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Main points of today’s lecture: Physics 231 Lecture 14 Multiple impulses and rocket propulsion. propellant f 0 F t mv mv Δ=Δ −Δ ( ) ( ) propellant f 0 m F v v Thrust t m Thrust v v Δ = =− Δ Δ = Things that oscillate Mass and Spring ( ) 0 f Thrust v v t = Δ Circular motion T 1/f; f 1/T; 2f ω π = = = 2 for mass and spring k/m ω = Simple harmonic motion 0 0 x A cos( t ) v A sin( t ) ω θ ω ω θ = + =− + x 0 2 x 0 0 v A sin( t ) a A cos( t ) and A are constants ω ω θ ω ω θ θ + =− +

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### Transcript of Physics 231 Lecture 14 - NSCLlynch/phy231_2011/lecture14.pdf · • Main points of today’s... • Main points of today’s lecture:

Physics 231 Lecture 14

• Multiple impulses and rocket propulsion.

propellant f 0F t m v m vΔ = Δ ⋅ − Δ ⋅

( )

( )

propellant f 0mF v v Thrustt

mThrust v v

Δ= − = −Δ

Δ= −

• Things that oscillate • Mass and Spring

( )0 fThrust v vt

= −Δ

• Circular motionT 1/ f ; f 1/ T; 2 fω π= = =

2for mass and spring k / m ω =• Simple harmonic motion

0

0

x A cos( t )v A sin( t )

ω θω ω θ

= += − +

p g

x 02

x 0

0

v A sin( t )

a A cos( t ) and A are constants

ω ω θω ω θ

θ

+

= − + Mid-term correction set open • Final Mid-term exam scores will be• Final Mid-term exam scores will be

sent by email on Monday after corrections set closes.

40

50Inclass score vs. total score

Did corrections set

20

30

Tota

l sco

re

set

Did not

• In the past, most students did the corrections set. – Allows you to learn how to do the

0

10do correctionsset

Allows you to learn how to do the problems you may have gotten wrong.

– Improves your score.0 10 20 30 40 50

In class score

• Some students didn’t do the corrections setcorrections set.– Why? Conceptual question

S i f ll ti ll i t t lli l• 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

) i– a) increases.– b) does not change.– c) decreases. cart

f cart 0mv v=f cart ,0

cart waterm m+

Example of slow accumulation of mass by ti f t t ll i l ti lli ia continuous sequence of totally inelastic collisions. RocketsRocket propulsion is an example of conservation of momentum:Rocket propulsion is an example of conservation of momentum:

The rocket doesn’t push on the environment, as in propulsion. It pushes on the exhaust gas, and the exhaust gas pushes the rocket forward. It is a continuous

f ll l isequence of small explosions. Newton’s third law, but seen more easily from the perspective of conservation of momentum. Slide 9-24 • An astronaut is motionless in outer space. Upon command, his propulsion unit strapped to his back ejects some gas with a velocity of

Examplep p pp j g y+14 m/s, and the astronaut recoils with a velocity of -0.5m/s. After the gas is ejected, the mass of the astronaut is 160kg. What is the mass of the ejected gas? If the mass is ejected in 2 s, what is the thrust force and the average rate of emitted gas Δm/Δt? (Hint treat thrust as the average force acting over Δt.)

ast ,0 gas,0 ast ,f gas,fp p 0 p p+ = = +

gas,f ast ,fp p= −

ast ,f ast ast ,f gas,f gas gas,fp m v ; p m v= =

take motion to be along x axisgas gas,f ,x ast ast ,f ,xm v m v= −

v 0 5m / sast,f ,xgas ast

gas,f ,x

v 0.5m / sm m 160kg 5.7kgv 14m / s

−= − = − =

ast ,ft f

p 160kg 0.5m / sF t Thrust t p Thrust 40NΔ ≡ Δ = = = − = −

av ast ,fF t Thrust t p Thrust 40N

t 2sΔ Δ

Δ

m 5.7kg 2.85kg / st 2s

Δ = =Δ Rocket propulsion

Th th t f k t b t d i th i l• The thrust force on a rocket can be computed using the impulse momentum theorem knowing the rate of mass Δm/Δt of propellant that is being emitted per second and its emission velocity relative t the rocket v vthe rocket vf - v0.

