are
complete, because the student has not properly labeled forces "ON
object BY object", but which is best
Clicker question 4 Set frequency to BA
7/26/2019 Momentum for Changing Mass
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A student is asked to sketch a force diagram for Rover II only, as
viewed from the ground just after launch ("N" represents "normal",
W represents "Weight") Note that none of the diagrams are
complete, because the student has not properly labeled forces "ON
object BY object", but which is best
D is correct. There is the normal force of the floor UP, the weight
DOWN, and the normal force of the sensor pushing DOWN.
Clicker question 4 Set frequency to BA
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Rocket Equations
The two states of the system are shown at
instants t and t + !t. Initially, whole mass M moves
with . After !t time, !M moves with relativevelocity while the rest (M-!M) moves
with . Note that absolute velocity of !M is
!
V!
Vrel
!
V + "!
V!
Vrel+
!
V + "!
V
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Rocket Equation
On the System
!
Pi=M
!
V!
Pf "!
Pi
#t=
M#!
V + #M!
Vrel
#t
"t# 0$d!
P
dt=M
d!
V
dt+
dm
dt
!
Vrel
!
Fext ="t#0lim
"P
"t=
"t#0lim
Pf $ Pi
"t
!
Pf= (M
" #
M)(
!
V+ #
!
V) + #
M(!
Vrel+
!
V+ #
!
V)
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Rockets cont.
!
Fext=d
!
P
dt=M
d
!
V
dt+dm
dt
!
Vrel
!
Fext =M
d!
V
dt+
dm
dt
!
Vrel =M
!
a +
dm
dt
!
Vrel
If there is no external force (ie gravity), then
M!
a = " dm
dt
!
Vrel=R
!
Vrel
Note R is the positivemass rate of fuel
consumption.
RVrelis call the thrustof the rocket engine!
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How does velocity change as fuel is
consumed?
M!
a =Mdv
dt= "
dM
dt
!
Vrel
dv = "VreldM
M
dvvi
vf
" = #Vrel dMMMiM
f
"
vf " v
i=V
relln
Mi
Mf
v f " v i = v0 " gt+VrellnMi
Mf
If in gravity and
with initial velocity
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Balancing the THRUST of a water hose
A firehose delivers 50kg/s at a speed of 30m/s. How many
75-kg firefighters are need to hold the hose on muddy
ground for which the coefficient of static friction is 0.35?
With their feet firmly planted on the ground, the firefighters
(suppose there are n of them provide a maximal force of
friction
First, we must enough firefighters to balance the THRUST of
the hose.
THRUST=VrdM
dt=(30m /s)(50kg /s) =1500N
Ff=
nsN=
nsmgIf they are to remain steady , soFf " THRUST
n "THRUST
smg=
1500N
(0.35)(75kg)(9.8m /s2)=5.8 So six is sufficient
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Space Shuttle
A space shuttles main engines develop a thrust of35 x106N as they eject gas at 2500 m/s. How
much fuel must the shuttle carry to permit a 5-
minute engine firing?
From the rocket equation, we find:
dM
dt=
Thrust
Vr=
35"106N
2500m /s=14,000kg /s
For a five minute burn, we need
(1.4 "104kg /s)(300s) =4.2"10
6kg
Space shuttles external tank contains 500,000 gallons of
super cold liquid oxygen and liquid hydrogen
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Varying Mass
Two large barges are moving in
the same direction in still water,
one with speed 10km/hr and
the other with a speed 20km/hr.
While they are passing each
other, coal is shoveled from theslower to the faster one at a
rate of 1000kg/min. How much
additional force must be
provided by the engines of the
faster barge and the slower
barge if neither is to change
their speed?
Force = Vreldm/dt
VrdM
dt=
10km /hr "1000kg /s
=10km
hr
1hr
3600s"1000
kg
min
1min
60s
= 46N For faster
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A hopper releases grain at a rate of dm/
dt onto a conveyer belt that moves atconstant speed v. What is the power of
the motor driving the belt?
The system is some arbitrary length of belt whose mass
we can call M. The mass of the system increases at thesame rate that the grain falls so dM/dt=dm/dt
Let Fext be the force needed to maintain constant speed
because the mass is increasing. Fext= v
dm
dt
The power required is P =!
F !
v = v2 dm
dt
Note :dK
dt
=
d
dt
1
2
mv2
"
#
$%
&
'=1
2
v2 dm
dt
Grain Hopper Problem
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Chain Falling
A vertical chain has a length L and a mass M. It isreleased with the bottom just touching a table.
a) Find the force on the table as a function of the
distance fallen by the top end.
F=dp
dt=
dm
dtv +m
dv
dt
If dm/dt=0, F is just mg. So if the mass was falling in one
piece, the force would still be equal to dp/dt, not dp/dt+mg.
Just before chain hits the table,
vdm
dt=v
M
L
dx
dt=v
M
Lv
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Chain Falling
F= dpdt
= dmdt
v +mdvdt
vdm
dt
=vM
L
dx
dt
=vM
L
v
Part of chain not on tableis in free fall:1
2mv
2= mgx
vdm
dt=
M
Lv
2=2
Mgx
L
F= 2M
Lgx +
M
Lgx = 3
M
Lgx
Note Force could be
3mg if the whole
chain falls