Special Homework for Chapter 18, HW11, Due Friday, Dec. 5 ... · 5 short answer, most with multiple...

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Physics 207 – Lecture 24

Physics 207: Lecture 24, Pg 1

Lecture 24, Nov. 24Review:Review:

•• Exam covers Chapters 14Exam covers Chapters 14--17 plus angular momentum17 plus angular momentum

•• Exam has 15 questions, Exam has 15 questions, �10 multiple choice dominated by conceptual ideas� 5 short answer, most with multiple parts

•• AssignmentAssignment� Special Homework for Chapter 18, HW11, Due Friday, Dec. 5� For Wednesday, Read through all of Chapter 18


Exam III Room assignments

� 613 Room 2223 Koki

� 601 Room 2241 Matt603 Room 2241 Heming608 Room 2241 Matt609 Room 2241 Heming607 Room 2241 Koki

� And all others in Room 2103602 604605606610611612614

� McBurney and special requests, Room 5310 Sterling

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Angular Momentum Exercise

� A mass m=0.10 kg is attached to a cord passing through a small hole in a frictionless, horizontal surface as in the Figure. The mass is initially orbiting with speed ωi = 5 rad / s in a circle of radius ri = 0.20 m. The cord is then slowly pulled from below, and the radius decreases to r = 0.10 m.

� What is the final angular velocity ?� Underlying concept: Conservation of Momentum

ri

ωi


Angular Momentum Exercise

� A mass m=0.10 kg is attached to a cord passing through a small hole in a frictionless, horizontal surface as in the Figure. The mass is initially orbiting with speed ωi = 5 rad / s in a circle of radius ri = 0.20 m. The cord is then slowly pulled from below, and the radius decreases to r = 0.10 m.

� What is the final angular velocity ?No external torque implies

∆L = 0 or Li = Lc

Ii ωi = If ωf

I for a point mass is mr2 where r is the distance to the axis of rotation

m ri2ωi = m rf2 ωf

ωf = ri2ωi / rf2 = (0.20/0.10)2 5 rad/s = 20 rad/s

ri

ωi

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Example: Throwing ball from stool

� A student sits on a stool, initially at rest, but which is free to rotate. The moment of inertia of the student plus the stool is I. They throw a heavy ball of mass M with speed v such that its velocity vector moves a distance d from the axis of rotation.

� What is the angular speed ωF of the student-stool system after they throw the ball ?

Top view: before after

d

MvM

I I

ωF

r


Example: Throwing ball from stool

� What is the angular speed ωF of the student-stool system after they throw the ball ?

� Process: (1) Define system (2) Identify Conditions

(1) System: student, stool and ball (No Ext. torque, L is constant)

(2) Momentum is conserved (check |r| |p| sin θθθθ for sign)

Linit = 0 = Lfinal = −−−− M v d + I ωf

Top view: before after

d

vM

I I

ωF

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Bernoulli Equation � P1+ ½ ρ v12 + ρ g y1 = constant

A 5 cm radius horizontal pipe carries water at 10 m/sinto a 10 cm radius. What is the pressure difference?

P1+ ½ ρ v12 = P2+ ½ ρ v2

2

∆P = ½ ρ v22 - ½ ρ v1

2

∆P = ½ ρ (v22 - v1

2 )

and A1 v1 = A2 v2

∆P = ½ ρ v22 (1 – (A2/A1 ) 2 )

= 0.5 x 1000 kg/m x 100 m2/s2 (1- (25/100) )= 37500 N/m2

y1

y2

v1

v2

p1

p2

∆∆∆∆V

Ideal Fluid


A water fountain� A fountain, at sea level, consists of a 10 cm

radius pipe with a 5 cm radius nozzle. The water sprays up to a height of 20 m.

�What is the velocity of the water as it leaves the nozzle?

�What volume of the water per second as it leaves the nozzle?

�What is the velocity of the water in the pipe?

�What is the pressure in the pipe?

�How many watts must the water pump supply?

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A water fountain

� A fountain, at sea level, consists of a 10 cm radius pipe with a5 cm radius nozzle. The water sprays up to a height of 20 m.

� What is the velocity of the water as it leaves the nozzle?Simple Picture: ½mv2=mgh � v=(2gh)½ = (2x10x20)½ = 20 m/s

� What volume of the water per second as it leaves the nozzle?Q = A vn = 0.0025 x 20 x 3.14 = 0.155 m3/s

� What is the velocity of the water in the pipe?An vn = Ap vp � vp = Q /4 = 5 m/s

� What is the pressure in the pipe?

