pwiki - User contributions [en]

Electricity and Magnetism

2024-08-26T16:54:25Z

Nuzhat: /* NUZHAT JILANI'S LECTURES */

*[[Electrostatics]]

*[[The Electric Field]]

*[[Electric Potential]]

*[[Capacitors]] 

*[[Magnetic Fields]]

*[[Notes on DC Circuits]]


*[[The role of a battery in a circuit]]



== NUZHAT JILANI'S LECTURE SLIDES ==

* [http://gauss.vaniercollege.qc.ca/~physics/NJ_LECTURES/NYB NYB]

Electricity and Magnetism

2024-08-23T01:12:28Z

Nuzhat:

*[[Electrostatics]]

*[[The Electric Field]]

*[[Capacitors]] 

*[[Magnetic Fields]]

*[[Notes on DC Circuits]]


*[[The role of a battery in a circuit]]

Electricity and Magnetism

2024-08-23T01:09:47Z

Nuzhat:

Electrostatics

2024-08-21T15:28:06Z

Nuzhat: /* Electrostatics Simulations */

= Textbook =
[https://openstax.org/books/university-physics-volume-2/pages/5-introduction University Physics Volume 2: Chapter 5]

= Electrostatics Videos =

== Theory ==
=== Charge, Conductors, and Insulators: Introduction to Electrostatics ===
<youtube>-Oq16ndKja8</youtube>
 

=== Coulomb's Law in Vector Form ===
<youtube>MwzwnhxoQh4</youtube>
 

=== Triboelectric effect/series or triboelectricity ===
<youtube>Fph08eKTVZM</youtube>
 

----

== Problem Solving ==
=== Calculation Example of Coulomb's Law in Vector Form ===
<youtube>7oYnrb89gmk</youtube>
 

----

== Demonstrations ==
=== Bending Water ===
<youtube>u-SIJSSBsjo</youtube>
 

=== Sticking a balloon to the wall ===
<youtube>bjU-Ll6U1ig</youtube>
 

= Electrostatics Simulations =
Check out these links for playing with charges:

*[http://phet.colorado.edu/en/simulation/travoltage Be careful John Travolta!] 

*[http://phet.colorado.edu/en/simulation/balloons Charge up a balloon] 

These and more links can be found at:



http://www.thephysicsteacher.ie/lcphysics19staticelectricity.html 



See how lightning strikes:

[http://regentsprep.org/regents/physics/phys03/alightnin/ Lightning applet]

STUDY DOCS

2019-09-18T22:15:03Z

Nuzhat: /* Pdf Format: */

=== Your Textbook ===

Openstax University physics is a free online book available in three volumes. See the links below to access the relevant topics:

* [https://openstax.org/details/books/university-physics-volume-1 Mechanics, Sound, Oscillations, and Waves] 

* [https://openstax.org/details/books/university-physics-volume-2 Electricity and Magnetism] 

* [https://openstax.org/details/books/university-physics-volume-3 Optics and Modern physics] 



=== Study Guides ===

* [http://gauss.vaniercollege.qc.ca/~physics/Study_Guides_Student_Docs Study Guides for NYA]



=== Other ===
* [http://stefan.bracher.info/files/physics/Formulas_and_Constants_for_College_Physics.pdf Stefan's Formulas and Constants for College Physics]

Conservation of Momentum

2017-08-29T21:28:16Z

Nuzhat: /* Some interesting VideoClips */

==Some Notes on Momentum==

''Karen Tennennhouse''

<ol>

<li>What is momentum? It is not that easy to describe. It is abstract in the sense that it is a calculated quantity.
The reason we calculate it (and give it a name) is that it turns out to obey an important conservation law.</li>
 
<li>Definition: An object of mass m, travelling at velocity <math>\vec v</math>, is defined to have momentum <math>\vec p = m\vec v</math>.
*Notice, Momentum is a vector
*Units are kg.m/s</li>
 
<li>When and how does the momentum of an object change?
 When a force acts on it (for some time). 
If the force is constant, the change in momentum <math>\Delta p</math> will obey <math>\Delta {\vec p} = \vec F \Delta t</math>. 
The quantity is called the Impulse delivered by this force. 
NB: Impulse is a vector, and this is a vector equation.</li>
 
<li>For a single object, the ideas of momentum and impulse don’t really give us any more information or convenience than just using Newton’s 2nd law. Momentum becomes more interesting, significant and useful when dealing with a group of objects (system). It is also very useful in handling a system where the details of the internal forces between the objects are complicated or changing.</li>
 
<li>Recall from the energy unit that:
*A system means any group of objects we choose to examine.
*For a given system, we say that a force is internal iff it is exerted BY an object IN the system (on another object in the system.)
*We say a force is external iff it is exerted BY an object which is NOT IN the system (on an object in the system.)</li>
 
<li>Law of Conservation of Momentum: 
Given any system, if the net external impulse is zero, then the (vector) total momentum of the system stays constant.</li>
 
<li>In many situations, the net external impulse might not be zero, but could still be small enough to ignore (negligible). This could happen if the external forces themselves are very small OR if the force is small-to-medium but acts for only a very short time interval.</li>
*What is “very small”, “small enough”, etc? It’s small enough if the impulse (hence the change in momentum, <math>\Delta{\vec p}</math> ) is much smaller than the momentum value <math>\vec p_{total}</math>
*This gives us a law like: 
:If the net external impulse on a system is zero OR negligible then the total momentum is “almost” constant, to as much precision as we need. 
i.e. <math>\vec p_{TOT_i} = \vec p_{TOT_f}</math>

*That is the law which applies to all our large, quantitative problems this semester.
*Some examples of situations where this is true (i.e. where net external impulse is negligible) include
**Collisions (where the system includes all the colliding bodies);
**Explosions;
**Events in a system on a level, frictionless surface.</li>
 
<li>As you see, the momentum law is in some ways very similar to the conservation of energy law. But there are important differences.
*The most important difference is that energy and momentum are different physical quantities. (They do not measure the same physical “stuff.”)
 
:In any given situation, the energy might be conserved, but not the momentum. OR momentum might be conserved, but not energy. Or both, or neither.
*For other differences and similarities, see the comparison chart below. Its purpose is to help you understand the various features.</li>
 

<li>'''Method for solving large 2-Dim Momentum Problems'''

 A. Choose the SYSTEM. You’ll need to be a little more careful here than you did for energy problems. 

 B. CHECK: Verify that the net external impulse (sum of ) is zero or “small enough”

*Notice, we’re checking the impulse due to external forces only (it does not matter whether they are conservative or not)

*In all our large, quantitative momentum problems this condition will be true. But you should still understand what you are checking for, and why it is true in a certain problem. 

 C. Write the LAW: If yes to B, then
(If no to B, then changes, but we will not do large problems using this.)
 
 D. Large, clear DIAGRAM must include:
*Choose and show AXES
*Before the event, show and label velocity of each object as a vector.
Label the unknowns with suitable symbols, for example, for initial velocity of second object.
*Do the same for the information after the event.
*Label angles and masses (with symbols, if unknown).
*Sketch, and label with symbols, all components. 

 E. To keep everything organized, it is strongly recommended to use the “table” layout.
If you prefer another method, it must be equally organized and explicit.
The overall strategy is to apply <math>\vec p_{TOT_i} = \vec p_{TOT_f}</math>
as vector sums, using the component method.
So, do all of the following separately for X and Y:
*Calculate the components of velocity and momentum.
*Calculate the total momentum before and the total momentum after.
*Apply the conservation law, i.e. set <math>\vec p_{TOT_i} = \vec p_{TOT_f}</math>

*Solve for desired unknowns. 

 F. You’re almost finished:
*If the question asks for them, you might need to find some magnitudes or directions, by combining the components in the usual ways. Otherwise, just leave your answers in î, ĵ notation.
*State final answers. 

</li>

 
<li>There are many other situations where the total momentum is not conserved, because the net external impulse is not small enough to ignore. In such a case the change in the total momentum will equal the net external impulse.
That is, for any system, <math>\vec p_{TOT_i} - \vec p_{TOT_f} = \Delta \vec p = \sum (\vec F \Delta t)</math> <math>\qquad </math> (vector equation).

</li>
 
<li>In general, in a collision, even when the total momentum is conserved, the total kinetic energy may or may not be.
*If the total KE before and after is conserved, then we say that the collision was “elastic”.
:If not, the collision was '''inelastic'''.
*The confusing term totally inelastic unfortunately does not mean that all KE was lost; it just means the objects stick together after collision

</li>

 

<li>Comparing what we know about Energy and Momentum</li>

</ol>
 

{|border="1"

! !! ENERGY !! MOMENTUM
|-
!
| is a SCALAR (number).
Has no direction.
We can use minus signs to indicate decrease, or less than some zero.
|| is a VECTOR !!

In 2 or 3 dims we must remember to use vector operations for it.
|-
! Forms
|Has many different forms
(for example, KE, GPE, heat,
Elastic PE, etc), and can transfer from one form to another.
|Always one form, <math>\vec p = m\vec v</math>
|-
! Measure/cause of changes
|For constant force,
energy change caused by <math>\vec F</math> will be
Work = <math>\vec F. \Delta \vec s </math> (dot product)

[ For changing forces, would use an integral, but not used much in our course. ]
|For constant force,
momentum change caused by <math> \vec F </math> will be
'''Impulse''' = <math>\vec F \Delta t</math>
[ For changing forces, NOT in our course, we would use an integral ]

|-
! The Main Conservation law
|If net work done by external forces is zero, then total energy will be conserved.
That is the total (scalar) of all forms present in the system.
|If net external impulse is zero, then total momentum ( vector total ! ) will be conserved.
That is, vector total of <math>m \vec v</math>‘s in the system.
|-
! What happens when it is NOT conserved
|The change in total energy of a system will equal to the net work done by external forces.

That is, change (final minus initial) in the number total of all forms.

|The change in total momentum of a system equals the net impulse by external forces.

Notice this is a vector equation: the “change” <math>\vec {p_f} - \vec {p_i}</math> is a vector subtraction, and the total (net) of the external impulses is a vector sum.

|-
! Other important laws for this quantity?
|'''Yes'''. For energy, we are also interested in knowing when certain forms (not just the whole total) will be conserved. We express this information with several other laws.
For example: 

*If the work done by non-conservative forces is zero, then the total mechanical energy (TME) is constant. That is, the total of just the kinetic plus all forms of potential energy,( but not other forms such as heat.)

*If work by nonconserv. forces is not zero, then the change in TME will equal the work by nonconservative forces.

*Change in Kinetic energy equals work done by the net force.
|'''Nope'''.

|}
 

==More on Momentum and Collisions==

*[[Linear Momentum and Collisions]]
 

==Some interesting VideoClips==
* Boy + Cart ''added by Scott Redmond''
*: <youtube>KL8-PbdRYY0</youtube>


* [http://www.youtube.com/watch?v=ExQUGk12S8U&feature=relmfu Safety consequences of vehicle size and weight] (2:34, youtube) ''added by Scott Redmond''
*: This shows very similar content to the CTV News clip, since the CTV News clip seems to be unavailable as of 12-Nov-2012.
* [http://www.youtube.com/watch?v=yUpiV2I_IRI&feature=related Understanding Car Crashes: It's Basic Physics] (22:15, youtube) ''added by Scott Redmond''

==Exercises==

*[[Exercises on Conservation of Momentum]]

Motion Along a Straight Line: Graphical Representation

2017-08-29T21:20:54Z

Nuzhat: /* Simulations */

''Helena Dedic''

==x-t graphs==
===Average velocity from the x - t graph===

'''The slope of the secant line is equal to the average velocity during an interval of time <math>\Delta t = t_2 - t_1</math>.'''

To determine the average velocity we find the two points on the curve corresponding to instants of time <math>t_1</math> and <math>t_2</math>.

