final review, day 1 - university of colorado boulderjcumalat/phys1110/lectures/lec42.pdf · final...

Final Review, Day 1

•  Final exam next Wednesday (5/9) at 7:30am in the Coors Event Center.

•  Recitation tomorrow is a review. Please feel free to ask the TA any questions on the course material.

•  Lecture on Friday is a free-for-all review. Please bring any questions you have. I will plan to ask clicker question after clicker question.

Web page: http://www.colorado.edu/physics/phys1110/phys1110_sp12/

Announcements:

Grading for the Course

3 midterm exams, 15% each: 45% Final exam: 23% CAPA homework 15% Tutorial participation (3%) and TA-graded written homeworks 7% In-lecture “clicker” participation: 4% “SmartPhysics” Prelecture participation 3%

Posted

Posted

Total Clicker Scores for Course

Tutorial Scores

Reminder to check your scores

Final information Final is on 5/9 from 7:30–10:00am in the Coors Event Center

The following constants are provided: G, g (10 m/s2), density of water, mass of Earth, radius of Earth, and atmospheric pressure.

You are allowed a calculator and two double sided sheets of paper with any handwritten notes you like.

The same moments-of-inertia as in exam 3 are provided

We may ask you to show your student ID if your TA doesn’t recognize you so make sure you bring it with you.

Final information We are using the south end of Coors Event Center. Sit every other seat and as close to the court as possible.

Exam questions •  4-5 fluid questions •  ~8 questions on oscillations and waves •  Problems on collisions, work, force and motion, conservation of

energy, static equilibrium, angular momentum, torque and angular motion, simple harmonic motion, and 2D kinematics.

•  4-5 questions of the type from the first midterm •  4-5 questions of the type from the second midterm •  4-5 questions of the type from the third midterm •  Carefully review the clicker concept questions from lectures

Studying materials Old exams can be found from the content section of the D2L link. Obviously you can ignore the thermodynamics questions.

Tutorial Homework Solutions can also be found in the content section of the D2L link.

Study suggestions (approximately in order) 1. CAPA: Make sure you understand and can do the CAPA problems. Think about possible modifications to the problems and what you would need to do different to solve it.

2. Lecture: Make sure you understand the lecture notes and can do all of the clicker questions. Think about modifications to the clicker questions that could be asked.

3. Tutorials: Review the tutorials including the homework to make sure you understand. For questions you got wrong, analyze where your thinking went astray and make sure you have corrected that aspect. For ungraded questions, go over them again to see if you have changed your mind about your answers.

4. End of chapter problems: These problems should help you setup and solve problems. Try to solve the problems using only your formula sheet and calculator as in the exam.

5. Old exams: The exams this time are not as representative because we have not covered gravity yet. Nevertheless, some problems are quite good. Try to solve the problems using only your formula sheet and calculator and impose a time limit.

6. Read the text: There are examples worked out, suggestions for how to solve problems, and interesting relationships explored that might help you remember the essential bits.

Study suggestions (approximately in order)

Material covered since last exam We covered three subjects since the last exam

Chapter 14: Fluid mechanics Sections 14.1 – 14.10 covered. 14.10 on proof of Bernouli’s Equation was not covered

Chapter 15: Oscillatiions Sections 15.1 – 15.7 covered. Sections 15.8 – 15.9 on damped harmonic motion and on forced oscillations were not covered.

Chapter 16: Waves Sections 16.1 – 14.10, 16.12-16.13 covered. 16.11 on phasors was not covered

Chapter 14: Fluid mechanics Defined density: Intrinsic quantity but equal to mass/volume.

Defined pressure: A location has a pressure (which is a scalar quantity) and a pressure can exert a force on an area. The force is perpendicular to the area. Pressure is force/area and has units of N/m2 or pascals (Pa).

A fluid doesn’t hold its shape and can flow. Liquids are (approximately) incompressible fluids

Gases are compressible fluids

Atmospheric pressure at sea level is 105 Pa.

Fluid mechanics: Hydrostatics Statics is when fluids are not moving

This also shows that for a connected liquid, the pressure at a given depth is the same no matter where you are.

