Physics 201: Lecture 30, Pg 1 Lecture 30 Review (the final is, to a large degree, cumulative) Review (the final is, to a large degree, cumulative) ~50%

Physics 201: Lecture 30, Pg 1

Lecture 30

• Review (the final is, to a large degree, cumulative)Review (the final is, to a large degree, cumulative) ~50% refers to material in Ch. 1-12 ~50% refers to material in Ch. 13,14,15

-- Chapter 13: Gravitation

-- Chapter 14: Newtonian Fluids

-- Chapter 15: Oscillatory Motion

• Today we will review chapters 13-15Today we will review chapters 13-15

• 11stst is short mention of resonance is short mention of resonance

• Order Ch. 15 to 13Order Ch. 15 to 13


Final Exam Details

Sunday, May 13th 10:05am-12:05pm in 125 Ag Hall & quiet room

Format: Closed book Up to 4 8½x1 sheets, hand written only Approximately 50% from Chapters 13-15 and 50% 1-12 Bring a calculator

Special needs/ conflicts:

All requests for alternative test arrangements should be made by today (except for medical emergency)


Driven SHM with Resistance Apply a sinusoidal force, F0 cos (t), and now consider what A and b do,

2220

2

0

)()(

/

mb

mFA

tm

Fx

m

k

dt

dx

m

b

dt

xd cos 2

2

Not Zero!!!

b/m small

b/m middling

b large

stea

dy s

tate

am

plitu

de


Dramatic example of resonance

In 1940, a steady wind set up a torsional vibration in the Tacoma Narrows Bridge


Dramatic example of resonance

Eventually it collapsed


Mechanical Energy of the Spring-Mass System

Kinetic energy:

K = ½ mv2 = ½ m(A)2 sin2(t+)

Potential energy:

U = ½ k x2 = ½ k A2 cos2(t + )

And 2 = k / m or k = m 2

K + U = constant

x(t) = A cos( t + )

v(t) = -A sin( t + )

a(t) = -2A cos( t + ) & F(t)=ma(t)


SHM

x(t) = A cos( t + )

v(t) = -A sin( t + )

a(t) = -2A cos( t + )

A : amplitude : angular frequency=2 f =2/T

: phase constant

xmax = A

vmax = A

amax = 2A


Recognizing the phase constant An oscillation is described by x(t) =A cos(ωt+φ).

Find φ for each of the following figures:

Answers

φ = 0

φ = π/2

x(t)= A cos (π)

φ = π


Common SHMs

mk

Lg

I


SHM: Friction with velocity dependent Drag force -bv

b is the drag coefficient; soln is a damped exponential

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

t

A

)(cos 2

expA )( )( ttmbtx mbo 2/if

22

2

m

bo

2

2

m

b

m

k


SHM: Friction with velocity dependent Drag force -bv

If the maximum amplitude drop 50% in 10 seconds,what will the relative drop be in 30 more seconds?

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

t

A

)()s 10( 21 txtx

)()s 30( tcxtx

2expA

)30(2

expA

)()30(

)(

)(

tmb

tmb

txtx

)(

202

exp)10(

)(

)30( )(

txmbtx

tx

tx

2

102

[exp

)()30(

2)](mb

txtx

8

1

2)(

)30( 221

tx

tx


Chapter 14 Fluids Density ρ = m/V Pressure P = F/A P1 atm = 1x105 N/m2

Force is normal to container surface Pressure with Depth/Height P = P0 + ρgh Gauge vs. Absolute pressure Pascal’s Principle: Same depth Same pressure Buoyancy, force, B, is always upwards B = ρfluid Vfluid displaced g (Archimedes’s Principle) Flow

Continuity: Q = v2A2 = v1A1 (volume / time or m3/s)

Bernoulli’s eqn: P1+ ½ ρv12 + ρgh1 = P2+ ½ ρv2

2 + ρgh2


Example problem

A piece of iron (ρ=7.9x103 kg/m3) block weighs 1.0 N in air.

How much does the scale read in water? Solution: In air

T1 = mg = ρiromV g In water: B+T2-mg = 0

T2 = mg-B

= mg – ρwaterVg

= mg – ( ρwater /ρiron ) ρiron Vg

= mg (1-ρwater /ρiron )

= 0.87mg = 0.87 N


Another buoyancy problem

A spherical balloon is filled with air (air 1.2 kg/m3). The radius of the balloon is 0.50 m and the wall thickness of the latex wall is 0.01 m (latex 103 kg/m3). The balloon is anchored to the bottom of stream which is flowing from left to right at 2.0 m/s. The massless string makes an angle of 30° from the stream bed.