[[image:Graph_xt_1.png|top]] 

Then we determine the "coordinates" of these points and then we are ready to compute the average velocity:

<math>v_{av} = \frac {x_2 - x_1}{t_2 - t_1}</math>

For example, we want to determine the average velocity during the last two seconds of motion of a particle. Its motion is illustrated by a graph below:

[[image:Graph_xt_2.png|top]] 

Therefore the average velocity is:

<math>v_{av} = \frac {1.5 - (-0.5)}{5 - 3} = 1 m/s</math>

===Instantaneous velocity from the x - t graph===

'''The slope of a tangent line drawn at a point (x,t) is equal to the instantaneous velocity at t.'''

To determine the instantaneous velocity we first draw a tangent line at (x,t).

[[image:Graph_xt_3.png|top]] 

Then we choose two points on the tangent line and then determine the "coordinates" of these two points. We can then compute the instantaneous velocity:
<math>v = \frac {x_2 - x_1}{t_2 - t_1}</math>

For example, we want to determine the velocity at 2 s of a particle. Its motion is illustrated by a graph below:

[[image:Graph_xt_4.png|top]] 

We draw the tangent line at (2, 0.9). Then we choose two points on the tangent line. In this case, we chose the points (1, 1.4) and (4, 0). Then we compute:

<math>v = \frac {0 - 1.4}{4 - 1} = - 0.47 m/s</math>

===v - t graph from x - t graph===

When sketching a v - t graph we follow these steps:

1. Move a ruler along the curve from left to right so that it shows the tangent line. Note when the ruler is horizontal, that is when the slope of the tangent line is zero.

For each such instance record the instant (or interval of) time on t-axis of the v - t graph. The v - t graph will cross t-axis at those points. The graphs below show the movement of the "ruler" until it reaches the first point of zero velocity.
[[image:Graph_xt_vt_1.png|top]] 

2. Move the ruler again along the curve from left to right so that it shows the tangent line. Note the intervals of time when the ruler does not change its slope.

For each such instance note whether the slope is positive or negative. Then draw for each such interval a horizontal line on the v - t graph. The horizontal line should be above t-axis when the slope is positive and below t-axis when the slope is negative. The graphs below show the movement of the "ruler" along a straight line segment of the x - t graph. Note that the "ruler" does not change its slope.

[[image:Graph_xt_vt_2.png|top]] 

3. Move the ruler again along the curve from left to right so that it shows the tangent line. Note the intervals of time when the ruler turns while moving along the curve. It indicates that the slope of the x - t graph is changing.

When the ruler turns clockwise then the slope of the x - t graph is decreasing and when the ruler turns counterclockwise then the slope of the x - t graph is increasing. For each such instance draw a straight line with a negative slope (clockwise) or positive slope (counterclockwise) on the v - t graph. The graphs below show the movement of the "ruler" along a curved line segment of the x - t graph. Note that the "ruler" does changes its slope from positive to negative and that it rotates clockwise. The incomplete v - t graph shows decreasing velocity.

[[image:Graph_xt_vt_3.png|top]] 

4. It is important to say that the resulting graph should be continuous and that any discrepancy that you encounter when drawing the graph are due to some error somewhere. Check your work.

The complete v - t graph looks like this:

[[image:Graph_xt_vt_4.png|top]] 

== Understanding v-t Graphs ==

===Table of Synonyms===

Note that the following sketches are examples of motion and cannot be generalized.

{| border="1"

! Synonymous statements !! v-t graph

|-

|
*the particle moves at constant velocity forward/backward
*the constant velocity is positive/negative
| The velocity graph is above or below t - axis

[[image:V-t_fig_1.png|top]]

|-

|
* the particle is at rest during an interval of time
* the particle has zero velocity over an interval of time
* the particle is stationary
* both velocity and acceleration are simultaneously zero during an interval of time

|The graph runs along t-axis.

[[image:V-t_fig_2.png|top]]

|-

|
* the particle is at rest for an instant
* the particle has a zero velocity at an instant
* the particle changes it's direction of motion at an instant

|The graph of velocity vs. time crosses the t-axis - see the dot in the graph.

[[image:V-t_fig_3.png|top]]

|-

|
* the particle has decreasing/increasing velocity
* the particle has negative/positive acceleration

|The velocity graph has a negative/positive slope
[[image:V-t_fig_4.png|top]]

|-

|
* the particle is slowing down/speeding up
* velocity and acceleration point in the opposite/same direction

|The velocity graph "approaches" the t-axis in the two graph on the left (slowing down). The velocity graph "goes away" from t-axis in the two graph on the right (speeding up).

[[image:V-t_fig_5.png|top]]

|}
 
===Important Points to Remember===

* The instantaneous acceleration is the slope of the v - t graph.
* A straight line v - t graph indicates motion with constant acceleration.
* A straight and horizontal v - t graph indicates motion with zero acceleration.
* Important notions:
** v = 0 does not imply that a = 0
** a = 0 implies that v is constant
** "speeding up" implies only that the acceleration and the velocity have the same direction while "slowing down" implies that they have the opposite direction
** The speed is equal to the absolute value of the velocity. When thinking about speed visualize a graph of absolute value of velocity versus t

 

==Relationship between x-t and v-t graphs==
===How to compute the displacement from the v-t graph===

The area under the graph tells us about the displacement during an interval of time Dt = t2 - t1. When we say "area under the graph" we mean an area bound by the segment of the t-axis (t1, t2), by the segment of the vertical line drawn from point t1 on the axis to the corresponding point on the curve, by another segment of the vertical line drawn from point t2 on the axis to the corresponding point on the curve and by the segment of the curve (v(t1, v(t2)). It is important to note that the displacement is positive if the area is above the t-axis and negative if it is below the t-axis. For example, the displacement during the interval of time (1 s,2 s) is equal to the shaded area shown below. The blue lines indicate the boundaries of this area.

[[image: Ex Graph xt vt 1.png|TOP]]

The area of this triangle is equal to ½x1 m/s x1 s = 0.5 m. Note that we have only found the displacement during this interval of time. To determine the position of the particle at the end of this interval of time, we have to add the displacement to the position of the particle at the beginning of the interval of time. Let say that the position of the particle was 0 at t = 0. During the first second of motion the particle traveled 1 m and therefore at the beginning of the interval of time we dealt with the position was 1 m. The position at t = 2 s is the sum of the initial position and the displacement or 1 + 0.5 = 1.5 m. Similarly, the displacement during the interval of time (3 s, 4 s) is shown by the shaded area in the graph below.

[[image: Ex Graph xt vt 2.png|TOP]]

In this case, the area is below the t-axis and so the displacement is equal to -1 m/s x 1 s = - 1 m. To find the position at 4 s, we would have to add - 1 m to the position at 3 s.

==Exercises==
*[[x-t graphs]]
*[[v-t graphs]]
*[[Relationships between Graphs]]

==Simulations==
Qi Luo


*[http://phet.colorado.edu/en/simulation/moving-man The moving man]

Vectors

2017-08-29T21:15:20Z

Nuzhat:

===Workshop on Vectors===

''Wissam Chaya''

====Overview of Vectors====

1 - For a start, An overview about vectors: 
http://www.physics.uoguelph.ca/tutorials/vectors/vectors.html 
What is the difference between <math>\vec{A}</math> and <math>-\vec{A}</math>? How do we perform vector subtraction of <math>\vec{A}- \vec{B}</math>? 

====Applications of Vectors====

For each of the following application, draw a diagram and show vectors acting on the object. Define these vectors and answer the questions related when applicable. 

• [http://www.physicsclassroom.com/mmedia/vectors/plane.cfm Plane and the Wind] 

• [http://www.physicsclassroom.com/mmedia/vectors/rb.cfm The River Boat] 

• [http://www.physicsclassroom.com/mmedia/vectors/bds.cfm Parabolic Motion of projectile] 

• [http://www.physicsclassroom.com/mmedia/vectors/mzng.html Throw at the Monkey in a Gravity Free environment] 

• [http://www.physicsclassroom.com/mmedia/vectors/mzs.html Throw at the Monkey at a slow speed with Gravity on] 

• [http://www.physicsclassroom.com/mmedia/vectors/hlp.html Horizontally Launched Projectiles] 

::What are the initial vx and vy speeds? 

::What are the final vx and vy speed (when it hits the ground)? 

• [http://www.physicsclassroom.com/mmedia/vectors/nhlp.html Non-Horizontally Lauched Projectile] 

::What are the initial vx and vy speeds? 

::What are the vx and vy at the peak? 

::What are the final vx and vy speeds (when it hits the ground)? 

• [http://www.physicsclassroom.com/mmedia/vectors/mr.html Maximum Range] 

::At what angle is the maximum range can be obtained? 

• [http://www.physicsclassroom.com/mmedia/vectors/pap.html The Plane and the package] 

::Why does the package fall just underneath the moving plane? 

• [http://www.physicsclassroom.com/mmedia/vectors/tb.html The Truck and the Ball] 

::Why does the fall back on the moving truck? 

::If you were sitting on the truck, how would you see the path of the ball: As a projectile or as a vertical up/down path? 

• [http://www.widro.com/throwpaper.html Throw paper !] 

::Can you throw the paper in the basket? 

::Play ! Consider the wind direction and speed as well as the velocity direction of your thrown paper. Can you score? how many? Finally draw a diagram of all vectors involved in this game! 



====Viewing 3D vectors====

[http://www.univie.ac.at/future.media/moe/galerie/vect1/vect1.html#vkenn 3D Vectors] 



====Math & physics puzzles====

[http://www.sci.sdsu.edu/mathtutor/demos.html Puzzles!!]
 

===Relative Motion===
''Helena Dedic''

[http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=140 Relative Motion]

===More on Vectors===

*[[Theory]]

*[[Exercises on Vectors]]

===More Links===

http://phet.colorado.edu/en/simulation/vector-addition

Magnetic Fields

2016-04-01T16:41:29Z

Nuzhat:

Understanding the earth's magnetic field:

* [http://phet.colorado.edu/en/simulation/magnet-and-compass Bar Magnet and compass]
* [http://www.windows2universe.org/physical_science/magnetism/earth_magnet_dipole_interactive.html Earth's magnetic Field]
* [https://www.youtube.com/watch?v=M9Gdm_OXKz4, demo of magnetic field lines with iron filings (one magnet)] 
* [https://www.youtube.com/watch?v=GCHJmMdHNPo, demo of magnetic field lines with iron filings (two magnets)] 

Vector Product:

* [http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Flash/Vectors/CrossProduct/CrossProduct.swf Cross product of two vectors]

Electricity and Magnetism

2016-03-30T17:03:22Z

Nuzhat:

*[[Notes on DC Circuits]]


*[[The role of a battery in a circuit]]

*[[Electrostatics]]

*[[The Electric Field]]

*[[Capacitors]] 

*[[Magnetic Fields]]

The role of a battery in a circuit

2016-03-22T20:25:18Z

Nuzhat: /* The emf of a battery */

===The emf of a battery===

A source of emf is a device that uses some form of energy (other than electrical energy) and coverts it to electrical potential energy. For example, in a battery, the chemical reactions taking place within the body of the battery creates a separation of charges at its two electrodes, so that one becomes positive and the other negative. This increases the electrical potential energy of the system of charges, and we can see that the battery does positive work in achieving this separation. The chemical energy is converted into electrical energy. In case of generators, which we shall see later in the course, mechanical energy is converted into electrical energy. 

When the two ends of the battery are connected, the electrons from the negative end can move through the circuit to reach the other positive end of the battery. As they pass through the circuit components they lose part of their potential energy in each, so that when they reach the positive electrode, they have zero potential energy. The batteries role is to continuously keep supplying the electrons with a certain potential energy so that the flow of electrons is maintained in the circuit. The amount of electrical potential energy that the battery can provide to the electrons is clearly related to the potential gradient that is created as a result of charge separation, and hence the battery “value” is written in terms of the potential difference that exists between its electrodes. This is called the emf of the battery.

“The emf (ξ) of a device is defined as the work done by the source of emf in moving the charge around a closed loop or circuit.”


[[image: battery.PNG|center]] 
<math>\Rightarrow</math> ξ = W/q (Units: J/C ≡ V) 

Thus, a source of emf supplies electrical energy in a circuit, whereas the other components in the circuit (example resistors) dissipate that energy. Hence we can say that the emf supplies electrical Power, and a resistor dissipates electrical Power (in case of simple resistors, the electrical energy is converted into thermal energy).