The weight of a fluid above causes an increase in pressure as the depth increases. At a depth d below the surface which has pressure p0 in a liquid of density ρ the hydrostatic pressure is

Since force is pressure times area, one can use a small area to increase the pressure which can exert a large force over a small area. Hydraulic lifts and hydraulic brakes use this principle.

Fluid mechanics: Buoyant force The buoyant force is always equal to the weight of liquid displaced by an object.

An object completely submerged in a liquid displaces an amount of liquid equal to the volume of the object so and therefore An object floating on the surface must have an upward (buoyant) force which exactly cancels the downward force of gravity (weight) so the buoyant force is

The buoyant force is still but

So which means the floating object displaces an amount of water equal to the weight of the object.

Fluid mechanics: Fluid dynamics

The equation is which has units of volume/time

The continuity equation simply says the amount of liquid entering a pipe is the same as the amount of liquid exiting the pipe.

Bernoulli’s equation is a restatement of conservation of energy. Forces associated with pressure can do work which adds a component to the normal potential and kinetic energy terms.

Bernoulli’s equation: or

is the angular frequency (rad/s). It sets how quickly the system oscillates.

The frequency is how many cycles are completed per second. .

The time it takes to increase by (a complete cycle) is the period

Frequency SI unit is hertz (Hz). Units of cycles/second or s-1 are also used.

Note that .

Chapter 15: Oscillations

Mass on end of spring 0 The force exerted by the spring

is .

The potential energy stored in a spring is .

Acceleration is not constant, but can use Newton’s second law

Dividing both sides by m:

We guess a solution

Velocity versus time:

Initial conditions Position versus time:

The initial (t=0) position and velocity are

Since is set by , we can solve these two equations for A and (assuming we have the initial position & velocity).

Thus, initial conditions (position & velocity) give us A & φ.

Energy considerations Our analysis of the mass on a spring tells us about the position, velocity, acceleration as a function of time.

Without friction, energy is conserved: Remember that when , so the total energy is

The maximum velocity occurs at so using conservation of energy we see so

Springs in Parallel

Force is same on both springs so

€

F = −keff x = −k1x1 = −k2x2

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x = x1 + x2

€

keff x = keff (x1 + x2) = k2x2

€

x1 =k2k1x2

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keff =k2x2x1 + x2

=k2x2

k2k1x2 + x2

=k2

k2k1

+1=

k1k2k1 + k2

If

€

k1 = k2, then keff =k2

2

We found the force on a pendulum along the θ direction is . Using the small angle approximation gives us .

Pendulum L

But note also that so

The angular frequency is

The period of a physical pendulum is

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T = 2π Imgl

Torsion Pendulum •  Consider a disk suspended from a torsion wire attached to its centre.

Called a torsion pendulum. A torsion wire is essentially inextensible, but is free to twist about its axis. As the wire twists it also causes the disk attached to it to rotate in the horizontal plane. Let be the angle of rotation of the disk, and equilibrium position is when the wire is untwisted.

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T = 2π Iκ

Note: I thru center of wire

k is the torsion constant > 0

Chapter 16: Waves

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y(x, t) = ym sin(kx −ωt + φ) Transverse Displacement

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v =τµ

€

τ is the tension in the stringµ is the mass per unit length

€

Pavg = 2 dKdt

avg

=12

µv ω 2ym2

Interference (Standing Waves)

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y(x, t) = [2ym sinkx] cosωt

In a standing wave the amplitude varies with position. The place where amplitude is zero is when

so

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kx = nπ for n = 0, 1, 2,.....

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k =2πλ

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x =nλ2

for n = 0, 1, 2, ... (nodes)

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λ =2Ln

for n =1, 2, 3,...

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f =vλ

= n v2L

for n =1, 2, 3,... n is called the harmonic number

Amplitude at position x

Oscillating Term

A long, taut string is shaken up and down at one end by a machine. This creates a right-going traveling wave on the string that is described by the equation y(x,t)=Asin(kx-ωt). The wavelength λ of the wave is 0.50 m, the period T of the wave is 0.10 seconds, and the amplitude of the wave is 0.0010 m (1.0 mm).