What is the magnitude of the drag force

on the balloon?

Key physics: Equilbrium and buoyancy.

Fx=0 & Fy=0


Another buoyancy problem

A spherical balloon is filled with air (air 1.2 kg/m3). The radius of the balloon is 0.50 m and the wall thickness of the latex wall is 1.0 cm (latex 103 kg/m3). The balloon is anchored to the bottom of stream which is flowing from left to right at 2.0 m/s. The massless string makes an angle of 45° from the stream bed.

What is the magnitude of the drag force on the balloon?

Key physics: Equilibrium and buoyancy. Fx=0 & Fy=0

Fx=-T cos + D = 0

Fy=-T sin + Fb - Wair = 0

Wair = air V g= air (4/3 r3) g with r=0.49 m

Fb= water V g= water (4/3 r3) g

Variation: What is the maximum wall

thickness of a lead balloon filled with He?

Wair

Fb

T

D


Pascal’s Principle

Is PA = PB ?

Answer: No!

Same level, same pressure, only if same fluid density

AB


Power from a river

Water in a river has a rectangular cross section which is 50 m wide and 5 m deep. The water is flowing at 1.5 m/s horizontally. A little bit downstream the water goes over a water fall 50 m high. How much power is potentially being generated in the fall?

W = Fd = mgh Pavg= W / t = (m/t) gh Q = Av and m/t = water Q (kg/m3 m3/s) Pavg= water Av gh

= 103 kg/m3 x 250 m2 x 1.5 m/s x 10 m/s2 x 50 m

= 1.8x108 kg m2/s2/s = 180 MW


Chapter 13 Gravitation Universal gravitation force

Always attractive Proportional to the mass ( m1m2 ) Inversely proportional to the square of the distance (1/r2) Central force: orbits conserve angular momentum

Gravitational potential energy Always negative Proportional to the mass ( m1m2 ) Inversely proportional to the distance (1/r)

Circular orbits: Dynamical quantities (v,E,K,U,F) involve radius K(r) = - ½ U(r)

Employ conservation of angular momentum in elliptical orbits No need to derive Kepler’s Laws (know the reasons for them) Energy transfer when orbit radius changes(e.g. escape velocity)


Key equations

Newton’s Universal “Law” of Gravity

2by1on12221

1by 2on r̂ Frmm

GF

Universal Gravitational Constant G = 6.673 x 10-11 Nm2 / kg2

The force points along the line connecting the two objects.

2surface

ERGM

g On Earth, near sea level, it can be shown that gsurface ≈ 9.8 m/s2.

Gravitational potential energy rmGmrU /)( 21“Zero” of potential energy defined to be at r = ∞, force 0

At apogee and perigee: mvrL


Dynamics of Circular Orbits For a circular orbit: Force on m: FG= GMm/r2

Orbiting speed: v2 = GM/r (independent of m) Kinetic energy: K = ½ mv2 = ½ GMm/r Potential energy UG= - GMm/r

Notice UG = -2 K Total Mech. Energy:

E = KE + UG = - ½ GMm/r


Changing orbit

A 200 kg satellite is launched into a circular orbit at height h= 200 km above the Earth’s surface.

What is the minimum energy required to put it into the orbit ? (ignore Earth’s spin) (ME = 5.97x1024 kg, RE = 6.37x106 m, G = 6.67x10-11 Nm2/kg2)

Solution:

Initial: h=0, ri = RE Ei = Ki +Ui = 0 + (-GMEm/RE )

= -1.25x1010J In orbit: h = 200 km, rf = RE + 200 km

Ef = Kf + Uf = - ½ Uf+ Uf = ½ Uf

= - ½ GMEm/(RE+200 km)

= -6.06x109J

E = Ef – Ei = 6.4x109 J


Escaping Earth orbit Exercise: suppose an object of mass m is

projected vertically upwards from the Earth’s surface at an initial speed v, how high can it go ? (Ignore Earth’s rotation)

)2( 2

22

iE

iE

vRGMvR

h

02 2 iEvRGM

implies infinite height

km/s 2.112

Escape ER

GMv

221

G )( iEi mvRUE

0)(G hRUE Ef


We hope everyone does well on Sunday

Have a great summer!

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Physics 201: Lecture 30, Pg 1 Lecture 30 Review (the final is, to a large degree, cumulative) Review (the final is, to a large degree, cumulative) ~50%