Let us look at a standard lead-acid battery:

Pb and <math>PbO_2</math> plates are submerged in <math>H_2SO_4</math>.

<math>H_2SO_4 \rightarrow 2H^{+} + SO_{4}^{2-} </math> 

Pb + SO42- -> PbSO4 + 2e-
These electrons move in the wire and when they reach the positive plate:
PbO2 + 4H+ + SO42- + 2e- -> PbSO4 + 2H2O
Thus in this process PbSO4 is deposited on both plates and the acid is used up, and because of this there is a life span of the battery, after which no further reactions can take place since all the active ingredients have been used up. 

(Rechargeable batteries are based on reactions that can be reversed by passing a current through them in the opposite direction with the help of another source of emf. So in this case, clearly the battery that is being charged is consuming Power that is being supplied by the other source of emf).

File:Battery.PNG

2016-03-22T20:14:12Z

Nuzhat: Nuzhat uploaded a new version of File:Battery.PNG

Basic Concepts 2

2016-02-29T21:24:11Z

Nuzhat: Created page with "===Testing==="

===Testing===

Dynamics of Circular Motion

2015-10-28T18:28:45Z

Nuzhat: /* THE DYNAMICS OF CIRCULAR MOTION */

''Kreshnik Angoni''

===THE DYNAMICS OF CIRCULAR MOTION===
 
* In kinematics we found that a particle moving at constant speed v around a circle with radius r has acceleration. This acceleration is directed towards the circle center at all times, and has constant magnitude: 

<math>a_c = \frac{v^2}{r}</math> 

So, in vector form, we have: 

<math>\vec{a_c} = -\frac{v^2}{r}\hat{r}</math> ......... (1) 

where <math>\hat{r} = -\frac{\vec{r}}{r}\hat{r}</math> 

is the unit radial vector (tail at center) defining the direction from origin. 

[[image: Kreshnik_Circ Dyn_Fig1.png|left]]

 The vector <math>\vec{a_c}</math> is called '''''centripetal acceleration''''' because it is directed all time versus the circle center. The 2nd law tells that a particle with mass <math>m_p</math> which is moving with centripetal acceleration is under the action of a net force directed along <math>\vec{a_c}</math> (fig.1) 

<math>\vec{F}_{NET} = \vec{F_c} = m_p \vec{a_c} = -\frac{m_p v^2}{r}\hat{r}</math> ........ (2) 

This force, with magnitude <math>\frac{m_p v^2}{r}</math>, is called centripetal force because it is directed
all time versus the circle center. Note that it is named only from its direction and
not from its physical origin. A force of gravity, friction, electric …or their sum
can be in the role of a centripetal force. 

'''''Note:''' '''You must not add a centripetal force in a free body diagram'''''. 

* The action of centripetal force consists in not leaving the particle to follow its path along the tangent “as required by its inertia”. As the third law affirms, the particle reacts on this action by a force of equal magnitude, opposite direction and applied on that source itself. 

'''Example 1''': A man has fixed an object at a string and rotates it in a horizontal plane.
The net force acting on the object is the sum of his hand force (transmitted by
string via its tension) plus the weight of object. This net force acts as centripetal
force exerted on the object. The object reacts (via the string tension) by a force
with equal magnitude and opposite direction applied on the man’s hand; this is
a centrifugal force acting on the man (who is not in circular motion).
 

'''Example 2''': A car travels at high speed around a horizontal curved path. The friction force exerted by the
road on the rotating tires plays the role of centripetal force (its source is the road). An equal force
with opposite direction, the centrifugal force, is exerted on the road.
Note: Sometimes, by error, one talks about the centrifugal force exerted on the rotating body. The
example of a passenger inside the rotating car that feels himself pushed “out of circle” is mentioned.
The truth is that the passenger is not hard bounded with the seat to get the centripetal force via the
action of friction straight away. So, due to his inertia he follows his way along the tangent….until
he touches the interior of the door. Then, the door transmits to him the centripetal action and after
that he follows the same circular motion as the car. 

Remember: There is no net outward (centrifugal force) exerted on the rotating body…only….centripetal force. 

===Loop the loop:===

 
:::::<youtube>y64JhPtpicA</youtube>
 

===SATELLITE MOTION===

* The satellite is a small body with mass m that turns around a central body with bigger mass M due to the gravitational interaction. In fact, being the only force exerted on the satellite, the gravitation force is the net force on the small body and it plays the role of centripetal force. 

:By equalizing the two expressions for the magnitudes 

<math>G\frac{mM}{r^2} = \frac{mv^2}{r}</math> ........(3) 

one finds out that: 

<math>v_{orb} = \sqrt{\frac{GM}{r}}</math>....... (4) 

'''So, the orbital speed of a satellite''': 

:::'''a)''' ''Depends only on central mass M and not on satellite mass'' 
:::'''b)''' ''Decreases with the increase of orbit radius''. 
 

* The period of satellite in a fixed (r value) orbit is 

<math>T = \frac{2\pi r}{v_{orb}} = \frac{2\pi}{\sqrt{GM}}r\sqrt{r}</math> ........ (5) 

:and 

<math>T^2 = \frac{4\pi^2}{GM}r^3 \equiv k r^3</math> ........(6) 

The expression (6) known as Kepler’s third law shows that the period of different satellites around the same central body depends only from their distance. 
 
This expression is very useful because by measuring the orbit radius and the satellite period one may calculate the mass of the central body. For example, one may calculate the sun’s mass by using the periods and distances of its satellites (planets) or the earth’s mass by using moon’s period and distance from the earth.

 
''See also'' 

*[[Newton's Law of Universal Gravitation]]

Friction

2015-08-25T20:50:53Z

Nuzhat: /* Related Videoclips */

''Karen Tennenhouse''

The actual reasons for friction are complicated, at the level of the atoms and molecules. However, the behaviour of large-scale (visible) objects is fairly simple and consistent, and we include some of it in our course, as follows.

There are three kinds of friction:

• Kinetic friction ( better called “sliding friction” )

• Static friction

• Fluid friction

==== Kinetic (Sliding) Friction====

<ol> <li>It acts only when an object IS sliding/slipping against a surface or other object. (Notice, I said “sliding”, not “moving”.)</li> 

<li> Direction of the kinetic friction force, on each object, is whatever will tend to make it stop sliding.
(Not necessarily to stop moving.)</li> 

<li> Magnitude of kinetic friction has an exact formula:</li>

*It depends only on the normal force of the surface pressing against the object, and on a coefficient called μ<math>_k</math>, the “coefficient of kinetic friction” . Specifically, 

<math>F_{fk}</math> = μ<math>_k
</math> N 

The value of the μ<math>_k</math> depends on the materials and roughness.
(for example there is a μ<math>_k</math> for rubber tires sliding on cement, a different one for copper on steel, and so on.) 

* Notice that the magnitude of kinetic friction does not physically depend on the object’s speed, acceleration, net force, nor on any other forces that may be acting at this instant (unless they affect the N.)

* Of course it may happen that we calculate <math>F_{fk}</math> from knowing 'a' or 'Net F' etc; but these things are not causing the friction to have this value. Also, in many problems we calculate <math>F_{fk}</math> from its formula μ<math>_k</math> N .</ol>

 
====Static Friction====

<ol> <li> It acts only when an object is in contact with a surface or other object; is not sliding/slipping; but something is “trying” to make it slide. [The object may be stationary or moving, but it is not sliding.]</li> 

<li>Direction of the static friction force, on each object, is whatever will prevent it from starting to slide.
[Not necessarily to prevent it from moving. For example, static friction being used to walk, to make a car drive forward, etc.
Also, object riding on a truck: the static friction makes the box accelerate forward, so it stays in contact with the same spot on the truckbed; if there was no friction the truck would slide out from under the box.]</li> 

<li> Magnitude of static friction does NOT have a formula
(In particular, it does not have the formula μ<math>_s</math> N . Really!)
* Magnitude of <math>F_{fs}</math> will physically adjust itself, (if it can) to whatever value is needed to prevent slipping.
Therefore the actual value (magnitude) of <math>F_{fs}</math> really, physically depends on all the other forces which have components parallel to the surface.
* In a problem, either <math>F_{fs}</math> must be given, or else we figure it out from the second law, known acceleration value, and the other forces.</li> 

<li> What did we mean by saying “<math>F_{fs}</math> will adjust itself (if it can)”, in point 3?</li>
It turns out that there is a maximum (limit) to how large static friction can become in each situation. There actually is a formula for that limit, as follows:
* Any pair of surfaces also has another characteristic coefficient, which helps to determine the largest possible static friction. It is called μ<math>_s</math>, the “coefficient of static friction”, and there is a true formula that 

<math>F_{fs}</math> <= μ<math>_s</math> N

::''(<= means “less than or equal to”.)'' 
* This allows us to solve various problems of the kind “will it start to slide?” or “what is the max/min value of (some physicsl quantity) for the object not to slide?” etc. (The Rotor ride is a cute example.) 
* Just notice that in most problems, and in most real-life situations, the magnitude of static friction (really acting) is not equal to μ<math>_s</math> N. 
* For most surfaces, μ<math>_k</math> is smaller than μ<math>_s</math> . That is, it is harder to make something start sliding than it is to keep it sliding.
</ol>
 
====Fluid Friction====
<ol> <li> This refers to the “drag” which tries to retard a body moving through a liquid or gas (eg water or air.) It may also be called “air friction” or “air resistance”.</li> 
<li>Its direction is such as to slow down the object, or to drag along the fluid.</li> 
<li>It has great practical importance, but because it is mathematically complicated, it is normally left until a university-level physics course. We do almost nothing with it in physics NYA.
[ except that, of course, it can be one of the forces acting in a problem. Its value could be given, or could be figured out from knowing the accel and all the other forces.]</li> 
<li>When a question says a body is falling “at terminal velocity”, it means that the upwards force of air friction is equal in magnitude to the weight, so Net <math> \bar{F} </math> = 0, which means that the body is falling at constant velocity.</li> 

<li> Because fluid friction gets larger with larger speeds, a real body in free fall does not continue to accelerate at 9.8 m/s2. Rather, as it falls, the upward air friction gets larger; the downward net force ( = W – F<math>_{air}</math> ) gets smaller, until F<math>_{air}</math> = W and the body falls at constant speed. The speed at which this happens is called the “terminal velocity”; it depends on the shape and density of the object, and the viscosity of the fluid. We are familiar with the fact that a penny falls much faster than a sheet of cardboard or aluminum foil, even if they have the same weight. Issues of shape affecting fluid friction are crucial to the design of cars, airplanes, etc,... and of birds, whales and fish! ]</li>
</ol>
 

==Related Videoclips==

[http://www.youtube.com/watch?v=LdyJLLumCEc The rotor]
 

:::::<youtube>LdyJLLumCEc</youtube>

Electrostatics

2013-01-22T22:52:33Z

Nuzhat:

Check out these links for playing with charges:

*[http://phet.colorado.edu/en/simulation/travoltage Be careful John Travolta!] 

*[http://phet.colorado.edu/en/simulation/balloons Charge up a balloon] 

These and more links can be found at:



http://www.thephysicsteacher.ie/lcphysics19staticelectricity.html 

http://www.regentsprep.org/Regents/physics/phys03/aeleclab/default.htm 

YouTube Videos:

[http://www.youtube.com/watch?v=eCLu4t12LdE MIT Electrostatics demo] 

:::::<youtube>eCLu4t12LdE</youtube>
 

See how lightning strikes:

[http://regentsprep.org/regents/physics/phys03/alightnin/ Lightning applet]

Circular Motion

2012-11-15T18:42:15Z

Nuzhat: /* Related Videoclips */

''Karen Tennenhouse''

==Some Notes and Cautions about Circular Motion Problems==

<ol><li>Consider an object travelling in a circle (radius R) at some instantaneous speed (v).

