PHYS 207.013

Key slides

all chapters

Instructor: Dr. Bianco

TAs: Joey Betz; Lily Padlow

 

University of Delaware - Spring 2021

https://slides.com/federicabianco/phys20713_5

  • always put units in physical quantities

KEY POINTS:

  • convert units by multiplying quantities by                           where 1 old units = N new units
    to cancel the old unit and appropriately modifying the quantity
  • adjust the significant digits the the minimum number of significant digits in your input data
\frac{N ~\mathrm{new~unit}}{1~\mathrm{ old~unit}}

H&R CH1 measuring things - including length

unit conversion

USE SI UNLESS YOU ARE INSTRUCTED OTHERWISE

  • check that the units on the left and right side of "=", "-", "+" sign are the same every time you write down an equation

DIMENSIONAL ANALYSIS

H&R CH1 measuring things - including length

whenever I have an equation

(always every single time I have an = sign!!!)

this will be super helpful starting next class!

\frac{\mathrm{[L]}}{\mathrm{[T]}} = \frac{\mathrm{[L]}}{\mathrm{[T]}} + \frac{\mathrm{[L]}}{\mathrm{[T^2]}} [T^2]
\mathrm{[L]} - \mathrm{[L]} = \frac{\mathrm{[L]}}{\mathrm{[T]}}\mathrm{[T]} + \frac{\mathrm{[L]}}{\mathrm{[T^2]}} [T^2]
\frac{\mathrm{[L^2]}}{\mathrm{[T^2]}} = \frac{\mathrm{[L^2]}}{\mathrm{[T^2]}} + \frac{\mathrm{[L]}}{\mathrm{[T^2]}} [L]
\mathrm{[L]} - \mathrm{[L]} = (\frac{\mathrm{[L]}}{\mathrm{[T]}} - \frac{\mathrm{[L]}}{\mathrm{[T]}}) [T]
\mathrm{[L]} - \mathrm{[L]} = \frac{\mathrm{[L]}}{\mathrm{[T]}} + \frac{\mathrm{[L]}}{\mathrm{[T^2]}}) [T^2]

H&R CH2 1D motion - velocity - acceleration

H&R CH2 1D motion - velocity - acceleration

KEY POINTS:

  • displacement is the distance between final and initial position
  • average velocity is displacement over time
  • average acceleration is change of velocity over time
  • instantaneous velocity is the derivative of displacement in time
  • instantaneous acceleration is the derivative of velocity in time
\Delta x
\bar{v} =\frac{\Delta x}{\Delta t}
\bar{a} = \frac{\Delta v}{\Delta t}
v= \frac{d x}{d t}
a= \frac{d v}{d t} = \frac{d^2 x}{dt^2}

1D motion

H&R CH2 1D motion - velocity - acceleration

motion in 1D

H&R CH2 1D motion - velocity - acceleration

Equations of motion in 1D

you do not need to memorize them necessarily but you should be extremely familiar with them and know how to use each one. 

 

When you try and remember them on the fly, think about the dimensional analysis. 

To equate to displacement, velocity needs to be multipled by time.

To equate displacement acceleration has to be multiplied by time-squared

 

In a few weeks, you should have solved enough problems that you will have them memorized without even trying!

HOW TO SOLVE 1D MOTION PROBLEMS

STEP 1. Put together all the information you have

STEP 2. Put the info in a frame of reference

STEP 3. Identify the unknown quantity

STEP 4. Choose the equation with gives you the unknown you need

A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side.