What is the maximum vertical speed vy of a piece of the string? (A) 0.013 m/s (B) 0.063 m/s (C) 5.0 m/s (D) 31.4 m/s (E) 0.0020 m/s

Clicker question 1 Set frequency to BA

A long, taut string is shaken up and down at one end by a machine. This creates a right-going traveling wave on the string that is described by the equation y(x,t)=Asin(kx-ωt). The wavelength λ of the wave is 0.50 m, the period T of the wave is 0.10 seconds, and the amplitude of the wave is 0.0010 m (1.0 mm).

What is the maximum vertical speed vy of a piece of the string? (A) 0.013 m/s (B) 0.063 m/s (C) 5.0 m/s (D) 31.4 m/s (E) 0.0020 m/s


€

vy = −ωAcos(kx −ωt)

€

vy = 2πfA =2πAT

= 2π (0.001) /.1= .0628m /s

Referring to the previous problem above, if the frequency with which the machine shakes the string is doubled, what happens to the speed of the wave, vwave? (A) the wave speed increases (B) decreases

(C) remains unchanged


Referring to the previous problem above, if the frequency with which the machine shakes the string is doubled, what happens to the speed of the wave, vwave? (A) the wave speed increases (B) decreases

(C) remains unchanged


In a traveling wave the speed is determined by

Tension and mass per unit length – not frequency

A mass m is thrown downward with an initial speed vo a height hi above a table top on which sits a spring with spring constant k. The mass compresses the spring by a maximum amount x and stops for an instant at a height hf-. There is no friction in this problem. Which of the following equations correctly expresses conservation of energy and allows one to solve for the compression x of the spring?

(A) mghi + ½ mv2 = ½ kx2 (B) mghi + ½ mv2 =1/2 kx2+mghf (C) mghi+1/2 kx2=mghf

(D) mgh(hf-hi) = ½ kx2+1/2 mv2 (E) None of these equations is correct.


A solid piece of plastic of volume V, and density ρp is floating in a cup of water. (The density of water is ρw, and ρp < ρw. ) What is the magnitude of the buoyant force on the plastic?

(A) Zero (B) ρp V g (C) ρw V (D) ρw V g (E) ρp V


The diagram shows a snapshot of a traveling wave at some given time. The frequency of this wave is 120 cycles/sec. What is the speed of the wave? A) 540 m/s B) 120 m/s C) 1.2 m/s D) 360 m/s E) not enough information given


The diagram shows a snapshot of a traveling wave at some given time. The frequency of this wave is 120 cycles/sec. What is the speed of the wave? A) 540 m/s B) 120 m/s C) 1.2 m/s D) 360 m/s E) not enough information given


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v = fλ = (120s−1)(3m) = 360m /s

Suppose the "A" string on a piano is one meter long, and has a mass of 0.008 kg. The frequency of the fundamental of this string is 440 Hz. What tension in the wire is needed? HINT: Piano strings are fixed at both ends. What is the wavelength of the fundamental vibration? (It is NOT 1 m!)

A) .03 N B) 3.5 N C) 390 N D) 6200 N E) 2.4E7 N


Suppose the "A" string on a piano is one meter long, and has a mass of 0.008 kg. The frequency of the fundamental of this string is 440 Hz. What tension in the wire is needed? HINT: Piano strings are fixed at both ends. What is the wavelength of the fundamental vibration? (It is NOT 1 m!)

A) .03 N B) 3.5 N C) 390 N D) 6200 N E) 2.4E7 N


€

µ = 0.008kg /1m€

v =τµ

€

v = fλ = (440Hz)2m = 880m /s

€

τ = v 2µ = (880)2(0.008)N

A pendulum consists of a mass m at the end of a string of length L=1.00 m. The mass is released from rest at an angle of θ = 60.0o What is the distance d through which the mass travels in moving from the initial position to the lowest point?

(A) 0.17 m (B) 0.71 m (C) 1.05 m

(D) 30 m (E) None of these.

L

θ

d


A pendulum consists of a mass m at the end of a string of length L=1.00 m. The mass is released from rest at an angle of θ = 60.0o What is the distance d through which the mass travels in moving from the initial position to the lowest point?

(A) 0.17 m (B) 0.71 m (C) 1.05 m

(D) 30 m (E) None of these.

L

θ

d


€

(1m)(60 × π180

) =1.05m

final review, day 1 - university of colorado boulderjcumalat/phys1110/lectures/lec42.pdf · final...

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