If the speed (magnitude of velocity) is constant, we say that this is '''Uniform Circular Motion'''.
Such an object does not have constant velocity; its direction is changing. In fact, it is accelerating with an acceleration <math> \bar a </math> whose direction is towards the center of the circle, and whose magnitude is</li> 

<math>a = v^2/R
</math>

<li>What causes the object to accelerate towards the center? 
The same thing that ever causes any object to accelerate in any way: The real forces acting on it (gravity, friction, tension or whatever) are adding up to some nonzero net force. The net force causes the object to accelerate, in the direction of Net<math>\vec{F}</math> , with </li> 

<math>\vec{a}</math> = (Net<math>\vec{F}) / m</math> 

Notice, the acceleration does not cause a force.
Rather, NET FORCE causes acceleration.

<li>Because it points radially towards the center, the acceleration is also called the “centripetal acceleration” and also sometimes called the “radial acceleration”.

Likewise, because the net force is pointing towards the center, it is also called the “centripetal force.”</li> 

Centripetal force is NOT a “separate” or “extra” force;
it is just another name for the NET force
∑<math>\vec{F}</math> in this situation.

<li>Treat every circular motion problem as a Newton’s 2nd Law problem, just with a special formula for the ‘a’ and with special care about the direction of <math>\vec{a}</math>.
*Set up the solution as for any dynamics problem, by drawing the isolation diagram.
*But, very carefully figure out where is the plane of the circle, where is the centre, where is the object, and thus carefully determine the direction of acceleration. Draw it as a small double arrow.
*It is strongly advised that you choose your positive axis in the same direction as the acceleration.
:[If you really prefer, you may choose the opposite axis; BUT beware: you must then remember to substitute ( – a) instead of (a) in the Net <math>\vec{F}</math> equation. Most people forget.]
*As usual, write: ∑<math>\vec{F}_{axis}</math> = ..... = m <math>\vec{a}</math> but knowing that magnitude a = (v 2 / R)

</li>

We never draw the net force in the isolation diagram,
and we’re not going to start now!

So, the isolation diagram should NOT have a <math>\vec{F}_{C}</math> (Centripetal Force)
in it. 
(If you have drawn a force in your diagram and cannot say by what object this force is exerted, then probably this is not a real force.)
 
<li>All the above notes deal with uniform circular motion, ie where the object is going in a circle but at constant speed.
But, what happens if an object is travelling in a circle (thus changing direction) and is also changing its speed? We will look at that situation very briefly near the end of our NYA course (see ''Rotational Motion''). For your interest, here is a mini-preview: 
In such a case, the object’s acceleration will have two (perpendicular) components: 
*One component, along the direction of motion, is called the tangential acceleration, aT . The tangential acceleration is related to the change in speed: it is in the same direction as velocity if the object is speeding up, or opposite to velocity if object is slowing down.) 
*The second component, called the radial acceleration aR , is exactly the centripetal acceleration, ie it points towards the center and has magnitude aR = (v2 / R) 
*The object’s acceleration is the vector sum of its two components.
</li>
</ol>
 

* [[Rotational Motion]] 

*[[#Some Notes and Cautions about Circular Motion Problems|TOP]] 

==Related Videoclips==

<ol><li>[http://www.youtube.com/watch?v=WKvhgkg2_dw Introduction to uniform circular Motion]</li>

:::::<youtube>WKvhgkg2_dw</youtube> 

<li>[http://www.youtube.com/watch?v=LdyJLLumCEc The rotor]</li>

:::::<youtube>LdyJLLumCEc</youtube> 

</ol>

*[[#Some Notes and Cautions about Circular Motion Problems|TOP]] 

==Exercises==
''Helena Dedic''

===Exercise 1===

True or False: When a particle moves in uniform circular motion its acceleration is constant.

'''Solution:'''
 

'''False.''' The radial acceleration has a constant magnitude but its direction changes as the particle moves around the circle. The acceleration always points towards the centre.
 

===Exercise 2===

The electron in a hydrogen atom has a speed of <math>2.2 x 10^6</math> m/s and orbits the proton at a distance of <math>5.3 x 10^{-11} m</math>. What is its centripetal acceleration? 

(b) A neutron star of radius 20 km is found to rotate once per second. What is the centripetal acceleration of a point on its equator?

'''Solution:''' 

a. It is given that the electron has speed <math> v = 2 * 10^6 m/s</math> and orbits the proton with radius <math>r = 5.3 * 10^{-11} m</math>. You are asked to find its centripetal or radial acceleration:

<math>a = (2 * 10^6 m/s)^2 / (5.3 * 10^{-11} m) = 9.1 * 10^{22} m/s^2</math> 

It is always interesting to think about one's results. It is particularly intriguiging in this case because the magnitude of the acceleration is such an awesome number (compared to <math>10 m/s^2</math> on Earth). Now, suppose the Earth were to shrink and become a Black Hole; then at a distance of 3 cm from the centre, acceleration would still be only <math> 4 * 10^{17} m/s^2</math>. Thus the large radial acceleration of the electron is testimony to the strength of the interaction between the electron and the proton in the atom.
 

b. You are told that a neutron star rotates once per second (i.e that it has period T = 1 s) and that it has a radius of <math>r = 2 * 10^4</math> m. You are asked to find the centripetal acceleration of a point on its equator:

<math>v = (2\pi) / T = (2\pi * 2 * 10^4 m) / 1 s = 4\pi * 10^4 m/s</math> 

<math>a = v^2 / r = (4\pi * 10^4 m/s)^2 / (2 * 10^4 m) = 7.9 * 10^5 m/s^2</math> 

 

===Exercise 3===

Can a particle move at a constant speed and yet be accelerating? If so, give an example.

'''Solution:''' 

Yes, a particle can move at constant speed around a circular or any other curved path and have an acceleration. Since the direction of the velocity changes the velocity vector is not constant and therefore the motion is an accelerated motion.

 
* [[Rotational Motion]] 

<center> *[[#Some Notes and Cautions about Circular Motion Problems|TOP]] </center>

Newton's Laws

2012-09-10T18:32:49Z

Nuzhat: /* Newton's Third Law of Motion */

===Newton's Three Laws of Motion===


====Newton's First Law of Motion====

''"A body continues in a state of rest, or of uniform motion in a straight line, unless acted upon by an external force."''
 
In other words, an object remains at rest or moves at constant velocity when

<math>\Sigma \vec{F} = 0</math> 

Newton’s first law is based on the principle of inertia. The tendency of a body to resist a change in its state of motion is called '''inertia'''. 

*[[Inertia|MORE ON INERTIA]]

 
:::::<youtube>Q0Wz5P0JdeU</youtube>
 

====Newton's Second Law of Motion====

''"An object of mass <math>m</math> moves with an acceleration that is defined by <math>\Sigma \vec{F} = ma</math>."''
 
Thus, the acceleration of a body is directly proportional to the net force acting on the body, and is inversely proportional to its mass.

 
:::::<youtube>WzvhuQ5RWJE</youtube>
 

====Newton's Third Law of Motion====

''"For every action there is an equal and opposite reaction."''
 
<math>\vec{F_{12}} = -\vec{F_{21}}</math>

In other words, if body <math>1</math> applies a force on body <math>2</math>, then body <math>2</math> applies an equal and opposite force on body <math>1</math>.
 
Convention for the subscripts: If a force is exerted by an object A on an object B then we write this force as follows:

<math>\vec{F_{AB}}</math>

 
:::::<youtube>cP0Bb3WXJ_k</youtube>
 
 

http://www.youtube.com/watch?v=KeNye0nTqmM
:::::<youtube>KeNye0nTqmM</youtube>
 

*Focus on notation: The first subscript <math>A</math> indicates the particle which exerts the force; the second subscript <math>B</math> indicates the particle on which the force is exerted. For example, the normal force exerted by the floor (<math>F</math>) on a box (<math>B</math>) in the diagram below should be labelled

<center><math>\vec{F_{FB}}</math> </center>

[[image:Newtons_Third_Law_Notation.png|TOP]] 

*Action/reaction pairs: The two forces <math>F_1</math> and <math>F_2</math> form an action-reaction pair if and only if the two forces are of the same type (a normal force and a gravitational force cannot be an action-reaction pair) and if and only if we can write that

<center><math>F_1</math> is exerted by an object B on an object A and</center>

<center><math>F_2</math> is exerted by an object A on an object B.</center>

For example: <math>{F_G}_{EM}</math> which is a gravitational force exerted by the Earth on the man and <math>{F_G}_{ME}</math> which is a gravitational force exerted by the man on the Earth are the action-reaction pair.

On the other hand, <math>{F_G}_{EM}</math> which is a gravitational force exerted by the Earth on the man and <math>{F_N}_{SM}</math> which is the normal force exerted by the surface of the Earth on the man are not the action-reaction pair. 

*Application of these concepts to find the acceleration of systems — problem solving strategy:

#Define a system of particles that you will study — each particle should be dealt with separately;
#List external forces acting on each particle use the notation which indicates the agent and then draw arrows to represent each external force in the diagram;
#Draw a free body diagram for each particle:
#*Select and draw an "appropriate" coordinate system for each particle so that
#**The x-axis is parallel to the direction in which a particle may possibly move or the direction of possible acceleration;
#**The orientation of the x-axis for each particle in the system must be in the same sense;
#**The positive y-axis is oriented +90º from the positive x-axis.
#*Transfer the arrows representing forces into this system (each arrow has the tail at the origin).
#*Determine the angle <math>\theta</math> each arrow makes with the positive x-axis.
#Write all forces in terms of their x- and y-components: for a given force of magnitude <math>F</math> and direction <math>\theta</math>, <math>F_x = F \cos\theta</math> and <math>F_y = F \sin \theta</math>
#Write equations for each particle <math>\Sigma F_x = 0</math> or <math>\Sigma F_x = ma</math>
#Write equations for each particle <math>\Sigma F_y = 0</math>
#Identify all the action-reaction pairs.
#Solve the system of equations. 

*An overview of the forces acting on a body in a mechanical system:

#Gravitational force: <math>F_G = {G m_1 m_2 \over r^2}</math>
#Special case of a gravitational force (close to the surface of Earth): <math>F_G = mg</math>
#Contact forces: Tension, Normal force and Friction.
#Force of a spring: restoring force <math>F = -kx</math>
#Tension is caused by a restoring force acting between particles of thin long objects: rope, rod, beam, string, etc.
#The direction of the tension is always parallel to the thin object.
#Normal force is caused by a restoring force acting between particles of surfaces; it is always perpendicular to the surface.
#Friction occurs when two surfaces are in contact. It is due to an interaction between surface particles of one object and surface particles of the second object.



===Exercises===

*[[Applications Of Newton's First and Second Laws]] 

*[[Applications Of Newton's Third Law]]

Preparation for Labs

2011-10-03T15:34:17Z

Nuzhat: /* Physics is an experimental science */

''Helena Dedic''

====The World of the Physicist====

'''Physicists only deal with quantities that they can measure.''' 


'''Physicists try to understand the world by observing phenomena of nature.''' '''When they observe phenomena, they confine'''
'''themselves to characteristics that are objectively measurable and do not deal with those that are subjective, such as'''
'''emotions.''' 
 
In the Olympics, a physicist would happily measure the time that it takes a speed skater to cover a 1000 m
distance, but would not participate in judging a figure skater's performance, since that judgement includes a subjective
element. 
 
For example, imagine that you are at the park and observe that there are swings with children swinging on them. You
notice that some swings go back and forth faster than others. 

Thinking about this phenomenon a physicist can choose to
describe the swings' motion quantitatively, possibly noting the time it takes to go back and forth as well as the properties of the moving object i.e. the length of the swing, the mass of the child sitting on it, how high the swing will go, the age of the child, etc. 

In doing this, the physicist has chosen to define the elements of what we call '''a system''' by describing its measurable characteristics while ignoring others, e.g. the appearance of the swing. The quantitative characteristics of a system can vary and are thus, called 
'''variables'''.
 

====Physics is an experimental science====


Physicists look for patterns. They are interested in seeing how one variable depends on another. 

