Physics 201, Review 4

Important Notes: v This review does not replace your own preparation efforts v  Exercises used in this review do not form a test problem pool. Please practice more with end of chapter problems.

About Final q  When and where

§  Thursday Dec 21st 2:45-4:45 pm §  Room Allocation: See my email on Dec 4th.

q  Format §  Closed book §  1+3 8x11 formula sheets allowed, must be self prepared. (Absolutely no photocopying/printing of sample problems,

examples, class lectures, HW etc.) §  30 multiple choice questions. (count as 200 points) §  Bring a calculator (but no computer). Only basic calculation

functionality can be used. q  Special needs/ conflicts:

§  All special requests should have been settled by now. (except for medical emergency) §  All alternative test sessions are in our lab room, only for

approved requests.

Chapters Covered q  The final exam is cumulative. q  ~50 % will be on old chapters (Ch 1-12) q  ~50 % will be on new chapters (Ch 13,14,15)

§  Chapter 13: Gravitation §  Chapter 14: Fluid Mechanics §  Chapter 15: Oscillation Motion.

Super Friday: December 15th. 10am-5pm in the lab room.

Basic Concepts and Quantities q  Concepts: Understanding all key concepts in the covered chapters.

(“key” summary of each chapter) q  Basic Quantities:

§  Gravitational Force §  Gravitational Field (conceptual only) §  Gravitational Potential Energy §  Orbital parameters (r, T, ω, v, KE, PE,…) §  Escape Speed §  Pressure, Density, Buoyancy, Flow Velocity, Continuity. §  Bernoulli’s Equation (qualitative level only). §  Parameters to a simple harmonic oscillator A, ω, φ, (f, T) §  Kinetic and Potential Energy (For orbital motion, and for SHM)

Exam Topics(1) q  Universal gravitation force

§  Always attractive §  Dependency on masses(m1m2) and distance (1/r2)

q  Gravitational potential energy §  Dependency on masses and distance (1/r).

q  Orbits §  Circular orbit: v, E, KE, U, F are all determined by r. Why and

How? §  General elliptical orbit: General picture, energy consideration. §  Do not need to derive Kepler’s Laws but need to know

conceptually the physics behind it. §  Energy transfer when orbit changes (launch, fall)

Review: Dynamics of Orbiting q  For a circular orbit:

§  Force on m: FG= GMm/r2

§  Orbiting speed: v2 = GM/r (independent of m) §  T= 2πr/v (ω=2π/T, f=1/T) §  Kinetic energy: KE= ½ mv2 = ½ GMm/r §  Potential energy UG= = - GMm/r (note: UG = -2 KE !) §  Total Mechanic Energy: E = KE + UG = - ½ GMm/r (= ½ UG)

q  For a generic elliptical orbit : It can be derived (not required): E = - ½ GM/a (a: semi-major axis)

Exercise 1: Conceptual Q q  Compare two satellites in circular orbits, A and B, of the same

mass. Satellite A is on a lower orbit than satellite B’s. §  Satellite A has higher kinetic energy but lower potential energy §  Satellite A has lower kinetic energy but higher potential energy §  Satellite A’s kinetic energy and potential energy are both

higher than satellite B’s. §  Satellite A’s kinetic energy and potential energy are both lower

than satellite B’s. §  None of above or not enough information to determine

q  Solution: §  higher orbit higher potential energy §  higher orbit lower speed lower kinetic energy.