(a) How long does the package take

KEY POINTS:

  • cross product of 2 vectors produces a vector with magnitude that points perpendicularly to the plane of the 2 vectors according to the right hand rule
  • vector multiplication with a scalar changes the magnitude of the vector
  • dot product of vectors produces a scalar with magnitude
\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \phi
|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \phi
| c \vec{a} |= c|\vec{a}|

H&R CH3 vectors

vectors

KEY POINTS:

  • the tangent of an angle is the ration of sin/cosin and this relation can be inverted to measure angles
  • the cosine of an angle is its projection on the x axis for a vector of unit magnitude
  • the sine of an angle is its projection on the y axis for a vector of unit magnitude
  • the tangent of an angle is the ration of sin/cosin and this relation can be inverted to measure angles
  • the sum of the square of cosine and sine is equal to 1
\cos(\theta)^2 + \sin(\theta)^2 = 1
\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}

H&R CH3 vectors

vectors

KEY POINTS:

  • almost always you can decompose and solve along each axis separately - turn the problem into a 1D motion problem!

motion in 2D and 3D

KEY POINTS:

  • the velocity along y is zero at max height
  • projectile motion is motion under some initial velocity in 2D and downward acceleration along the y axis (no acceleration in x)
  • I can calculate how far in x a projectile starting from y=0 would land (i.e. second y=0 solution) if I know the initial velocity and angle at which it was thrown (range) 

H&R CH4 motion in 2D and 3D

R = \frac{v_0^2}{g} \sin{2\theta_0}
v_x = \mathrm{const}: x - x0 = v_{0,x} t\\ v_y = \mathrm{accelerated}: y - y0 = v_{0,y} t - \frac{1}{2}a t^2

parabolic motion

KEY POINTS:

  • the acceleration is toward the center of the circle and constant in magnitude
  • uniform circular motion is the motion of an object with velocity v constant in magnitude and uniformly changing in direction
  • we define the period T frequency F (inverse of the period) and angular velocity omega

H&R CH4 motion in 2D and 3D

v = \frac{2\pi r}{T} = \omega r
\vec{v} = -v \sin{\theta} \hat{i} + v \cos{\theta} \hat{j}
a = -\frac{v^2}{r}

uniform circular motion

  • the velocity vector always points tangentially to the circle

KEY POINTS:

  • Forces are vectors and all vector rules apply
  • Newton I: A body's velocity does not change unless a force is applied: v=const =>
  • Newton III:  For every action, there is an equal and opposite reaction.
\vec{F}_{1->2} = - \vec{F}_{2->1}
\vec{F} = m\vec{a}
\vec{F}_{net} = \sum{\vec{F_i}}

Forces and Newton Laws

  • Newton II: A force is proportional to the acceleration and the mass of a body

H&R CH5 Newton Laws

\sum\vec{F}_i = 0

how to solve problems with force

4) identify if there system is initially at rest or in motion and if you need to choose the appropriate Chap2 equation of motion now that you have the acceleration from the forces.

1) make your Free Body Diagram identifying all forces acting on the body

3) identify if the system is in equilibrium                       or the forces are unbalanced and cause acceleration

\vec{F} = m\vec{a}

Forces and Newton Laws

2) identify all angles and where to use cosine and sine and the component of all forces

H&R CH5 Newton Laws

\sum_i \vec{F}_i = 0

KEY POINTS:

  • The force of friction between two surfeces that are not moving is larger than the force of friction between two bodies that are already moving (sliding) fk<fs   
  • fs is a force that acts between two body in contact against any force that tries to displace (slide) one of them
  • there is a maximum value for fs : fmax,s 

Force of friction

  • fs is not constant, grows with the force applies that tries to slide 

H&R CH6 Friction

KEY POINTS:

H&R CH6 Friction

drag

KEY POINTS:

  • the Kinetiuc energy and the work are related by

  • Kinetic energy K is energy associated with the state of motion of an object.

     

  • Work is the application of a force over a distance.