The theories and principles of physics are statements about patterns in the behaviour of systems, e.g., the law of gravity comes from the observation that any object regardless of its characteristics (mass, shape) gains speed in the same way when dropped in a vacuum. 

To discover patterns in the behaviour of various physical systems physicists perform experiments. They vary the
characteristics of the system, and then observe the changes in its behaviour. 

Since any system has a number of variables, physicists systematically investigate the effect of one variable at a time. Otherwise even a simple pattern may be difficult to deduce. Here is the rule for conducting good experiments: 

Change only one variable (which is called the '''independent variable'''),
at a time and record the impact of its change on
another variable (which is called the '''dependent variable'''), in the system. 


For example, physicists want to see how the distance that a ball travels depends on its mass, diameter and amount of time
travelled. 

Distance, mass, diameter, and time are the variables under consideration. 

The distance travelled, the variable whose changes are observed, 
is called the dependent variable. 

They will have to perform three experiments in this study. 

In the first experiment, they would investigate the effect of changing the mass of the ball on the distance travelled keeping
the diameter and time period constant. 

The mass, the variable which is being systematically changed, 
is called the independent variable. 

The diameter and the amount of time travelled, the variables that are kept constant, 
are called the control variables (these variables are kept constant). 

The distance travelled by the ball is called the dependent variable. 
 
In the second experiment, they would investigate the effect of changing the diameter on the distance travelled keeping the
mass and time period constant. 

The diameter of the ball would be the independent variable while its mass and the
amount of time travelled would be control variables (these variables are kept constant). The distance travelled by the ball
is called the dependent variable. 

In the third experiment, they would investigate the effect of changing the amount of time for travel on the distance travelled
while keeping the mass and the diameter constant. 

The amount of time would be the independent variable while mass
and diameter of the ball would be control variables (these variables are kept constant). The distance travelled by the ball
is called the dependent variable. ( Answer the question 3 )
Independent variable: the variable (one only) which is being systematically changed
Dependent variable:
the variable (one only) whose changes are observed
Control variables:
the variables (any number) that are kept constant.

====The nature of measurement====

Although physicists cannot ever make an exact measurement, they are confident that anyone else who made the same
measurement would obtain the same result within a small uncertainty. W hen they make a measurement, x, they indicate
its uncertainty as ± Äx. The interval of uncertainty is the interval (x - Äx, x + Äx). For example, acceleration due to
gravity on the surface of the earth 'g' is quoted as 9.81 ± 0.01 m/s 2. The uncertainty ± 0.01 m/s2 is an estimate of how
accurate the measurement is. The symbolic statement means the following: While the most likely value of 'g' is 9.81 m/s2,
it is almost certain that the value of 'g' is within the interval of uncertainty (9.80, 9.82) m/s2.

File:Uncertainties last.png

2011-10-01T01:26:47Z

Nuzhat:

Potential Energy

2011-08-16T03:47:47Z

Nuzhat: /* THE CONCEPT OF POTENTIAL ENERGY */

==THE CONCEPT OF POTENTIAL ENERGY==

''Kreshnik Angoni''

Consider an object at rest on the floor. As long as the application point of exerted forces (<math>\vec{W}</math>, <math>\vec{N}</math> ) is at the same location there is no work production. If one push the object at height ‘h’ and leaves it free, the object will fall, i.e. it will move vertically to h = 0. The force of “weight” will produce (positive) work
because its application point (Fig.1) is shifted by <math>\vec{s}</math> 

[[image: Kreshnik_PE_1.png|left]]

<math>W_G = \vec{W}.\vec{s} = mgh</math> ....... (1)

One can bring the object at the height ‘h’ by hand or by throwing it upward with the right initial velocity. No matter what way the object goes to the height ‘h’, once there, it is able to produce mechanical work. 

One says that the object possesses a mechanical energy just because of the location. Essentially, this kind of energy is due to the gravity force which origin is the interaction earth – object. 

So, when talking about this energy, it is more precise to refer to the configuration of earth - object system
(instead of object location). 

In general terms: 

Any kind of energy which is due to a system configuration is called potential energy. 

-Another example; A block tied at the free end of an elastic spring. If we extend the spring end by “x”, the restoring force produced by the spring will be directed towards the equilibrium position. When the block is returned at equilibrium position, the restoring force has produced the work: 

[[image: Kreshnik_PE_2.png|left]]

<math>W_{el} = \frac{1}{2}kx^2</math> ........ (2)
 
If the spring is left at rest, at its equilibrium, it does not possess any capacity to produce work. But, when it is extended (or compressed) , the same spring “is able” to produce work. 

As the spring produces work this means that it does possess energy. As its ability to produce work depends only on its configuration (it has to be extended or compressed) this energy is due to configuration; So, it is a potential energy.
 

- In the two cases, initially, an external force (say hand’s force) does a positive work <math>W_ext</math> to shift the
system from an initial configuration ‘i’ to a final configuration ‘f’. Then, just because of being at “f”
configuration the system can provide work. 

This capacity to produce work depends only on system configuration. So, it can be calculated by use of a configuration function (''Function that depends only on the location (coordinates)''); let’s call it <math>U = U_{conf}\,\!</math>. If we refer to values of this function at initial-final configurations <math>U_i</math> and <math>U_f</math> and remember that work <math>W_{ext}</math> brings the system from “i” to “f ” the simplest logical relation would be 

<math>W_{ext} = U_f - U_i = \Delta U\,\!</math> ........... (3) where <math>U_f > U_i \,\!</math>

[[image: Kreshnik_PE_3.png|left]]

This relation fits perfectly with the logic that the positive external work (<math>W_{ext} > 0</math>) goes to increase the energy of the system on which this work is done (figure 3). 

Note that this definition means that all the exterior work goes only for configuration changes and not for kinetic energy change of particles. 

So, the system must be moved from<math> U_i</math> configuration to the considered one <math>U = U_f</math> by a constant speed (in practice very slowly) so that its kinetic energy remains unchanged. 

- To avoid the ambiguity related to the external force, one prefers to refer the potential energy definition to the work by internal force. 

The third law tells that during system transfer from <math>U_i</math> to <math>U_f</math> , the work by internal forces <math>W_{int} = - W_{ext}</math>. 

From relation (3), one gets to the actual definition for potential energy as: 

<math>W_{int}</math> = - ΔU =<math> -(U_f - U_i) \,\!</math> ........... (4) 

- Note that the equation (4) is based on the difference ΔU = <math>U_f - U_i</math> and not on the U- values. 

This means that only ΔU has physical meaning (not U values). 

The definition of '''U''' leaves “free choice” for the selection of configuration where '''U = 0'''. 

Actually, one fixes <math>U_i = 0</math> to a selected initial configuration which depends from the considered problem. 

Then, <math>W_{int} = -(U_f - U_i ) = - U_f</math> and we get to 

<math>U = U_f = - W_{int} \,\!</math> ....... (5) 

'''Example:''' One selects <math>U_i = U_{floor} = 0\,\!</math> when studying the displacement of an object from the floor up to a certain height, and <math>Ui = U_{earth} = 0\,\!</math> when shifting it from the earth up. In both cases, the system is the
same “'''earth-object'''”, the internal force is the weight and 

<math>W_{int} = W_G = \vec{W}.\vec{s} = - mgh</math> 

So, U = mgh 

In case of the “spring block” system, one selects<math> U_i = 0</math> for undistorted spring (x = 0), the Hook’s force
is internal and 

<math>W_{int} = W_{el} = -\frac{1}{2}kx_f^2 \equiv -\frac{1}{2}kx^2</math> and 

<math>U = - W_{int} = \frac{1}{2}kx^2</math> 

'''''REMEMBER:''''' ''In any case one has to define a '''potential energy function''' one refers to in equations (4,5)''. 

'''NOTES:''' 

- In general terms, a potential energy is due to interaction between system’s constituents. 

- So, it does not make sense to talk about potential energy of a single object or particle alone. 

- When one says that the potential energy of an object with mass ‘m’ at height ‘h ‘ is U = mgh , actually, this means
the energy of the ‘system object – earth due to their gravitational interaction.’ 

-The mechanical potential energy is due to gravitational and restoring (elastic) interactions. 

- From a general point of view, one may define a potential energy only when the involved forces at
origin of this energy are conservative. In the following we explain the meaning of a conservative force. 

 

==TIPS AND TECHNIQUES==

''Helena Dedic''
 

• The work done be conservative forces depends only on the initial and final position of the particle. This is why the potential energy is called the energy of position. 

• The work-kinetic energy theorem can be also stated as follows: 
<math>\Delta K - W_C = W_{NC}\,\! </math> 

• Definition of potential energy: The change of potential energy is defined as: 
<math>\Delta U = U_f - U_i = - W_c\,\!</math> 
 

===='''Systems of two interacting particles'''====
 
{|border="1"
!2-body systems!! potential energy!!<math>\Delta U = U_f - U_i</math>!!Choice of potential U = 0!!U

|-

|mass m and earth near surface of earth 
[[image:Helena_Pot_Energy_Table_1.png|TOP]]
|Gravitational <math>U_g</math>
|<math>mg(h_f - h_i)</math>
|at h = 0, Ug = 0||mgh

|-
|Spring-mass 
[[image:Helena_Pot_Energy_Table_2.png|TOP]]

|Spring <math>U_sp</math>
|<math>\frac{1}{2} k (x_f^2 - x_i^2)</math>
|at x = 0, Usp = 0
| <math>\frac{1}{2}k x^2</math>

|-

|two masses <math>m_1</math> and <math>m_2</math> 
[[image:Helena_Pot_Energy_Table_3.png|TOP]] 

examples: rockets and satellites
|Gravitational <math>U_g</math>
| <math>-Gm_1m_2(\frac{1}{r_f}-\frac{1}{r_i})</math>||as <math>r\rightarrow \infin, U_g\rightarrow 0</math>
|<math>\frac{-Gm_1m_2}{r}</math>

|}

 

===='''Systems of three or more interacting particles====
(Total potential energy is equal to the sum of potential energies for each interaction''' <math>U = \sum{U_i}</math>)
 

{|border="1"
|-
|spring-mass and earth 
[[image:Helena_Pot_Energy_Table_4.png|TOP]]
|<math>U = U_{sp} + U_g</math>
|-
|Gravitational interactions between three masses <math>m_1, m_2 and m_3 \,\! </math> 
[[image:Helena_Pot_Energy_Table_5.png|TOP]]
|<math>U = U_{12} + U_{23} + U_{13} = -G \frac{m_1m_2}{r_{12}} -G \frac{m_2m_3}{r_{23}} -G \frac{m_1m_3}{r_{13}} </math>
|}
 

'''Why is gravitational potential energy negative?''' 

<ul>
<li>convention: gravitational potential energy is equal to zero when the gravitational force is equal to zero. </li>
<li>When a system is held together by attractive forces the potential energy U is negative</li>
</ul>
 

'''Computing the gravitational potential energy''' 

<ul>
<li>for any problem involving rockets and satellites use formula</li> 

<math>\Delta U_g = -Gm_1m_2(\frac{1}{r_1} - \frac{1}{r_2})</math> 

<li>for any problems involving motion close to earth surface (<math>h << R_E</math>) use formula</li> 

<math>\Delta U_g= mg(h_f - h_i)</math> 

</ul>

 
'''Determine the change of potential energy - problem solving strategy''': 

a. draw a diagram of the system to be considered 

b. determine which conservative force(s) act in this system 

c. determine the initial position and the final position of the system 

d. if change in gravitaional potential energy, decide which formula to use for computation of gravitational potential energy 

e. if there is a spring in the system, determine how is the displacement of a particle related to the extension of the spring 

f. compute the change of the total potential energy of the system <math>\Delta U = \Delta Ug + \Delta Usp</math>, where <math>\Delta Ug = Ugf - Ugi</math> and <math>\Delta Usp</math> = ½ <math>k (x_f^2 - x_i^2)</math>. 