Exercise 2: Change orbit q  A satellite of mass m=200kg on a circular orbit at h1=200km above the earth surface. What is the minimum energy required to put it into a circular orbit of h=300km? (ignore earth spin, ME=5.97x1024kg, RE=6.37x106m, G=6.67x10-11 Nm2/kg2) Ø  Solution: initial: h=200km, ri = RE +200km, Ei = - ½ GMEm/(RE +200km) = -6.06x109J on orbit: h=300km, rf = RE + 300km Ef = - ½ GMEm/(RE+300km) = -5.97x109J Energy required = Ef – Ei = 9.1x107J

Exercise 2’: Launch Satellite q  A satellite of mass m=200kg is launched into a circular orbit at height h=200km above the earth surface. What is the minimum energy required to put it into the orbit ? (ignore earth spin) (ME=5.97x1024kg, RE=6.37x106m, G=6.67x10-11 Nm2/kg2) Ø  Solution: initial: h=0, ri = RE , Ei =KEi +Ui = 0 + (-GMEm/RE ) = -1.25x1010J on orbit: h=200km, rf = RE + 200km Ef = KEf + Uf = - ½ GMEm/(RE+200km) = -6.06x109J Energy required = Ef – Ei = 6.4x109J

Extra Exercise: Sun-Earth Distance q  Estimate the distance between the Earth and the Sun. (ME=5.97x1024kg, MSun= 1.989x1030 kg , G=6.67x10-11 Nm2/kg2) Ø  Solution : You can do it !

(Check with me or your TA if you are still puzzled after today)

Exam Topics(2) q  Density ρ = m/V q  Pressure P = F/A

§  Always normal to container surface q  Patmosphere = 1.05x105 N/m2 q  Pressure Variation on Depth/Height P=P0 + ρgh

§  Pascal’s Principle: Same depth Same pressure q  Buoyancy

§  Always upwards §  B = ρfluid Vin_fluid g (Archimedes’s Principle)

q  Fluid continuity: v2A2 = v1A1 q  Bernulli’s principle : P1 + ½ ρv1

2 + ρgh1 = P2 + ½ ρv22 + ρgh2

Exercise: Hydraulic Jack q  The radius of the car side cylinder of a hydraulic press is r2= 0.5m

while the press side radius r1=2cm. §  How much force is required to lift a 1500kg car?

§  If the car is to be lifted by 1.0m, how much work is required

€

P2 = P1 → F2

A2

=F1

A1

→F2

F1

=A2

A1

= r22

r12 = 625

→F1 = F2 /625 = 23.5N(="2.4kg")

€

W = F • Δx = F2Δx2 =1500x9.8x1m =14.7x103 J (= F1Δx1)

Exercise: Weight in Water q  A piece of iron (ρ=7.9x103 kg/m3) block weighs 1kg in air, How much does the scale read in water? Solution: In water: B+T2-mg =0 T2 = mg-B = mg – ρwaterVg = mg –( ρwater /ρiron ) ρiron Vg = mg (1-ρwater /ρiron ) = 0.87mg = “0.87 kg”

Exercise: Pascal’s Law

B A

q  Is PA > PB , PA = PB , or PA < PB ? Ø  Answer: PA= PC – ρHggh < PD - ρh2ogh= PB

D C

h

Exercise: Pascal’s Principle q  A swimming pool has a total bottom area of 100 m2. When

filled with 1.0 m of water, what is the total force on the bottom by the water? (ρh2o=103 kg/m3, Patm= 1 x 105 Pa, g = 10 m/s2))

1.0x106 N, 1.1x107 N, other (10% error)

p0 =1atm

p = p0 + ρgh F= pA = (105 + 103x10x1.0)x100

= 1.1 x107 N

Another Simple but Non-trivial Exercise Ø  A legendary Dutch boy saved Holland by plugging a hole in a

dike with his finger. The area of the hole was A=1.0x10-4 m2. and the hole was 3.00 m below the surface of the North Sea

(ρwater =1.0x103 kg/m3, 1ATM=105 Pa, g=10 m/s2) q  what was the total force on his finger? 3.0 N, 0.0 N, 10 N, 13 N, other q  what was the total force on his finger by water? 3.0 N, 0.0 N, 10 N, 13 N, other Please do spend time to understand this problem. Answer offered in the text may not be correct.