1atm + ½ ρ vn2 = 1 atm + ∆P + ½ ρ vp

2 � 1.9 x 105 N/m2

� How many watts must the water pump supply?

� Power = Q ρ g h = 0.155 m3/s x 103 kg/m3 x 9.8 m/s2 x 20 m = 3x104 W (Comment on syringe injection)


Fluids Buoyancy

� A metal cylinder, 0.5 m in radius and 4.0 m high is lowered, as shown, from a masslesrope into a vat of oil and water. The tension, T, in the rope goes to zero when the cylinder is half in the oil and half in the water. The densities of the oil is 0.9 gm/cm3 and the water is 1.0 gm/cm3

� What is the average density of the cylinder?

� What was the tension in the rope when the cylinder was submerged in the oil?

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Fluids Buoyancy

� r = 0.5 m, h= 4.0 m

� ρoil = 0.9 gm/cm3 ρwater = 1.0 gm/cm3

� What is the average density of the cylinder?When T = 0 Fbuoyancy = Wcylinder

Fbuoyancy = ρoil g ½ Vcyl.+ ρwater g ½ Vcyl.

Wcylinder = ρcyl g Vcyl.

ρcyl g Vcyl. = ρoil g ½ Vcyl.+ ρwater g ½ Vcyl.

ρcyl = ½ ρoill.+ ½ ρwater

What was the tension in the rope when the cylinder was submerged in the oil?

Use a Free Body Diagram !


Fluids Buoyancy� r = 0.5 m, h= 4.0 m Vcyl. = π r2 h� ρoil = 0.9 gm/cm3 ρwater = 1.0 gm/cm3

� What is the average density of the cylinder?When T = 0 Fbuoyancy = Wcylinder

Fbuoyancy = ρoil g ½ Vcyl.+ ρwater g ½ Vcyl.

Wcylinder = ρcyl g Vcyl.

ρcyl g Vcyl. = ρoil g ½ Vcyl.+ ρwater g ½ Vcyl.

ρcyl = ½ ρoill.+ ½ ρwater = 0.95 gm/cm3

What was the tension in the rope when the cylinder was submerged in the oil?

Use a Free Body Diagram!

Σ Fz = 0 = T - Wcylinder + Fbuoyancy

T = Wcyl - Fbuoy = g ( ρcyl .- ρoil ) Vcyl

T = 9.8 x 0.05 x 103 x π x 0.52 x 4 .0 = 1500 N

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A new trick

� Two trapeze artists, of mass 100 kgand 50 kg respectively are testing a new trick and want to get the timing right. They both start at the same time using ropes of 10 meter in length and, at the turnaround point the smaller grabs hold of the larger artist and together they swing back to the starting platform. A model of the stunt is shown at right.

� How long will this stunt require if the angle is small ?


A new trick

� How long will this stunt require?

Period of a pendulum is just

ω = (g/L)½

T = 2π (L/g)½

Time before ½ periodTime after ½ period

So, t = T = 2π(L/g)½ = 2π sec

Key points: Period is one full swingand independent of mass

(this is SHM but very different than a spring. SHM requires only a linear restoring force.)

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Example

� A Hooke’s Law spring, k=200 N/m, is on a horizontal frictionless surface is stretched 2.0 m from its equilibrium position. A 1.0 kg mass is initially attached to the spring however, at a displacement of 1.0 m a 2.0 kg lump of clay is dropped onto the mass. The clay sticks.

What is the new amplitude?

km

km

-2 20(≡Xeq)

M

M


Example

� A Hooke’s Law spring, k=200 N/m, is on a horizontal frictionless surface is stretched 2.0 m from its equilibrium position. A 1.0 kg mass is initially attached to the spring however, at a displacement of 1.0 m a 2.0 kg lump of clay is dropped onto the mass.

What is the new amplitude?Sequence: SHM, collision, SHM½ k A0

2 = const.½ k A0

2 = ½ mv2 + ½ k (A0/2)2

¾ k A02 = m v2 � v = ( ¾ k A0

2 / m )½

v = (0.75*200*4 / 1 )½ = 24.5 m/sConservation of x-momentum:mv= (m+M) V � V = mv/(m+M) V = 24.5/3 m/s = 8.2 m/s

km

km

-2 20(≡Xeq)

M

M

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Example

� A Hooke’s Law spring, k=200 N/m, is on a horizontal frictionless surface is stretched 2.0 m from its equilibrium position. A 1.0 kg mass is initially attached to the spring however, at a displacement of 1.0 m a 2.0 kg lump of clay is dropped onto the mass. The clay sticks.