==EXERCISES==

*[[Exercises on Potential Energy]] 

==WORK AND ENERGY==

* [[Work and Energy]]

File:Kreshnik Measuremment Fig 7.png

2011-08-15T23:21:38Z

Nuzhat:

File:Kreshnik Measuremment Fig 6.png

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File:Kreshnik Measuremment Fig 3.png

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File:Kreshnik Measuremment Fig 2.png

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File:Kreshnik Measuremment Fig 1b.png

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File:Kreshnik Measuremment Table 2.png

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File:Kreshnik Measuremment Fig 1.png

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File:Kreshnik Measuremment Table 1.png

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Thermal and Other Types of Internal Energy

2011-08-11T03:28:41Z

Nuzhat: /* GENERAL PRINCIPLE OF ENERGY CONSERVATION */

''Kreshnik Angoni''

===FRICTION AND THERMAL ENERGY===

- Till here we neglect that objects (system components) are constituted by atoms and molecules which
“interact and move inside the object” and this means some “energy inside the object”. 

Molecules and atoms are strongly linked inside a solid, slightly linked inside a liquid and free in a gaz.
But, no matter what is the physical state of object, they are in a continuous random motion in space. 

This motion means some average speed and this means some average kinetic energy. When this energy is
considerable the object has high temperature. The temperature of the object is a macroscopic parameter
related to the microscopic average kinetic energy of particles inside it. In energy terms, we say that, there
is a thermal energy <math>E_{th}</math> inside any object. 

-The experiments have proven that the magnitude of work by friction goes completely to increase the
thermal energy of two objects that rub on each other, i.e. 

<math>\left | W_f\right| = f * d = E_{th2} - E_{th1} = \Delta E_{th}</math> ......(1) 

If the friction is one of external forces acting on the system, one may put aside the (negative) work by
friction in the expression of energy conservation principle and get 

<math>W_{ext-net} = W_{ext} + W_f = \Delta E_{mech}\,\!</math> .......(2) 

Then, 

<math>W_{ext} = \Delta E_{mech} - W_f = \Delta E_{mech} + |W_f|\,\!</math> and finally 

<math>W_{ext} = \Delta E_{mech} + \Delta E_{th}\,\!</math> ....... (3) 

-Note that <math>W_{ext}</math> in this expression does not include the work by friction. 

The expression (3) tells that the external work can go for the change of pure mechanic energy and thermal energy of a system. 

'''Example:''' A block with mass m = 10kg moving with speed <math>v_1</math> = 5m/s on a horizontal frictionless plane
stops after distance “d” when moving on a surface with friction where f = 5N. 

a) Find the distance “d”. 

1) Apply the work-energy theorem ; <math>W_{net} = W_f = K_2 - K_1\,\!</math>; i.e.-5*d = 0 – 10*<math>5^2\,\!</math>/2 and d = 125/5 = 25m
2) Apply the principle of mechanical energy conservation (10) for earth - block non-isolated system

(state_1)<math> E_1 = K_1 + U_1\,\!</math> = (10*<math>5^2\,\!</math>)/2 + <math>mgh_1\,\!</math> ; 

(state_2) <math>E_2 = K_2 + U_2 = 0 + mgh_2 = mgh_1\,\!</math>; 

“Friction external force” . So, <math>E_{2-mech} - E_{1-mech} = W_{ext-net} = W_f\,\!</math> = -5*d 

[(0-125)+( <math>mgh_2 - mgh_1\,\!</math>)] = - 5*d and d=125/5=25m 

b) From the experience, we know that the temperature of block and the plan increases. Another way of
asking questions on this example is: 

b-1) Find the increase of thermal energy of block - plan system then 

b-2) Find “d” by using the change in thermal energy when the block is stopped. 
 
If we see the words temperature or thermal we must figure out that we have to exclude the friction from external forces and refer to the principle energy form containing the thermal energy. For this system: 

- The friction work is not part of external work and <math>W_{ext} = 0\,\!</math> (Isolated system) 

-The principle of mechanic energy conservation tells that 

<math>\Delta E_{mech} + \Delta E_{th} = W_{ext} = 0 [(0 + mgh) - (125 + mgh ) +( E_{th2} - E_{th1})] = 0\,\!</math> 

So, 

b-1)<math> E_{th2} - E_{th1} = 125 J\,\!</math> and 

b-2) <math>\Delta E_{th}</math> = f*d i.e. 125 = 5*d and d = 25m 

'''Remember:''' If the heat or thermal energy is not mentioned one refers to a system where friction is external force and uses the principle of mechanic energy conservation in its simple form. 

<math>W_{ext-net} = E_{mech2} - E_{mech1} = [K_2 +U_2] - [K_1+ U_1]\,\!</math> .......(4) 

But, if the thermal energy is required one must refer to the total mechanic + thermal energy of system 

<math>E_{mech} = U + K + E_{th}\,\!</math> and apply the principle of energy conservation in the form 

<math>W_{ext} = \Delta E_{mech} + \Delta E_{th} = [(K_2 + U_2 ) - ( K_1+ U_1)] +[ E_{th2} - E_{th1}]\,\!</math>......(5) 

'''Important Note:''' The experiments show that K and U can transform completely into each other or into
<math>E_th\,\!</math> but <math>E_th\,\!</math> cannot transform completely into K or U. 

A disorganized motion cannot be converted naturally (by itself) into an oriented motion. 

'''Ex:''' A hot block cannot start moving along one direction just because its temperature decreases. That’s why the thermal energy is not considered as a pure mechanical energy. In general, one uses '''“calorie”''' a particular unit to measure heat. ('''1 cal = 4.18 J''') 

===OTHER TYPES OF INTERNAL ENERGY===

-In exothermic chemical reactions, the molecules of a sample_ A interact with molecules of a sample_ B at room temperature and produce molecules of the sample A-B with higher temperature (may be even an explosion i.e. high kinetic energy of particles). 

As the system of two samples (A, B) is isolated, the principle of energy conservation would tell that 

<math>\Delta E_{mech} + \Delta E_{th} = 0\,\!</math> but the experiment shows that <math>\Delta E_{mech} + \Delta E_{th}\,\!</math> > 0. 

This contradiction can be solved easily if we consider that there is a molecular energy <math>E_{mol}</math> inside the
molecules of two objects which remains unchanged as long as there is no chemical reaction. 

So, the total energy of two samples is <math>E_{tot} = E_{mech} + E_{th} + E_{mol}\,\!</math>. 

Then, the principle of energy conservation for an isolated system would be written 

<math>\Delta E_{mech} + \Delta E_{th} + \Delta E_{mol} = 0\,\!</math> and <math>\Delta E_{mech} + \Delta E_{th} = - \Delta E_{mol}\,\!</math> > 0 because <math>\Delta E_{mol} = E_{mol-2} - E_{mol-1}\,\!</math> < 0 

which explains that the molecular (or chemical energy) is decreased and transformed into mechanical and
thermal energy of the system. 

-In a battery the molecular energy is transformed (via chemical reaction) into electrical energy<math> E_{el}\,\!</math> which is another type of internal energy. 

There is a set of other internal energies( <math>E_{struct}\,\!</math> in a solid or fluid structure, <math>E_{at}\,\!</math> atomic energy inside the atoms, <math>E_{nucl}\,\!</math> nuclear energy inside the nuclei and <math>E_{rad}\,\!</math> radiation energy) which cannot transform completely and naturally into pure mechanic energy (K or U). 

So, they are all non-mechanical forms of energy. If we group all them into a single term E_{int} , we get 

<math>E_{int} = E_{el} + E_{struct} + E_{mol} + E_{at} + E_{nucl} + E_{rad}\,\!</math> ...... (6) 

and the total energy of an object can be expressed as: 

<math>E_{tot} = E_{mech} + E_{th} + E_{int}\,\!</math> ...... (7) 

===GENERAL PRINCIPLE OF ENERGY CONSERVATION===

The results of many experiments show that: 

- If we include inside a system all the objects that can affect each other, we create an isolated system.
The total energy of an isolated system remains constant in time i.e. 

<math>\Delta E_{tot} = \Delta E_{mech} + \Delta E_{th} + \Delta E_{int} = 0\,\!</math> .......(8) 

'''Note:''' This does not exclude motion and energetic exchanges inside the system; but if they happen, the
energy is transformed from one form to another in such a way that their sum remains unchanged. 

- If the system is not isolated the amount of energy it exchanges with adjacent space regions E_{exchange},
is equal to the change of its total energy 

<math>\Delta E_{tot} = E_{tot-2} - E_{tot-1} = E_{exchange}\,\!</math> ......(9) 

If the system exchanges only work, then<math> E_{exchange} = W_{ext}\,\!</math> and <math> W_{ext} = \Delta E_{mech} + \Delta E_{th} + \Delta E_{int}\,\!</math>......(10) 

'''Remember:'''

*The positive work done by forces outside system ( <math>W_{ext}\,\!</math> > 0) increases the total energy of system. 
* The negative work done by forces outside system ( <math>W_{ext}\,\!</math> < 0) decreases the total energy of system. 
* The work by friction is not included into <math>W_{ext}\,\!</math> at expression (10). 
*The total energy conservation principle (10) transforms into (3) when <math>\Delta E_{int} = 0\,\!</math>. 

-A “source” can transfer different types of energy (work, heat, irradiation, electric..), to the adjacent
regions of space. So, the general expression for the delivered power [watt] from a “source” must take into
account all the possible energy transfers in “ a second” 

<math>P = - \frac{dE}{dt}\,\!</math> ......(11) 

where E [joule] stands for the total energy(all type included) of the “source” that delivers energy. 

The work done by the source is a portion of energy transferred from the “source” to the adjacent regions. So, the
delivered mechanic power is only one of the components of expression (11). 

If only the mechanical energy of “source” changes during the time it produces work then the delivered mechanical power [watt] is due to only mechanical changes of energy. 

'''Note:''' The sign “ - “ in expression (11) is related to the fact that the energy of the source is decreased when
providing power while this power is positive in the sense that the force originated from the system does
positive work.

Linear Momentum and Collisions

2011-08-10T21:00:20Z

Nuzhat: /* ELASTIC COLLISION IN ONE DIMENSION */

''Kreshnik Angoni''

===LINEAR MOMENTUM===

* The observations on different situation of collision between two objects showed that one may model and explain the experimental results by use of a particular physical concept; the linear momentum 

<math>\vec{p} = m\vec{v}</math> .....(1) 
[[image: Kreshnik_Mom_Fig0.png|left]] 

:m is mass and <math>\vec{v}</math> is the velocity of considered object. 

* The linear momentum offers a very useful and more general way, to state the Newton’s second law: 

<math>\vec{F}_{NET} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt}</math> .... (2) 

'''Newton's 2nd Law:''' “The net force acting on a particle is equal to the rate of change of its linear momentum” 

If the mass of object remains constant during the observation (common situation) the eq.(2) transforms to
the classic form 

<math>\vec{F}_{NET} = m\frac{d(\vec{v})}{dt} = m\vec{a} </math>......(3) 

But the equation (3) '''''cannot''''' describe the motion '''''if the mass of object changes''''' during the observation
(reactive motion of rockets). Meanwhile the equation (2) does include those situations, too. 

===CONSERVATION OF LINEAR MOMENTUM===

* In one of the first relevant experiments about the linear momentum, it was found that when two isolated bodies ( net external force over them is zero) collide and stick together (figure 1), the total linear momentum before and after collision is the same, i.e 

[[image: Kreshnik_Mom_Fig1.png|left]] 

'''(before collision)''' <math>\vec{p_1} + \vec{p_2} = \vec{p}_{1-2} \,\!</math> '''(after collision)''' ...... (4) 
 
 
Further experiments proved that this is a particular application of a general mechanics principle: '''the conservation of linear momentum'''. 

[[image: Kreshnik_Mom_Fig2.png|left]] 

* One considers the collision between two particles with masses <math>m_1\,\!</math> and <math>m_2\,\!</math> with velocities
<math>\vec{u_1}</math>, <math>\vec{u_2}</math> before the collision and <math>\vec{v_1}</math>, <math>\vec{v_2}</math>
after the collision (Fig.2). 

Note that the collision (type of interaction) happens during a real touch of particles (like billiard balls) or during a no-touch repeal of particles (like two electrical charges of the same sign). 