Review: Fluid Dynamics

Bernoulli’s Equation: P+ ½ ρv2 + ρgy=constant

Continuity Equation: A1v1= A2v2 (=constant)

Exercise: Venturi’s Tube q  Consider a Venturi tube as shown below. Initially, when no air flow through the top (horizontal) tube, the fluid levels of the three vertical tubes are the same, as shown in the figure at right.

Now, when air is blown (from left to right) through the top tube, what can we say about the fluid levels in the vertical tubes? a. The middle tube has highest fluid level b. The middle tube has lowest fluid level c. All three tubes still have the same fluid level.

The Venturi Tube q 

same height: P+ ½ ρv2 = constant lower v higher P

Larger A, lower v

higher P lower liquid height

Exam Topics(3) q  Simple harmonic motion (SHM). x(t) = A cos(ωt + φ)

§  Basic parameters A,ω,φ. •  how to identify them (given an expression or given a

graph); §  v, a, F, E, KE, U in SHM. §  Spring-Block and simple pendulum in SHM. No derivation is

required but you are expected to know what determines the period (or angular frequency).

§  Conversion among ω, f, T §  Damped oscillation

•  Energy consideration •  Frequency change

§  Forced oscillation •  resonance condition

Review: Harmonic Oscillation q  Motion described by expression x(t)=Acos(ωt+φ) is called harmonic oscillation

A: amplitude

x(t)=Acos(ωt+φ)

ω: angular frequency

φ: phase constant

T= 1/f = 2π/ω

x(0)=Acos(φ)

Exercise: Recognize Phase Constant

q  An oscillation is described by x=Acos(ωt+φ). Find out φ for each of the following figures:

x

x

x

t

t

t

Answer

φ= 0

φ= π/2

φ= π

π/2 π

3π/2 2π

Or, one can use cosφ=x(0)/A

Determine φ Angle: A Real Test Problem q  A simple harmonic motion in the form x(t)=Acos(wt+φ) is shown

in graph below. Estimate the φ angle from the graph.

0.2 π , 0.8 π , 1.2 π, 1.8 π, other (more than 0.1π from above)

One Simple Exercise q  A simple harmonic oscillation for a particle is described by

x(t)= 2.7cos(4.3t+0.8). (in SI units). What is the period of the oscillation?

2.0 s , 1.5 s , 3.5 s , 4.3s, none of above is within 5%. Solution: ω=4.3 T = 2π/ω = 1.5 s

Common SHM’s

€

ω =km

€

ω =gl

€

ω =κI

SHM: Conceptual Quiz q  Compare two simple pendulum A and B. Pendulum A has

longer string than pendulum B. §  Pendulum A has longer period but smaller amplitude. §  Pendulum A has shorter period but larger amplitude. §  Pendulum A has longer period and larger amplitude. §  Pendulum A has shorter period and smaller amplitude. §  None of above

q  Solution: §  longer L smaller ω longer T §  but amplitude is independent of L. (determined by initial condition only)

SHM: More Exercises q  For a particle of mass m in harmonic motion x(t)=Acos(ωt+φ) :

§  what is the force on it as function of t? Ø F(t)=ma(t)= -Amω2cos(ωt+φ) i.e. F = - mω2x = - kx; k=mω2 or ω2=k/m

§  what is the kinetic energy as function of t? Ø K(t) = ½ mv2 = ½ mA2ω2sin2(ωt+φ)

§  Is the force on it a conservative force? Ø Yes. (recall: spring force is conservative)

§  If yes, what is the potential energy associated? Ø U(t) = ½ kx2 = ½ mA2ω2cos2(ωt+φ)

§  what is total mechanic energy as function of t? Ø E(t) =K(t)+U(t) = ½ mA2ω2

§  Is total mechanic energy a constant? Ø Yes, E(t) = ½ mA2ω2 independent of t.

Good Luck

(and Thanks for a Wonderful Semester!)