     

W = K_f-K_i [J = kg \frac{m^2}{v^2} = Nm]
K = \frac{1}{2}m v^2~ [J=kg\frac{m^2}{s^2}]

Kinetic energy and work

W = \vec{F}\cdot\vec{d}=Fd \cos{\theta} [J = kg \frac{m}{v^2} m = Nm]

H&R CH7 kinetic energy and work

KEY POINTS:

  • Earth gravitational potential energy: 

 

  • Hooks potential energy

  • The potential energy relates to the force

  • in absence of non conservative forces, for an isolated system, the mechanical energy is conserved

U_s = kx
\Delta U =- \int_{x_i}^{x_f} F(x) dx

potential energy and

conservation of energy

E_{mec} = U + K = \mathrm{const}

H&R CH7 kinetic energy and work

U_g = mgh

work and rotational K

For translation

\Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = W
W = \int_{x_i}^{x_f} F dx
P = \frac{dW}{dt} = F\frac{dx}{dt} = Fv
W = \int_{\theta_i}^{\theta_f} \tau d\theta
P = \frac{dW}{dt} = \tau\frac{d\theta}{dt} = \tau\omega
K = \frac{1}{2}mv^2
K = \frac{1}{2}I\omega^2
\Delta K = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2 = W

For rotation

rotational energy

key points

Center of Mass

r_{CoM} = \frac{1}{M}\sum_i m_i r_i
F_{net} = m a_{CoM}

CoM

key points

\vec{y}_{CoM} = \frac{1}{M} \int y~dm = \frac{1}{V}\int y ~dV
\vec{x}_{CoM} = \frac{1}{M} \int x~dm = \frac{1}{V}\int x ~dV
\vec{z}_{CoM} = \frac{1}{M} \int z~dm= \frac{1}{V}\int z ~dV

a system of N particles

\vec{x}_{CoM} = \frac{1}{M_{tot}}\sum_i m_i \vec{x}_i
\vec{x}_{CoM} = \frac{m_1~\vec{x}_1 + m_2~\vec{x}_2}{m_1 + m_2}

momentum

key points

Momentum is conserved in collisions between particles

p_f = p_i
K_f = K_i
K_f = K_i

Inelastic Collision

If energy is not conserved

Elastic Collision

If energy is conserved

K_f < K_i

Completely Inelastic Collision

If energy is not conserved

K_f < K_i

particles stick together

Momentum p = mv

J = \Delta P

momentum

key points

Momentum is not always conserved. When its not, "impulse" is the difference in momentum

Momentum is conserved in collisions between particles

p_f = p_i
K_f = K_i
K_f = K_i

Inelastic Collision

If energy is not conserved

Elastic Collision

If energy is conserved

K_f < K_i

Completely Inelastic Collision

If energy is not conserved

K_f < K_i

particles stick together

Momentum p = mv

key points

rotational forces and momentum

key points 

\vec{F}_{net} = 0 => \frac{d\vec{P}}{dt} = 0
\vec{\tau}_{net} = 0 => \frac{d\vec{L}}{dt} = 0

equilibrium

H&R CH12 equilibrium & elasticity

translational equilibrium

rotational equilibrium

 

must be 0 with respect to any point

Two conditions must be satisfied for equilibrium: net force AND net torque must be 0

key points 

H&R CH12 equilibrium & elasticity

elasticity

\frac{F}{A} = E \cdot\mathrm{deformation}
  • There are 3 types of stress: stretch (acting along the length) shear (acting perpendicular to the length in one direction hydraulic (compressing perpendicularly to the length

 

  • Stress is measure as force per unit area where the area is the area of the cross section following                               

      with e proportionality constant of the material

  • Stretch

 

  • Strain

 

  • Hydraulic stress 
\frac{F}{A} = E \frac{\Delta L}{L}
\frac{F}{A} = G \frac{\Delta x}{L}
\frac{F}{A} = B \frac{\Delta V}{V}

Simple Harmonic Oscillator

\frac{d^2x}{dt^2} \propto - x
T = \frac{2\pi}{\omega}

H&R CH15 oscillations

key points 

\frac{dx}{dt} = -A\omega \sin{(\omega t + \phi)}
x = A \cos{(\omega t + \phi)}
\frac{d^2x}{dt^2} = -A\omega^2 \cos{(\omega t + \phi)}
  • The kinematics of a SHO are defined by 