In general, there is a change of each particle linear momentum after a collision type interaction. 

In order to calculate these changes, one uses the '''''Conservation of Linear Momentum Principle''''', which states that: 

'''The total linear momentum of a system of particles remains constant''' 
'''if the external force <math>\vec{F}_{Sys-EXT}</math> exerted on the system is zero'''. 
<math>\Delta \vec{p} = \vec{p}_{fin} - \vec{p}_{in} = 0</math>, or <math>\vec{p}_{fin} = \vec{p}_{in}</math> .....(5) 

The application of this law in the case of two particles’ collision gives: 

<math>m_1\vec{u_1} + m_2\vec{u_2} = m_1\vec{v_1} + m_2\vec{v_2}</math> ........ (6) 

The vector equation (6) includes three conservation conditions, one for each velocity component 

<math>m_1u_{1x} + m_2u_{2x} = m_1v_{1x} + m_2v_{2x}\,\!</math> 

<math>m_1u_{1y} + m_2u_{2y} = m_1v_{1y} + m_2v_{2y}\,\!</math> ......(7)
 
<math>m_1u_{1z} + m_2u_{2z} = m_1v_{1z} + m_2v_{2z}\,\!</math> 

'''''The equations (7) show that when the linear momentum is conserved, each of its components is independently conserved.'''''
 

* When does the principle of linear momentum conservation apply? One must remember that, in order to apply the principle of linear momentum conservation, the net external force acting on the system must be zero. To explain “why” we will consider the system of two particles with masses <math>m_1</math> and <math>m_2</math>. 

[[image: Kreshnik_Mom_Fig3.png|left]] 

:The internal forces acting on this system are <math>\vec{F}_{12}</math> exerted on particle (1) from the particle (2) and <math>\vec{F}_{12}</math> exerted on particle (2) from particle(1). The third law of mechanics states that: 

<math>\vec{F}_{21} = - \vec{F}_{12}</math> .....(8) 

Applying the equation (2) for the motion of particle (1) we get 

<math>\vec{F}_{1-NET} = \vec{F}_{1-EXT} + \vec{F}_{1-INT} = \vec{F}_{1-EXT} + \vec{F}_{12} = \frac{d\vec{p_1}}{dt}</math> .....(9) 

and similarly for the particle (2) 

<math>\vec{F}_{2-NET} = \vec{F}_{2-EXT} + \vec{F}_{2-INT} = \vec{F}_{2-EXT} + \vec{F}_{21} = \frac{d\vec{p_2}}{dt}</math> .....(9) 
Taking the sum of equations (9) and (10) we have 

<math>\vec{F}_{SYST-NET} = \vec{F}_{1-NET} + \vec{F}_{2-NET} = \vec{F}_{1-EXT} + \vec{F}_{12} + \vec{F}_{2-EXT} + \vec{F}_{21} = \vec{F}_{1-EXT} +\vec{F}_{2-EXT} = \vec{F}_{SYST-EXT} = \frac{d\vec{p_1}}{dt} + \frac{d\vec{p_2}}{dt}</math> 

which gives: 

<math>\vec{F}_{SYST-EXT} = \frac{d\vec{p_1} + d\vec{p_2}}{dt} = \frac{d\vec{p}_{SYST}}{dt}</math> .... (11) 

where <math>\vec{F}_{SYST-EXT}</math> is the sum of external forces acting on each of particles of system and <math>\vec{p}_{SYST} = \sum_{i}\vec{p_i}</math> is the sum of linear momentums for all system particles. 

Now, if the net external force acting over the system is zero, from eq(11) we get: 

<math>\frac{d\vec{p}_{SYST}}{dt} = 0 \Rightarrow \vec{p}_{SYST} = \sum_{i}\vec{p_i}</math> = constant ..... (12) 

In the case of two particles, the equation (11) becomes: 
<math>\vec{p_1} + \vec{p_2} = constant</math> 
 

'''The eq(12) holds on as long as there is no external action over the system.''' 

* For cases when the <math>\vec{F}_{SYST-EXT}</math> ≠ 0, we deal with a vector along a direction in space. 

:This vector has no components (i.e. = 0 ) over a plane perpendicular to its direction. 

:So, one may select the Ox, Oy axes on this plane and apply the two first equations of system (6) because they hold on independently. 

It comes out that, even when <math>\vec{F}_{SYST-EXT}</math> ≠ 0,
the components of linear momentum of system along the directions perpendicular to <math>\vec{F}_{SYST-EXT}</math>
are conserved. So,this principle has a wide range of applications.
 

===TYPES OF COLLISIONS===

'''-'''The principle of linear momentum conservation is very general. It applies in all fields of physics,
mechanical collisions, explosions, reactive motion, light emission/absorption, nuclear decay and
nuclear reactions. So, it is important to clarify some basic issues related to term “collisions”. 

'''-''' In physics, one uses the term '''collision''' when talking about '''“ a brief and strong interaction between
two or more bodies”'''. 

What does one considers as a brief time of interaction? 

This parameter ( Δt ) depends on the considered phenomena; for mechanical collisions (like balls, cars, people, objects) a
collision lasts (Δt ) from 0.001s to 1 s. A collision between elementary particles lasts (Δt ) for about
10-23s. In the case of galaxies a collision lasts (Δt =) about several millions of years. 

'''-''' There are two types of collisions: '''''elastic''''' and '''''inelastic'''''. 

In both cases the total linear momentum of the system is conserved if no external forces are present. 

The distinction concerns the total '''KINETIC''' energy of the system. 

In''' ELASTIC''' collisions, '''the total KINETIC energy of the system is conserved too.''' 

So, there are two equations that apply for elastic collisions. 

In the case of two particles’ elastic collision one gets: 

<math>m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\,\!</math> 

<math>\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_1u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_1v_2^2</math> ....(13) 

Note that during an elastic collision, the kinetic energy of the system is transformed (partially or
completely) into potential elastic energy but after the collision (after Δt) it is completely recovered into
kinetic energy of the system. The elastic collisions are common phenomena in atomic or nuclear
physics. In everyday life there are no really elastic collisions; even the collision between hard steel
balls is close but not exactly elastic. 

- In an '''inelastic collision''' '''the total kinetic energy of the system is not conserved'''. In a completely
inelastic collision, the bodies stick together after collision. 

In general, during an inelastic collision a part of kinetic energy of the system is transformed into potential elastic energy and the other part into thermal, internal energy, sound and even light. 

It is important to underline that this second part does not recover into kinetic energy of system particles after the collision. 

===ELASTIC COLLISION IN ONE DIMENSION===

-The figure 4 presents the collision of two particles with masses <math>m_1\,\!</math>, <math>m_2\,\!</math> . They have initial velocities <math>\vec{u}_1</math> ,<math>\vec{u}_2</math> along the same space direction that we select as '''''Ox''''' axis. 

We consider “a central collision” that leaves the motion of particles along the same space direction, i.e. their velocities after collision <math>\vec{v}_1</math>, <math>\vec{v}_2</math> are along '''''Ox''''' axis too. 

[[image: Kreshnik_Mom_Fig4.png|left]] 

So, <math>\vec{u}_1 = u_1 \hat{i}</math>, <math>\vec{u}_2 = u_2 \hat{i}</math>, <math>\vec{v}_1 = v_1 \hat{i}</math>, <math>\vec{v}_2 = v_2 \hat{i}</math> ..... (14) 

To simplify the picture, all velocities are shown directed at one sense but for the collision to happen one must have <math>u_1</math> > <math>u_2</math> > or <math>u_1</math> > 0 and <math>u_2</math> < 0 which is something that appears at algebraic values of components. 

As this is an elastic collision, the two conservation laws apply. By projecting the equations (13) on '''''Ox''''' axis, one gets: 

<math>m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \rightarrow\rightarrow m_1(u_1 - v_1) = - m_2(u_2 - v_2) \,\!</math> ...... (15) 

<math>m_1u_1^2 + m_2u_2^2 = m_1v_1^2 + m_2v_2^2 \rightarrow\rightarrow m_1(u_1^2 - v_1^2) = - m_2(u_2^2 - v_2^2) \,\!</math> ...... (16) 
We rewrite eq. (16) in form 

<math>m_1(u_1 - v_1)(u_1 + v_1) = - m_2(u_2 - v_2)(u_2 + v_2)\,\!</math> ..... (17) 

Then we divide eq.(17) by eq.15 and get: 

<math>(u_1 + v_1) = (u_2 + v_2) \Rightarrow )(v_2 - v_1) = -(u_2 - u_1)</math> .......(18) 

The relation (18) shows that in 1-D elastic collisions, the amplitude of relative velocity of second particles versus first particle remains the same but its direction is reversed. 

* '''Ex_1.''' Find <math> v_1\,\!</math> ,<math>v_2\,\!</math> for the central collision of two particles with equal masses( i.e. <math>m_1 = m_2\,\!</math> ≡ m ). 
In this case eq.(5) becomes: 

<math>\vec{u}_1 + \vec{u}_2 = \vec{v}_1 + \vec{v}_2</math> 

On '''''Ox''''' <math>\Rightarrow (u_1 + u_2) = (v_1 + v_2)\,\!</math> ....... (19) 

From two eq.(18-19) one can find easily that 

<math>u_1 = v_2\,\!</math> and <math>u_2 = v_1\,\!</math> ..... (20) 

If before the collision the target (particle 2) is at rest, <math>u_2\,\!</math> = 0; <math>u_1\,\!</math> ≠ 0, 

it follows that, after collision 

<math>v_1 = u_2 = 0\,\!</math> 

i.e. the first particle stops moving and 

<math>v_2 = u_1\,\!</math> 

i.e. the second particle moves with the velocity of the first particle before collision. 

One sees this situation often in billiard games.
 

* '''Ex_2.''' Find <math>v_1\,\!</math>, <math>v_2\,\!</math> for the central collision of two particles with unequal masses (<math>m_1\,\!</math> ≠<math> m_2\,\!</math>) when the target is at rest(<math>u_2 = 0\,\!</math>; <math>u_1\,\!</math> ≠ 0). 

In this case eq.(5) gives on '''''Ox''''' axis 

<math>m_1u_1 = m_1v_1 + m_2u_2\,\!</math> ........ (21) 

while the equation (18) transforms to 

<math>u_1 = v_2 - v_1\,\!</math> ......(22) 

The solution of system (21-22) gives 

<math>v_1 = \frac{(m_1 - m_2)u_1}{(m_1 + m_2)}\,\! </math> and <math>v_2 = \frac{2m_1u_1}{(m_1 + m_2)}\,\! </math> ......(23) 

The expressions (23) simplify essentially in the following situations: 

• When <math>m_1\,\!</math> >> <math>m_2\,\!</math> , it comes out that <math>v_1\,\!</math> ≈ <math>u_1\,\!</math> and <math>v_2\,\!</math> ≈ <math>2u_1\,\!</math> i.e. 

the first particle maintains its initial velocity but the second one imparts with the double of u1. 

This is a common situation when a golf club (<math>m_1\,\!</math>) gives a central hit to a golf ball (<math>m_2\,\!</math>). 

• When <math>m_1\,\!</math> << <math>m_2\,\!</math>, it comes out that <math>v_1\,\!</math> ≈ - <math>u_1\,\!</math> and <math>v_2\,\!</math> ≈ 0 

i.e. the first particle reverses the sense but keeps the same magnitude of velocity. 

This would correspond to the situation when a ping-pong ball hits centrally a bowling ball. 

===IMPULSE===

- Often we have to consider “impulsive actions” i.e. actions that last for a very short interval of time. A
specific physical quantity, the impulse vector is used to study these situations. 

If the linear momentum of a particle (body) changes from <math>\vec{p}_i</math> to <math>\vec{p}_f</math>,
, one says that the '''impulse''' <math>\vec{I}</math> is exerted on the particle: 

<math>\vec{I} = \vec{p}_f - \vec{p}_i = \Delta p</math> ...... (24) 

Note that impulse is a vector and its unit is [kg*m/s]. 

