Upload
rosa-manning
View
218
Download
0
Tags:
Embed Size (px)
Citation preview
Announcements
No Math Review Night Next Week
Pset 7 Due Tuesday Oct 25 at 9 pm
Test Two Next Week Thursday Oct 27 7:30-9:30 Section Room Assignments on Announcements Page
Test Two Topics: Circular Motion, Energy, Momentum, and Collisions
Wednesday and Thursday Classes will be Reviews for Test Two
Juno Lift-Off
http://anon.nasa-global.edgesuite.net/HD_downloads/juno_launch_1080i.wmv
Worked Example: Stream Bouncing off Wall
A stream of particles of mass m and separation d hits a perpendicular surface with speed v. The stream rebounds along the original line of motion with the same speed. The mass per unit length of the incident stream is = m/d. What is the magnitude of the force on the surface?
Category 1: Adding Mass But Not Momentum
There is a transfer of material into the object but no transfer of momentum in the direction of motion of the object. Consider for example rain falling vertically downward into a moving cart. A small amount of rain has no component of momentum in the direction of motion of the cart.
Concept Question
Suppose rain falls vertically into an open cart rolling along a straight horizontal track with negligible friction. As a result of the accumulating water, the speed of the cart
1.increases.
2.does not change.
3.decreases.
4.not sure.
5.not enough information is given to decide.
Concept Question
Answer 3. The water, because it falls vertically, does not change the cart's horizontal momentum. The mass of the cart increases, however, and so its speed decreases.
Category 2: Losing Mass But Not Momentum
The material continually leaves the object but it does not transport any momentum away from the object in the direction of motion of the object. For example, consider an ice skater gliding on ice holding a bag of sand that is leaking straight down with respect to the moving skater.
Category 3: Impulse
The material continually hits the object providing an impulse resulting in a transfer of momentum to the object in the direction of motion. For example, suppose a fire hose is used to put out a fire on a boat. The incoming water continually hits the boat impulsing it forward.
Category 4: Recoil
The material continually is ejected from the object, resulting in a recoil of the object. For example when fuel is ejected from the back of a rocket, the rocket recoils forward.
Worked Example: Rocket Motion
A rocket at time t = 0 is moving with speed vr,o in the positive x-direction in empty space. The rocket burns the fuel at a rate dmf /dt =b > 0. The fuel is ejected backward with speed u relative to the rocket.
a) What is the relationship between the time rate of change of exhaust mass dmf /dt, and the time rate of change of rocket mass dmr /dt?
b) Find an equation for the rate of change of the speed of the rocket in terms mr (t) ,u, and dmr
/dt and solve for v.
c) Find the differential equation describing the motion of the rocket if it is in a constant gravitational field of magnitude g
Strategy : Rocket Motion
• Goal: Determine velocity of rocket as function of time as mass is continuously ejected at rate dmf /dt with speed u relative to rocket.
• System: consider all elements that undergo momentum change: rocket and fuel
• Using Momentum flow diagram, apply
to find differential equation that describes motion.
total total
ext0
( ) ( )lim
t
t t t
tΔ →
+Δ −=
ΔP P
F
r rr
Rocket Motion: State at time t
1. Rocket with total mass mr(t) moves with speed vr(t) in positive x-direction according to observer
2. Total mass consists of mass of rocket mr,0 and fuel mf (t)
3. Fuel element with mass Δmf, moves with speed of rocket vr (t) at time t, is ejected during interval [t, t+Δt]
4. x-component of momentum at time t
P
sys,x(t) =(mr (t) + Δmf )vr (t)
State at t + Δt
• Rocket is propelled forward by ejected fuel with new rocket speed
• Fuel is ejected backward with speed u relative to rocket. Relative to observer’s frame, ejected fuel element has speed
• x-component of system’s momentum at time t+Δt
vr(t + Δt) =vr (t) + Δvr
vr+ Δvr −u
P
sys,x(t + Δt) =mr (t)(vr + Δvr ) + Δmf (vr + Δvr −u)
Rocket EquationAre there any external forces at time t? Two cases:
(1) Taking off
(2) Negligible gravitational field
Apply Momentum Principle:
F
ext ,x=−mr (t)g
F
ext ,x=0
F
ext ,x= lim
Δt→ 0
Psys,x(t+ Δt) −Psys,x(t)
Δt
Fext ,x
= limΔt→ 0
mr (vr + Δvr ) + Δmf (vr + Δvr −u) −(mr + Δmf )vr
Δt
Fext,x = limΔt→ 0
mrΔvr −Δmfu
Δt+ lim
Δt→ 0
ΔmrΔvr
Δt
Fext,x =mr
dvr
dt−
dmf
dtu
Rocket Equation: Thrust
Fuel ejection thrust force:
Vector equation version:
F
ext ,x+
dmf
dtu=mr
dvr
dt⇒ Fext,x + Fthrust,x =mr
dvr
dt
F
thrust ,x≡
dmf
dtu
rF
ext−
dmf
dtru=mr (t)
drvr
dt⇒
rFext +
rFthrust =mr (t)
drvr
dt
rF
thrust≡−
dmf
dtru
Rocket Equation: Conservation of Mass
Conservation of mass: Rate of decrease of mass of rocket equals rate of ejection of mass
Rocket equation:
Vector version:
dm f
dt=−
dmr (t)dt
F
ext ,x=mr
dvr
dt+
dmr
dtu
Solution: Rocket Motion in Gravitational Field
• Rocket Equation:
• Fuel ejection term can be interpreted as thrust force
• Relative fuel ejection velocity
• External force
• Rocket equation
• Integrate with respect to time
• Solution:
Fext (t) mrgk̂
mrg mr
dvr ,z
dt
dmr
dtu
ru =−uk̂
dvr ,z
dt
1
mr
dmr
dtu g
dvr ,z
dtdt
t0 0
t f
1
mr
dmr
dtudt
t0 0
t f
gdtt0 0
t f
vr ,z (t f ) u lnmr (t 0)
mr (t f ) gt f
rF
thrust=
dmr
dtru
Concept Question: Rocket Fuel Burn Time
When a rocket accelerates in a constant gravitational field, will it reach a greater final velocity if the fuel burn time is
1. as fast as possible?
2. as slow as possible?
3. The final speed is independent of the fuel burn time?
4. I’m not sure.
Answer: Rocket Equation in Gravitational Field
• Fuel ejection term can be interpreted as thrust force
• Solution: shorter the burn time, the greater the velocity
rF
thrust
dmr
dtru
Concept Question: Rocket with Constant Thrust
If a rocket in gravity-free outer space has a constant thrust at all times while burning fuel, is its acceleration
1. constant?
2. increasing?
3. decreasing?
Concept Question: Rocket with Constant Thrust
Answer 3. Acceleration is increasing because fuel is decreasing.
dmr
dtru =mr (t)
drvr
dt⇒
rFthrust =mr (t)
drvr
dt⇒
drvr
dt=
rFthrust
mr (t)
Table Problem: Rocket SledA rocket sled ejects gas backwards at a speed u relative to the rocket sled.
The mass of the fuel in the rocket sled is equal to one half the initial total mass mr,0
(including fuel) of the sled. The rocket sled starts from rest on a frictionless track. You may ignore air resistance.
• Derive a relation between the differential of the speed of the rocket sled, dvr , and the differential of the total mass of the rocket, dmr
• Integrate the above relation to find the speed of the rocket sled as a function of mass, vr(m), as the rocket sled speeds up.
• What is the final speed of the rocket sled after all the fuel has been burned? Express your answers in terms of the quantities u, and mr,0 as needed.