ACCELERATIONS

There exist three types of accelerations:

- linear acceleration
- angular acceleration
- centripetal acceleration

Linear acceleration:

Symbol: a

Units: m/s²

Linear acceleration represents how quickly the speed of an object is changing. In physics jargon, we say it is the time derivative of the linear velocity.
That is the type of acceleration that you get when you slam on the accelerator or on the brake pedals in your car.

Angular acceleration:

Symbol: α , pronounced "alpha".

Units: rad/s²

Angular acceleration represents how quickly the angular velocity of an object is changing. In physics jargon, we say it is the time derivative of the angular velocity.
That is the type of acceleration that you get if you spin an object faster and faster, or slower and slower.
Conceptually it is more similar to linear acceleration than centripetal acceleration, because in both cases you it is about the change in how quickly an object moves: linear acceleration is about the change in how quickly an object moves along its path, angular acceleration about the change in how quickly it spins (about an axis inside the object) or rotates about some other point or axis, outside the object.

The relation between the angular acceleration and the linear acceleration is the same as the relation between the angular velocity and the linear velocity, and the same as between an angle and the corresponding arc length. As the diagram illustrate, for the same angle, the larger the radius is, the large the arc length is:

As you increase R, s increase, and it does so in the same proportion. So the relation between angle and arc length is:

s = R . θ

Linear velocity and angular velocity have the same relation (which one gets by just differentiating the equation above with respect to time):

v = R . ω

And so do linear and angular accelerations:

a = R . α

Centripetal acceleration:

Symbol: a_c

Units: m/s²

This is the one which is equal to v²/r.

Centripetal acceleration is the acceleration that represents a change in the direction in which an object is moving. It is the acceleration needed to deflect an object from moving in a straight line path.
How much an object is moving is represented by its velocity, and that has both magnitude (the speed) and direction. The change in speed is accounted for by the linear acceleration. The change in direction by the centripetal one.

CONSERVATION OF ENERGY MADE EASY

Wouldn't it be nice if there was a way to make using conservation of energy easy? Well the good news is there is.

The other good news is: just doing this correctly will earn you a lot of partial credit.

As for the bad news...:

- what I am going to show you helps to apply the law of conservation of mechanical energy (kinetic energy as well, but you might not even need the trick in this case). However a problem that involves conservation of energy may involve other things as well. Or it may involve applying only this law, but more than once (i.e. perhaps applying it between two moments and then between another two, different ones).

- what it makes easy is to write down the law, the equation that represents conservation of energy for your particular problem. That does not mean that this equation is always going to look pretty and simple by your standards. Nor that what you'll have to do with that equation is going to be something you'll find easy if you don't get along with algebra. 

THE SECRET

So you've decided that you need to use conservation of (mechanical) energy, but gosh your problem looks complicated (not an unlikely scenario if last test is anything to go by!). Here's the secret:

Don't waste your time staring at the problem in horror. Instead, use that time to DRAW A TABLE! 

This is what the table should look like if you had time to make it pretty:

But we all know, you don't want to take the time to make it pretty in an exam. So just draw this in a corner somewhere, knowing what KL, etc... stand for (and if you draw it in pen and fill it up in pencil, you can even reuse it for several problems ;-):

where:
- KL: linear kinetic energy
- KR: rotational kinetic energy
- UG: gravitational potential energy
- US: spring potential energy

Notice the structure: 

- you have 2 main columns, one for initial values of the energies, one for the final ones.

- Each of the main columns is split into more columns: one for each object in your system. Name them after the objects themselves, like, block, wheel, etc...

- There are 4 rows: 2 for the kinetic energies, 2 for the potential energies. These are the forms of energy that make up mechanical energy.

Once you've drawn your table, which should take you about one minute, fill it up. 

For each cell, ask yourself whether the object has this kind of energy in this situation (i.e. initial or final). 

If it doesn't, write 0.

If it does, insert a check mark.

Don't leave anything blank: you wouldn't be sure whether you considered all the cells or not. Yes, that will take you 2 long minutes. So all in all, you will have spent 3 minutes on this table. You have over 12 minutes per question, spending 3 of those 12 making sure that you don't waste them all by doing the problem wrong is well worth it ;-)

Also, as you know, for gravitational energy, you need to choose a reference height, where you take h = 0 m. You can choose a different one for each object.