 

  • The solution of this differential equation is

where A is the amplitude of the oscillation, ω is the angular frequency, and Φ the phase

  • The period of the oscillator is

 

  • From                                                            it follows that
x = A \cos{(\omega t + \phi)}

spring

H&R CH15 oscillations

\omega^2 = \frac{k}{m}

pendulum

\omega^2 = \frac{g}{l}

spring

H&R CH15 oscillations

T = \frac{2 \pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}

pendulum

T = \frac{2 \pi}{\omega} = 2\pi\sqrt{\frac{l}{g}}

SHO energy

\frac{dx}{dt} = -A\omega \sin{(\omega t + \phi)}
x = A \cos{(\omega t + \phi)}
\frac{d^2x}{dt^2} = -A\omega^2 \cos{(\omega t + \phi)}
U = \frac{1}{2}{kx^2} = \frac{1}{2}kA^2 \cos^2(\omega t + \phi)
\omega = \sqrt{\frac{k}{m}}
K = \frac{1}{2}{m\frac{dx}{dt}^2} = \frac{1}{2}kA^2 \sin^2(\omega t + \phi)
M.E. = U + K =
=\frac{1}{2} k A^2 (\cos^2({\omega t + \phi}) + \sin^2({\omega t + \phi}))=
=\frac{1}{2} k A^2

exam instructions

EXAM PREP

you have 3 hours - which is a long time. Take breaks!

exam instructions

What to have handy: remember that these exams are open-book, so also have whatever helps you solve problems handy! Also water and snacks - its 3 hours!!!

EXAM PREP

Think whether the problem lets you use:

  • Conservation of Energy and the equations of some common potential energies ME = U+K
  • Equations of conservation of energy for the rotational case, e.g. K = Krot + K CoM
  • Momentum and conservation of momentum

exam instructions

Draw the system: Take your time with this: it is going to save time when you work on solving the problem!

  • Identify the most convenient axis system to solve your problem: where does the y axis point (e.g. up or down with gravity?), what vectors are the x and y axis aligned with?

  • Identify with respect to what reference point the angles are given to you: if it is with respect to the axis or to a preferred direction in the problem (e.g. the forward direction, or the East-West directions). If you get this wrong the whole problem will be different and wrong.

  • Draw all the vectors that are described in the problem and those that are not but are implicitly known to be there (e.g. acceleration due to gravity). I recommend you use different colors if you can here, so that you do not get confused (you can raw each vector in a different color and its components with the same color along the axes for example)

Remember dimensional analysis: lots of people are losing points in the quiz cause they have the answer in the wrong units, and therefore the wrong answer!

  • All trigonometric functions on your calculator will give you/expect as input angles in radians unless you specify degrees. Remember 1 rad = 180/3.1415 degrees

  • If you input in your problem g=9.8 then you are using meters/second! That is the right unit to use - but remember: all other values will have to be input as meters, seconds, etc, and the result will be in meters, seconds, etc, not in feet or miles!!

  • When you figure out what equation to use, do a quick check if the dimensions are coming up to be what you expect when you solve it for the relevant quantity

EXAM PREP

exam instructions

EXAM PREP

Strategy:

Look at what the problem tells you (your data) and the equations that we studied (your tools) and

figure out which equations can give you the quantity you are asked to find.  You need as many equations as unknowns.

It may take more than one step: the equations that contain the quantity you are after may also contain other unknown quantities: find an equation that will give the other unknowns and solve the equations one after the other.

 

Example:  you may need to solve an equation of motion to find the terminal velocity, then plug that in the conservation of momentum to find the initial velocity

so long as the first equation is solved in terms of the relevant unknowns you can substitute and obtain an equation that has one fewer unknown

phys207 key points

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