What is the new amplitude?Sequence: SHM, collision, SHMV = 24.5/3 m/s = 8.2 m/s½ k Af

2 = const.½ k Af

2 = ½ (m+M)V2 + ½ k (Ai)2

Af2 = [(m+M)V2 /k + (Ai)2 ]½

Af2 = [3 x 8.22 /200 + (1)2 ]½

Af2 = [1 + 1]½ � Af

2 = 1.4 m

Key point: K+U is constant in SHM

km

km

-2 20(≡Xeq)

M

M


Fluids Buoyancy & SHM

� A metal cylinder, 0.5 m in radius and 4.0 mhigh is lowered, as shown, from a rope into a vat of oil and water. The tension, T, in the rope goes to zero when the cylinder is half in the oil and half in the water. The densities of the oil is 0.9 gm/cm3 and the water is 1.0 gm/cm3

� Refer to earlier example� Now the metal cylinder is lifted slightly

from its equilibrium position. What is the relationship between the displacement and the rope’s tension?

� If the rope is cut and the drum undergoes SHM, what is the period of the oscillation if undamped?

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� Refer to earlier example

� Now the metal cylinder is lifted ∆y from its equilibrium position. What is the relationship between the displacement and the rope’s tension?

0 = T + Fbuoyancy – Wcylinder

T = - Fbuoyancy + Wcylinder

T =-[ρog(h/2+∆y) Ac+ρwgAc(h/2-∆y)] + Wcyl

T =-[ghAc(ρo +ρw)/2 + ∆y gAc(ρo -ρw)]+ Wcyl

T =-[Wcyl + ∆y gAc(ρo -ρw)]+ Wcyl

T = [ g Ac (ρw -ρo) ] ∆yIf the is rope cut, net force is towards equilibrium position with a proportionality constant

g Ac (ρw -ρo) [& with g=10 m/s2]

If F = - k ∆y then k = g Ac (ρo -ρw) = π/4 x103 N/m



� A metal cylinder, 0.5 m in radius and 4.0 m high is lowered, as shown, from a rope into a vat of oil and water. The tension, T, in the rope goes to zero when the cylinder is half in the oil and half in the water. The densities of the oil is 0.9 gm/cm3

and the water is 1.0 gm/cm3

� If the rope is cut and the drum undergoes SHM, what is the period of the oscillation if undamped?

F = ma = - k ∆y and with SHM …. ω = (k/m)½

where k is a “spring” constant and m is the inertial mass (resistance to motion), the cylinder

So ω = (1000π /4 mcyl)½

= (1000π / 4ρcylVcyl)½ = ( 0.25/0.95 )½

= 0.51 rad/secT = 3.2 sec

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Underdamped SHM

If the period is 2.0 sec and, after four cycles, the amplitude drops by 75%, what is the time constant?

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

ωωωωt

A

)(cos expA )( )(2

φω +−= ttxm

bt mbo 2/>ωif

Four cycles implies 8 sec

So

0.25 A0 = A0 exp(-4 b / m)

ln(1/4)= -4 1/τ

τ = -4/ ln(1/4) = 2.9 sec


Angular Momentum

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Hooke’s Law Springs and a Restoring Force

� Key fact: ω = (k / m)½ is general result where k reflects a constant of the linear restoring force and m is the inertial response

(e.g., the “physical pendulum” where ω = (κ / I)½


Simple Harmonic Motion

Maximum kinetic energy

Maximum potential energy

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Resonance and damping

� Energy transfer is optimal when the driving force varies at the resonant frequency.

� Types of motion� Undamped� Underdamped� Critically damped� Overdamped


Fluid Flow

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Density and pressure


Response to forces

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States of Matter and Phase Diagrams


Ideal gas equation of state

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pV diagrams


Thermodynamics

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Work, Pressure, Volume, Heat

T can change!

In steady-state T=constant and so heat in equals heat out


Gas Processes

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Have a good Thanksgiving break!

Read all of chapter 18 for Wednesday

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Special Homework for Chapter 18, HW11, Due Friday, Dec. 5 ... · 5 short answer, most with multiple...

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