Its magnitude and direction are defined by the '''''difference of vectors''''' <math>\vec{p}_f</math> and <math>\vec{p}_i</math>. 

-As impulse and the force exerted on the particle are expressed through the changes of particle linear momentum, there is a relation between them, too. From the “modern expression” of the second law 

<math>\vec{F} = \frac{d\vec{p}}{dt}</math> we get: 

<math>d\vec{p} = \vec{F}dt</math> .......(25) 

which is a differential expression. 

The finite change of linear momentum Δ<math>\vec{p}</math> during time interval [''ti, tf''] is calculated as the sum of
elementary vectors <math>d\vec{p}</math> and for this type of calculations one uses the integral procedures. 

<math>\vec{I} = \Delta \vec{p} = \int d\vec{p} = \int_{t_i}^{t_f}\vec{F}dt</math> ..... (26) 

[[image: Kreshnik_Mom_Fig5.png|left]] 

If we consider 1-D problems, <math>\vec{F}</math>,<math> \vec{p}_{fin}</math>, <math>\vec{p}_{in}</math> have the same direction and the expression (26) transforms to: 

<math>I = \Delta p = \int dp = \int_{t_i}^{t_f}Fdt</math> ..... (27) 

-Without going into details, the integral (27) is equal to the surface under the graph F = F(t) shown in figure 5. 
The relation (27) is valid for all possible form of force evolution in time but it is mainly used in case of impulsive
forces. 
'''An impulsive force acts during a short interval of time''' 
'''during which it increases and decreases abruptly.''' 

'''''During this short interval, the action of impulsive force is much bigger than other forces exerted on the particle. So,
one neglects the effect of other forces when an impulsive force is in action'''''. 

In these circumstances, although F in (27) stands for the net force exerted over the particle, only the
impulsive force is counted under integral at formula (26). '''This is known as the impulsive approximation.''' 

'''Example:''' The impulsive force exerted on a ball during the short time interval a player 
kicks the soccer ball is much bigger than its weight and it is this force , 
that decides about the change of linear momentum of the ball. 

Only this force is included under the integral (26). 

-The problem with impulsive forces is that, in general, we do not know the analytic function F = F(t)
that describes its evolution in time. That’s why, instead of the real force, one operates with an average
(constant) force <math>F_{Av}</math> acting during the time interval '''Δt''' (fig.5) such that its impulse be I . 

We get then 

<math>I = \Delta p = F_{AV} \Delta t\,\!</math> .......(28) 

Note that the equity of expressions (27, 28) graphically means that the area under the graph F = F(t) is
equal to the area of rectangle with base '''Δt''' and height <math>F_{Av}</math>. 

The graphical visualisation of the impulse serves as a good reference to understand that a given change of particle linear momentum needs the application of a given impulse and this can be achieved in two different ways (see figure 6) ;
by use of a large force applied briefly or by use of a weaker force applied for a long interval of time.

[[image: Kreshnik_Mom_Fig6.png|left]]

''This kind of information is important to keep in consideration in situations when'': 

• The exertion of a very large force for a short interval of time can create problems for the stability of the system. 

• There is a limited flexibility about the production of large forces. 

• A little bit longer Δt interaction time does not affect essentially the expected effect.

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Rotational Motion: Newton's Second Law

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''Kreshnik Angoni''

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Conservative and Non-Conservative Forces

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Nuzhat: /* Non-conservative forces */

''Kreshnik Angoni''

===CONSERVATIVE FORCES, POTENTIAL FUNCTION, SYSTEM : FORMAL DEFINITIONS===
 
Some forces posses a specific quality: '''''their work equals zero''''' ''if the object on which they are exerted
moves along a '''closed path'''''. 

These are known as '''conservative forces'''.

 

Note that a conservative force does a work different from zero when the object on which it acts is shifted from point “1” to point “2” (fig.1) but: 

<math> W_{1-2} = - W_{2-1}\,\!</math> so that <math>W_{1-1} = W_{1-2} +W_{2-1} = 0\,\!</math>. 

Three main examples of such forces are the '''gravitational''', '''restoring (elastic)''' and '''static electric forces'''.
[[image: Kreshnik_CONS_NON CONS_Fig1.png|right]]
 

'''''Not all the forces are conservative'''''.

The '''net work by friction forces''' is '''negative''' (≠0) when object is shifted along a closed path. 
The normal force is always perpendicular to displacement and its work is always zero, no matter what is the path. 

Forces that depend on the velocity are not conservative
(the drag force in fluids, the magnetic force,..). 

In the case of a conservative force, it can be concluded that:

'''If the object path is not closed, the work done by the force depends only on initial and final
locations of object and not on the shape of the path to get from initial to final location'''. 

 

One defines a potential U(x,y,z) or potential energy which is a function of space coordinates.
This function is very useful for solving Physics problems. 

Remember that a force is produced by its “source”. 

If the force is conservative, one can use efficiently the concept of a "'''system'''" consisting of the “object that receives the force action AND the object (source) of this force”
and then define the potential energy of the system. 

'''Related Notes:''' 

a) Any two or more objects experience their gravitational attraction forces. As gravitational forces are conservative,
one may always define a system and a gravitational potential energy of this system. 

If other conservative forces(ex. restoring forces) act between system’s components, one must add the corresponding terms to potential energy of the system. 

(''When the secondary forces are much smaller, at a first approximation, one may neglect them and
consider only the main term of the potential''). 

b) The definition of a system cannot be tied to non-conservative forces because for them it is not possible to
define a potential function. So, in mechanics, '''''a non conservative force''' is an '''external force'''''. 

===INTERNAL FORCES, TOTAL MECHANIAL ENERGY OF AN ISOLATED SYSTEM===

-If one defines a system in upper terms, the source of the considered conservative force is part of the
system and one deal with an internal force of this system. (If the “system” contains several “sources” and
“objects”, the potential function will depend on a set of coordinates and there will be several internal forces.)
If the system contains one “source” and one “object”, we fix the frame origin on the “source” and select
Ox axe along the direction “source - object”. In this case, the potential function depends on xcoordinate
and the force acting on the object, which is an internal force of the system, can be derived
from the potential function U as: 

<math>f = - \frac{dU}{dx}\,\!</math> ............. (6)

The U- function (potential energy) is a system characteristic and it is often denoted Ep. 

'''Exemple:''' An aeroplane at high h from earth surface. The gravitational force acting on the plane (weight)
is a conservative force. The source of the weight is the earth. We define the system earth-aeroplane. We
select Ox-axe along vertical direction and origin O is at earth center. 

The potential energy of this system (we speak about aeroplane potential energy) is: 

<math>U = E_p = m_a*g*x\,\!</math> 

Then, the plane weight ( internal force of system) can be calculated by use of formula: 

<math>f_x = - \frac{dU}{dx} = - \frac{d(m_a*g*x)}{dx} = - m_a*g</math> 

The “–“ shows the weight direction opposite to selected Ox positive direction. 

-If a single conservative force is applied on the object of study, one may define a potential U, a system,
tie a reference frame Ox to the other component of the system (source of force), calculate the kinetic energy
of the object K (with respect to this frame), apply the work-energy theorem and find that the “sum of kinetic
energy K plus potential energy U of the object remains constant in time” as follows: 

<math>W_{net} = K_{fin} - K{in} = \Delta K \,\!</math> 

For a small shift “dx” it will be produced a small work: 

<math>dW_{net} = dK\,\!</math> ........ (7) 

As <math>dW_{net} = dW_{int} + dW_{ext} = dW_{int}\,\!</math> (because <math>dW_{ext} = 0\,\!</math>) and <math>dW_{int} = f_{int}*dx\,\!</math> we get: 

<math>dW_{net} = dW_{int} = f_{int}*dx = - \frac{dU}{dx}*dx = - dU = dK</math> 

<math>\Rightarrow dK + dU = 0</math> 

<math>\Rightarrow K + U = CONST</math> ........ (8) 

The potential energy concerns the object and the “source” of force. 

As the kinetic energy of the “source” is zero (reference frame is tied to it), it comes out that the kinetic energy of the system is equal to that of the object under study. 

So, actually the sum K + U presents the total mechanical energy of the system <math>E_{mech}</math> and the expression (8) tells that 

<math>E_{mech} = K + U = CONST\,\!</math> ........ (9) 

Remember that we assumed no action from outside, i.e. an isolated system. So, the expression (9) shows
that the total mechanical energy of an isolated system is conserved. 

===CONSERVATION OF MECANICAL ENERGY FOR NON ISOLATED SYSTEMS===

How to deal with situations where any kind of forces (conservative and non-conservative) act on the same
object while it is shifted from location “1” to location “2”? 

'''1.''' One identify the conservative forces acting on the object under study. 

'''2.''' One defines a common U potential for these forces and a common system. 

'''3.''' One divides the set of all acting forces on the object into internal (that do contribute to U) and
external (do not contribute to system potential function). 

'''4.''' One notes that if there are no external forces the total mechanical energy of the system. 
remains constant; <math>E_{mech-fin} = E_{mech-in}\,\!</math>. 

So, if the external forces are doing a positive work <math>W_{ext-net}\,\!</math> on the system (actually on “the object”
which is part of the system) the total energy of the system will be increased by this amount and 

<math>E_{mech-fin} = E_{mech-in} + W_{ext-net}\,\!</math> or 

<math>\Delta E_{mech} = E_{2-mech} - E_{1-mech} = W_{ext-net}\,\!</math> ......... (10) 

'''Notes:''' - The relation (10) is valid even for the case of a negative work by external forces. In this case
it shows a decrease of total mechanical energy of the system. 

- Expression (10) presents the general form of the principle of mechanical energy conservation.
It is valid for any kind of system (isolated <math>W_{ext-net} = 0\,\!</math> or not isolated <math>W_{ext-net}\,\!</math> ≠ 0). 

'''REMEMBER:''' The kinetic, potential and total mechanical energies of a system are mathematical
functions defined at a reference frame. 

‐ While the definition of kinetic energy does not need the concept of system, potential and total
energy functions cannot be defined without going through system concept. 

The value of any of functions (K, U, E) by itself does not have a precise physical meaning for the system because they
depend on the reference frame choice, too.

 

A simple change of the reference frame changes the value of those functions. But, the change of those functions has a very precise physical meaning. It is related to the achieved work and this quantity does not depend on the selected
frame for calculations. 

This is the basic issue one must not forget when dealing with energy
related problems. The zero value of energy function can be selected following the convenience of
problem requirements but the difference between its values must equal the net external work
done on the system. 

===COMPARISON OF CONSERVATIVE AND NON-CONSERVATIVE FORCES===
''Helena Dedic''
 

'''Comparison of the work done on a closed path (a closed path begins and ends at the same point) by the gravitational force and by the force of air resistance: a vertical flight of a ball with and without air resistance''' 

[[image: Helena_Cons_Non Cons_Fig 1.png|TOP]] 

'''Comparison of path dependence of work done by gravity and by friction for block sliding on incline''' 

'''path 1: from A directly to B, path 2: from A to B via C''' 

[[image: Helena_Cons_Non Cons_Fig 2.png|TOP]] 

• Conclusion: The work done by gravity is indpendent of the path while the work done be friction depends on the path. 

====Conservative forces====

o <math>W_c\,\!</math> , the work done by a conservative force, around any closed path is zero 
o <math>W_C\,\!</math> , the work done by a conservative force is independent of the path taken 
o Examples: gravitational force , spring force 

====Non-conservative forces====

o <math>W_{NC}\,\!</math> , the work done by a non-conservative force around any closed path is different than zero 

o <math>W_{NC}\,\!</math> , the work done by a non-conservative force depends on the path taken 

o Examples : resistive forces, friction, tension, applied forces 

• Given the distinction between these two types of forces we will rewrite the statement of the work-kinetic energy theorem as follows: 

<math>\Delta K = W_C + W_{NC}\,\!</math> 

where <math>W_C\,\!</math> is the sum of the work done by conservative forces and<math> W_{NC}\,\!</math> is the sum of the work done by non-conservative forces. 

===EXERCISES===

*[[Exercises: Conservative and Non-Conservative Forces]]

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