### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Intermediate Classical Mechanics PHYS 3550

Utah State University

GPA 3.6

### View Full Document

## 15

## 0

## Popular in Course

## Popular in Physics 2

This 10 page Class Notes was uploaded by Talon Thompson on Wednesday October 28, 2015. The Class Notes belongs to PHYS 3550 at Utah State University taught by Tsung-Cheng Shen in Fall. Since its upload, it has received 15 views. For similar materials see /class/230475/phys-3550-utah-state-university in Physics 2 at Utah State University.

## Similar to PHYS 3550 at Utah State University

## Reviews for Intermediate Classical Mechanics

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/28/15

Chapter 1 Statics From Newton s second law if all forces and torques are balanced the acceleration will be zero and the object is moving with a constant velocity More often than not in our daily life the constant velocity is actually zero and we like to know how much forces are needed to reach such a situation We will focus on friction tension and torques in various mechanical systems Key words static friction coefficient tension normal force freebody diagram center of mass torque dot product and cross product of two vectors right hand rule for cross product Newton s second and third law Key points to discuss Make sure you will include all possible forces in a freebody diagram Define a most convenient coordinate system Static friction has no xed direction but kinetic friction always opposes to the motion Friction is determined by normal force and the maximum friction force is uN Normal force and friction force is determined by equations of motion The positive and negative sign of physical quantities depends on the coordinate choices For a exible rope tension always follows the direction of the string Tension along a rope may not be the same if friction is included Torque can be calculated by setting any point as the origin 0 In statics all forces must pass through one point concurrent so that the torque can be zero H gt P gtP Nt Problems Due 9408 Thursday ch2 20 26 33 38 Exercises ch2 l 4 6 91415 23 Chapter 2 Fma Mass is a strange intrinsic property of any object from elementary particles to macroscopic objects Newton s laws relate acceleration to mass Freebody diagram is a useful technique to analyze the motion of an object We will discuss Atwood s machines projectile motion and circular motion Key words Atwood s machine momentum conservation subsystems centripetal acceleration cycloid polar coordinates Key points to discuss 1 Newton s equation is only for external forces However an internal force can be an external force for a subsystem Newton s third law is equivalent to momentum conservation It is not possible to have only one particle accelerating in the universe If an object is con ned to move in a certain trajectory one equation relates to the coordinates of the orbit will be generated For example the xy coordinates of a plane is needed to solve a block moving on an incline or the motion of masses in an Atwood machine is related to keep the string length constant If the pulley is massless moment of inertia of the pulley is not affecting the motion In a polar coordinate even the vector r has no 6 component but the change of r derivative of r will have a 6 component This is how centripetal and Coriolis force are revealed 7 Centripetal acceleration is needed to make a circular motion bUJN 0939 Suggested problems to practice not due ch3 2 6 8 9111315 21 22 Exercises Due 91608 Tuesday ch3 31 33 37 46 57 Chapter 3 Oscillations Oscillation is omnipresent in our world 7 from le building vibration to 38 X1013 Hz vibration of C02 molecules The basic equation is a second order differential equation The general behavior of oscillators will be dictated by damping Furthermore most of the vibration is a result of coupled oscillators We will introduce the concept of normal modes for coupled systems Key words oscillator superposition principle natural frequency underdamping overdamping critical damping resonance coupled oscillators normal mode eigenvalues and eigenfunctions Key points to discuss The force is always opposite to the displacement in an oscillator 2 Solving inhomogeneous differential equations 3 Amplitude and phase of a harmonic oscillator solution can be determined by the position and velocity at certain time The linear combination of two solutions in a linear differential equation is another solution The frequency of a damped oscillator is less than its natural frequency With sufficient damping the oscillator will not oscillate but reaches equilibrium monotonically Critical damping is the fastest way to stop the motion of an oscillator The number of normal modes equals to degree of freedom in a system If there are N atoms in a solid state crystal there will be 3N normal modes In a driven oscillator the phase between the motion and the driver depends on the natural frequency and damping The range is from 0 to TE 10 In a driven oscillator the homogeneous solutions are transients as in the damped oscillators 11 When the driving frequency equals to the natural frequency a system can resonate which generates the largest amplitude but the motion is TE 2 behind the driver 12 The fact that the amplitude of a driven oscillator increases with time when it approaches resonance can be viewed as energy feeding Thus without damping the amplitude will diverge Suggested problems to practice not due ch4 2 3 7 9 10 994 00gt O Exercises Due 92508 Thursday ch4 17 18 30 31 Mid term exam will be on 10208 Thursday Close book close notes Ch2 to 4 Chapter 4 Conservation of energy and momentum Revised Potential is de ned from energy point of view Potential has the advantage over force in some applications because it is not a vector However it also has less information than the force equation Fma Momentum is conserved if there is no external force acting on the system We will discuss momentum change due to mass change and collisions in center of mass frame Key words potential energypotential spherical symmetry center of mass frame momentum conservation energy conservation elastic collision impact force Key points to discuss 1 Force is the negative gradient of potential energy field is the negative gradient of potential The potential energy difference is the relevant quantity in dynamics Thus potential energy at one point is not clear unless the reference the other point is defined For gravity or Coulomb force the potential at infinity is usually defined to be the reference point Most wellbehaved potential function can be approximated by a harmonic oscillator potential near its minimum Thus one can model small oscillations of a molecule by an oscillator The gravitational force due to a sphere even with density pr is identical to that of a point mass The gravitational