Some rows you can fill up quick confidently: if there's not spring in the problem, the last row is 0 all the way. BUT, in general, be VERY careful in filling up the table. It's a lot like a free body diagram for Newton's second law: if you don't have all the correct forces on the diagram, you can forget about getting the problem right. Same here if you forget an energy.

Once you've got the table filled up, write down the equation: initial energies on the left of the equal sign and final ones on the right.

WARNING: 

Use subscripts whenever needed: look carefully at the equation after you wrote it down, and if any two letters/symbols are the same, ask yourself whether they do represent the same thing physically. If they don't, add subscripts to both of them, so you can tell them apart. Together with forgetting an energy term, this is the most common source of mistakes.

Example 1:

For the following situation of a mass attached to a string that is wrapped around a wheel:

The table, and the resulting equation, would look like:

Only one symbol is repeated in this equation: m. In both cases, it represents the same thing (the mass of the block), so we don't need to add any subscript.

Example 2:

For the situation illustrated below, where the block now falls on an initially uncompressed spring. The block  stops when the spring is completely compressed, but at that moment the wheel is still turning, just with a constant angular velocity now that the torque on it has disappeared - the string is about to go slack and is no longer pulling on the wheel. Pretty nasty-looking problem, right?!:

The table would look like the one below, and the equation that you would obtain for it would be the one under that:

See? No so bad after all!

Only one symbol is repeated in this equation: m. In both cases, it represents the same thing (the mass of the block), so we don't need to add any subscript.

Note that the spring does not have gravitational potential energy because it is assumed to have negligible mass.

CONSERVATION LAWS

CONSERVATION LAWS AND WHEN THEY APPLY

A conservation law is a law that tells you that some quantity (momentum, some energy, etc...) remains the same for a certain system (remember that "system" in physics is just a jargon way of saying an object or a group of objects).

When expressed in equation form, a conservation law always has the structure:

initial quantity = final quantity

Initial and final just refer to the two instants of time that you've decided to care about (the quantity in question is also conserved at other moments in between, but for the purpose of solving a problem you need to pick the two that you are given information about).

CONSERVATION OF MOMENTUM:

CONSERVATION OF LINEAR MOMENTUM:

p_i = p_f

It is always true for a system that is "isolated", that is on which there is no net force at any time between the moments you consider.

Example: think about the carts that you were colliding during the lab. If you consider just one cart, and you consider only what happens before the collision, the linear momentum of this cart does not change, it is conserved. The mass remains the same, and the cart moved at the same speed in the same direction. Until it hit the second cart...

Once it hits the other cart, that other cart exerts a force on it and changes its momentum (that's what you were calculating in one of the labs too: you were checking that the impulse due to the second cart was equal to the change in momentum of the first). So if you compare the momentum of the cart before the collision, to its momentum after, you will find that its momentum has changed. It is not conserved. HOWEVER, the momentum of the two carts considered together as one system is not changed by the collision. That "total" momentum was conserved at times before the collision, and it is also conserved at times before and after the collision.

This is why when it comes to momentum, whether or not it is conserved depends on which system you are talking about. In order to be able to apply this law, you've got to pick a system that is "isolated". In the context of collisions, that means that you need to apply the law to a system made up of all the objects that interact with one another between the times you care about.

CONSERVATION OF ANGULAR MOMENTUM:

L_i = L_f

Same idea and rules as above, except that instead of an external net force, what would change the momentum is an external net torque.

Conservation of angular momentum and linear momentum are separate laws: each form of momentum is conserved independently of each other, so it has its own conservation law, so that we have two different laws, expressed by two equations:

p_i = p_f

L_i = L_f

but NOT p_i +L_i = p_f + L_f, unlike you do when dealing with energy: in the case of energy, linear and angular kinetic energies all get to sit in one equation.

CONSERVATION OF ENERGY:

If by "energy" we mean the total energy, talking into account all possible forms it can take, then this law would be always true. But that's not very useful in practice: in the situations described in the problems given to you, you either don't need to take into account all these forms of energy, or you just cannot. 

The other conservation of energy laws that you are given are special cases of this one. The idea is this: 

- energy, like momentum, can be transferred between objects. So here as well, you need to apply the law to a system made up of all the objects that interact with one another between the times you care about.

Often there is only one object you need to care about, but sometimes there's more: for example in the collisions we did in the lab, you had to consider the energy of both carts: your system was the two carts.

- energy can change form. 

example: one of many! Kinetic energy can be transformed into potential gravitational energy. Linear kinetic energy can be transformed into angular kinetic energy. This is why linear and angular kinetic energies get to sit in one equation.