( )0 fmThrust v vt

Δ= −Δ

Pr oof :

( )propellant f 0

gas f 0propellant

F t m v m v

P m v vF

t t

Δ = Δ ⋅ − Δ ⋅

Δ Δ −= =

Δ Δ

( )

p p

propellant

t tF Thrust; by Newton's third law

mTh t

Δ Δ= −

Δ ( )0 fThrust v vt

= −Δ Periodic motion

P i di ill t ti i ti th t tl t it lf ft• Periodic or oscillatory motion is motion that exactly repeats itself after a certain amount of time T has elapsed.

• This time T is called the period. • The number of times the motion repeats itself in a second is call the

frequency.• Examples of systems that display periodic motion.

– a system of masses and springs– a pendulum or swing. – the second hand on a clock, which repeats its motion in exactly 60the second hand on a clock, which repeats its motion in exactly 60

seconds.– circular motion with constant angular velocity.

The vibrating surface that makes the sound in a musical– The vibrating surface that makes the sound in a musical instrument.

• Piano stringB ll• Bell.

• Simple Harmonic Motion is a common form of periodic motion. Springs and simple harmonic motion

• Consider the mass and spring system to the right. The system is at equilibrium at x=0.

• This is an example of stable equilibrium. If an s s a e a p e o s ab e equ b u . aobject is in equilibrium then the net force on it is zero. If you displace an object away from a stable equilibrium there will be a force p shing it back in the opposite direction of thepushing it back in the opposite direction of the displacement.

• When the equilibrium is stable, the force around equilibrium often depends linearly onaround equilibrium often depends linearly on displacement. For the spring, this gives rise to the Hooke’s law force: kxFs −=

• If you displace the mass away from equilibrium by an amount x0 =A, the mass will oscillate about the minimum.

• This oscillation is an example of Simple Harmonic Motion (SHM)• This oscillation is an example of Simple Harmonic Motion (SHM)

x Acos tω= Ω has units rad/s or Hz1/s is sometimes called Hz Review problem: Hooke’s law

I th t i 2 44 hi h i ( t i d l th 0 30 )• In a room that is 2.44 m high, a spring (unstrained length=0.30 m) hangs from the ceiling. From this spring, a board hangs so that its 1.98 m length is perpendicular to the floor, the lower end just extending to, but not touching the floor The board weighs 102 N What is thebut not touching, the floor. The board weighs 102 N. What is the spring constant of the spring?

total0 F ky mg= = − − ( )k d mg= − − − kd mg= −total y g ( )k d mg kd mg

mgk = ( )102N 638N / m

2 44 1 98 0 3= =

d=(2.44-1.98-0.3)md ( )2.44 1.98 0.3 m− −

If displaced from d and released, the board will oscillate up and down (ifboard will oscillate up and down (if the floor were not in the way). Vertical vs. horizontal oscillations

• Whether the motion is vertical or horizontal: k mω =

• The only thing gravity does is displace the equilibrium point a distance ΔL=mg/k downwards.

A• Mass moves with Simple Harmonic Motion:– Not that increasing the amplitude A only increases the

displacement from the equilibrium point at x=0.

x Acos tω= Conceptual question

A bj t h i l h i ti ( ill ti ) b t• An object can have simple harmonic motion (oscillation) about– a. any equilibrium point.– b. any stable equilibrium point.– c. around certain stable equilibrium points where Hooke’s

law is obeyed.– d. any point, provided the forces exerted on it obey Hooke’s y p p y

law.– e. any point. What is the acceleration in Simple Harmonic Motion (SHM)

F H k ’ l• From Hooke’s law

xF kx= −and from Newton’s second law:

x xF ma=and from Newton s second law:

P tti it t thPutting it together

xma kx= −

xka xm

= −

Simple harmonic motion 2 kx Acos t; m

ω ω= =kmeans 2

xka Acos t Acos tm

ω ω ω= − = −