potential of athin spherical shell is a constant inside and lr outside The two functions match their values at the surface The most general result of an elastic collision between two equal masses in 2D is moving apart at 900 from each other but the precise direction depends on the impact position In the CM frame the momentum before and after an elastic collision remains 0 In 1D the relative velocity before an elastic collision is negative of the relative velocity after that The total kinetic energy in the lab frame is the sum of the total kinetic energy in the CM frame and kinetic energy of the CM N L 4 UI ON 1 00 Suggested problems to practice not due ch 5 9 10 ll 22 23 27 28 92 93 94 Exercises Due 101408 Tuesday ch 5 40 63 68 81 95 Ref F Herrmann and P Schmalzle Simple explanation of a wellknown collision experiment Am J Phys 49 761 1981 Help sessions will be at 530 on 108 Wed and 1013 Mon students midterm 1 Average 59 Chapter 5 The Lagrangian Method Lagrange s equation is convenient to consider systems with constraints like a particle moving along a ramp or a pendulum It also involves only scalar quantities such as kinetic and potential energy instead of vector quantities such as forces The cleverness of choosing the generalized natural coordinates is the key to use Lagrangian method Furthermore the stationary action principle can be used to determine many natural phenomena including Snell s law and bubble surfaces Key words Lagrangian EulerLagrange equation action functional cycloid Fermat s Principle Key points to discuss Lagrange s equation is equivalent to Newton s equation 2 Express the kinetic and potential energy in Cartesian coordinate first then change to general coordinates comply with the constrained motion If there are more than one particles then the kinetic and potential energy will be the sum of all the particles The coordinates will be like X1 y1 21 X2 yz zz There will be 3Nr equations for 3N particles where r is the number of constraint equations 4 The coupling between particles will appear naturally in the Lagrange s equations LA Suggested problems to practice not due ch6 1 2 3 1114 15 17 18 19 21 22 23 24 Exercises Due 102308 Thursday ch6 27 43 Ref cycloid motion can be visualized at httparchivesmathutkeduvisualcalculusOparametric5indeXhtml httpwww ies on in umtl 39 V l Ldc Help sessions will be at 530 on 1022 Wed Chapter 6 Angular momentum 1 revised For an extended object the center of mass moves like a point particle subject to external forces and the rest of the particles have relative motion with respect to the CM How easy the object can rotate depends on its moment of inertia The angular momentum of an object changes if there is a net external torque acting on it We extend the collision problems to include rotation and angular momentum Key words Moment of inertia Conservation of angular momentum parallelaxis theorem perpendicularaxis theorem angular impulse Key points to discuss Angular momentum and torque are defined relative to a speci c origin The center of mass of two particles is always on the line joining the two particles Tricks to calculate moment of inertia Rotation without slipping is caused by friction but without energy dissipation One can treat an extended object by a pointmass at the center of mass CM and relative motion linear and angular of all the points relative to CM Without external torque the angular momentum is conserved Without external force the linear momentum is conserved Elastic and inelastic collision involving rotation Calculate rotation with torque equation Elk WP 509080 Suggested problems to practice not due ch 8 1 4 71318 24 26 31 38 47 58 Exercises Due 110608 Thursday ch 8 28 32 41 42 61 Second mid term exam will be on 111308 Thursday Close book close notes Ch5 6 8 students midterm 2 grades Chapter 7 Central forces It is remarkable that if the force between two particles is central and conservative a twobody problem can be reduced to a onebody problem with a reduced mass The three Kepler s laws on planets orbits can be derived in just one differential equation r6 The shape of the orbit is determined by energy and angular momentum of the planet A positive energy gives an unbounded orbit while a negative energy leads to a bounded orbit Key words Central force Reduced mass Effective potential Bounded and unbounded orbit Eccentricity 3 Kepler s laws Key points to discuss A particle s angular momentum is conserved subject to a central force A particle s motion will be con ned to a plane subject to a central force Two masses motion is equivalent to the motion of a center of mass and a reduced mass Without any external force the center of mass behaves as a free particle We can choose our reference frame to be at the center of mass so the motion is zero In the formalism of reducing a 2body problem to 1body problem the fictitious particle with reduced mass acts according to the force between the two real masses The total angular momentum in CM frame is the angular momentum of a single particle of reduced mass 7 One can use effective energy equation to determine the nature of orbits and we can prove Kepler s laws 8 By the 1r2 force the CM is one of the foci in the conical orbits 9 The second law of Kepler can be understood by angular momentum conservation 10 Eccentricity classifies the 4 Kepler s orbits 11 For a given angular momentum the lowest energy is manifested by a circular orbit 12 Geometry of ellipse 59 V39 O Suggested problems to practice not due ch7 8 9 14 17 Exercises Due 112508 Tuesday ch7 12151618 Chapter 8 Accelerated frames of reference If we continue to use Newton s equation in a noninertial frame we must add fictitious forces The general motion of the accelerated frame can be divided into translation and rotation If the rotation is constant there will be a Coriolis force and a centrifugal force If the rotation is time dependent there will be an azimuthal force Key words inertial frame Coriolis force centrifugal force azimuthal force Faucult pendulum tidal force Key points to discuss 1 Pendulum motion in a translational accelerated frame accelerated car or elevator 2 Fixed vectors in a rotating frame viewed in an inertia frame 3 Timevarying vectors in a rotating frame viewed in an inertia frame 4 Coriolis force and centrifugal forces are generated in the inertia frame 5 Effective gravity due to centrifugal force 6 Coriolis force viewed from inertial frame and rotating frame 7 Analysis of a pendulum in a rotating frame 8 Projectile motion in a rotating frame 9 If the earth were a rigid body then the tidal force will have no effect 10 Tides are a comparative effect Suggested problems to practice not due ch 10 1 2 3 5 6 71619 20 22 The nal exam will be on 121108 Thursday at 930 am All the topics discussed in this semester could be included

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.