This is what gives rise to the different energy laws: each one takes into account only some forms of energy. Consequently, they are only true when these forms of energy are not transformed into yet different ones (see below).

CONSERVATION OF KINETIC ENERGY:

K_i = K_f

This one is true when the kinetic energy is not transformed into any other kind of energy. It is useful when kinetic energy is merely transferred from one object to another. 

This happens for example in what we call elastic collisions. So if a problem tells you that a collision is elastic, then this law applies. 

Note that as with momentum, there are TWO kinds of kinetic energies: linear, and angular.

CONSERVATION OF MECHANICAL ENERGY:

E_Mi = E_Mf 

Many problems involve this one. 

"Mechanical energy" is the energy that takes into account, that contains, the following forms of energy:

- kinetic energy K, which can be linear or angular.

- mechanical potential energy U, which can be gravitational or spring potential energy.

So there are FOUR kinds.

So whenever you have a situation when kinetic energy is transformed into potential energy, or vice versa, that's the one to pick. 

HOWEVER, it does NOT apply when kinetic and/or potential energy is transformed into yet other forms of energy, like heat. This usually happens when friction is involved, or an object is permanently deformed (not elastically). When it does not apply, you may be able to use the work energy theorem instead (it all depends on the problem of course, the information given in it, the question...)

There is a trick to apply conservation of mechanical energy easily (and correctly!) to complicated problems. To find out which, go to this tutorial.

WORK ENERGY THEOREM

This rule does not have the form of a conservation law, but you need it in your tool kit too, it is related to the issue of conservation of energy. 

It state that the change in kinetic energy is equal to the work done by the net force. 

Another way to say this is that the change in kinetic energy is equal to the sum of the works done by all the forces.

This rule is very similar to the conservation of total energy (not just mechanical), and indeed you'll find that calculating the work done by the weight gives you the same expression as the gravitational potential energy. Like energy, work is expressed in joules. So the two laws are closely related, they're almost different versions of the same thing. Concretely what this implies for you is two things:

- it always applies (whether there is friction involved or not).

- it is most useful when friction takes energy out of the system. 

This is because you have no way to handle this situation with a conservation of energy law, which in turn is because we don't have a neat expression for that form of energy. And that, in turn, is due to the fact that how much energy is transformed by friction depends not just on the situation of the object at the two moments you want to consider, but also on what has happened to the object in between. That's what we mean by friction is a non-conservative force: the work it does can't be expressed as a change of energy that depends only on the situation at two different moments, and not what happens in between, as it can be done for weight and spring force.

HOW TO FIND (EXPERIMENTALLY) THE VALUE OF A CONSTANT PHYSICAL QUANTITY, WHEN YOU ARE GIVEN AN EQUATION THAT CONTAINS IT.

HOW TO FIND (EXPERIMENTALLY) THE VALUE OF A CONSTANT PHYSICAL QUANTITY, WHEN YOU ARE GIVEN AN EQUATION THAT CONTAINS IT.

In order to do this you need to draw a graph using data collected in a lab. However not any graph will do: you need to make sure that what you graph will lead to data points that form a straight line.

So you have to use a "trick".

Imagine that you were asked to determine experimentally a constant called m, and told that it obeys this awful looking equation:

I⁷ = 3 m a D⁵

What you need to do is:

1) First, identify what in this equation are variables, and what are constants. 

About the variables: Identify which symbols represent the physical quantities that you can and want to make vary in the lab. There should be two of those.

Then identify which one of the two you will be controlling yourself: that's your independent variable.

The other one, whose value will change automatically when you vary the independent variable, is the dependent variable.

About the constants: Constants can be just numbers, or they can be things like π;, or physical constants like g.

OR they can also be things that could in principle vary. You have to watch out for those for two reasons: 

- you will have to make sure that in fact do not, when you do the experiment. You need to keep them constant.

- you need to record their (constant) value.

In the example equation above for instance, I stands for the independent variable, D for the dependent one, m could in principle vary but we don't want it to, because we've decided we'd rather vary I instead for practical reasons. Of course in real life the variables are not conveniently called I and D for you. You have to figure out what plays the part of I and D in the equation given to you, from your knowledge of the experimental set-up.

For the sake of clarity, let me color coding the variables as red in the equation:

I⁷ = 3 m a D⁵

Then what we call the functions of these variables are the expressions involving them indicated in orange below:

                                                        I⁷ = 3 m a D⁵

The rest in white is an expression made up purely of constants, so it is something constant.

