Momentum for Changing Mass

7/26/2019 Momentum for Changing Mass

1/21

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:


2/21

Clickers Not Synched with Students

00153C29

0CCDF332

0E8D199A

0F8C6AE926657437

32AD801F

33505A3933506B08

33B10587

33BC9817

I have scores for these

clickers, but they are not

registered. Check D2L


3/21

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


4/21

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!



5/21

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.


6/21


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.


7/21


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.


8/21


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.


9/21

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)


10/21

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



11/21

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.



12/21

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


13/21

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)


14/21

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!


15/21

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


16/21

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


17/21

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


18/21

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


19/21

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


20/21

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


21/21

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

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Momentum for Changing Mass