2) The first thing you need to do is to consider the expression that contains your dependent variable, and solve for it. Here it is D⁵ .(If you're lucky the equation may already be in that form, but that's not always true of course). In our example this results in:

D⁵ =  1⁄₍3 m a₎ I⁷

3) Now comes the crucial step: decide what you need to graph on the y-axis and on the x-axis of your graph, in order to get a straight line.
Clearly, graphing D vs I won't work.
So you may be tempted to try and "get rid" of what seems to be causing a problem by, say, taking the power 1/5, or 1/7, to "get rid of" the powers of 5 and 7. The equation will look different, sure. But in fact that doesn't help either, because the shape of a graph (like it being straight), depends on the relationship between what you graph on the x axis and what you graph on the y-axis. And putting the equation in a different form doesn't change the relationship between the variables (D and I here).
So in order to know what you need to graph, you must understand what truly defines a straight line. Graphing one thing vs another will give you a straight line only if those two things are "linearly related" to each other, or "proportional" to one another - the two phrases mean the same thing.
"proportional" means: to differ from one another by only a constant factor; that the one is equal to the other, just multiplied by something constant. That is, we say that:

expression A     is      proportional to expression B

when 

expression A     =     something constant    x     expression B

And when that's the case, graphing expression A vs expression B will give you a straight line, with as its slope the value of that constant something.
That's what we mean by the equation of a straight line is:

Y                  =              slope              x              X

But Y and X don't have to be something simple, they can be complicated expressions; only it is these complicated expressions that must be graphed then. Also, the slope must be constant!

Now that we know this, let's go back to our example (with variable and constant expressions).

D⁵ =  1⁄₍3 m a₎ I⁷

Comparing it to the equation of a straight line, we see that what corresponds to:

Y                  =              slope              x              X

is:

D⁵ =  1⁄₍3 m a₎ I⁷

So IF you graph D⁵ on the y-axis, and I⁷ on the x axis, then (and only then!):

slope = 1⁄₍3 m a₎

Where "slope" stands for the slope of the line of best fit to the data points.

At this point, all you have to do to find m, is solve  this last equation! You get:

m = 1/(3 a slope) 

Then to find the value of m, which remember is what you were asked to find, you need to substitute the value of a (which hopefully you will have recorded in lab!) and the value of the slope (which on a graph by hand you need to find by circling 2 points on best-fit line and using them to calculate rise vs run; on a graph done in Excel is given by excel in the equation of the best fit line).
To sum up, here is what YOU would need to do BEFORE YOU START to write the lab report asking for you to find m, on a separate sheet of paper for yourself:

OTHER EXAMPLE:

In the example above the relationship between the variables was a power law. But this method works whatever the relationship is, as the next example shows:
You are told that:

                                                                e^r = ^{(7 π)}⁄_c . V .   ln (H)

You are asked to find c. 

1) From knowing what all these symbols actually mean physically in the context that they are discussed, you can tell that r and H are the only physical quantities that can be varied, and from the set-up you can tell that H is the independent variable, and r the dependent one i.e. you can vary H yourself and r will vary as a result.

2) You solve for the expression that involves the dependent variable. That is e^r. But, it is already solved for, so in this case you don't have to do anything.

3) You compare this to the form of the equation for a straight line:

 e^r = ^{(7 π)}⁄_c . V .   ln (H)

Y                  =              slope              x              X

and conclude that you need to graph e^r   on the y-axis, ln (H) on the x axis, and that in this case:

slope of the best fit line = ^{7 π V}/c

Therefore you will find the value of c with:

 c = ^{7 π V}/slope

and substituting the value you found for the slope of the best fit line.


To sum up, here is what YOU would need to do BEFORE YOU START to write the lab report asking for you to find m, on a separate sheet of paper for yourself:

PRACTICE EXAMPLES:

1)  You are told to measure the value of some constant quantity named k and you know that:

z  = (3/2) k m⁵

where z and m are variables that you can measure. In the experiment, you can control the value of z, and m varies as a result. Write down what the following must be:


- what you need to graph on your y axis:        m⁵

m is your dependent variable, so solve the equation for the expression that contains it: m⁵ = (2/(3 k)) z
That way the dependent variable ends up on the left hand side of your equation - ready to appear on the y-axis rather than the x-axis, as tradition requires.
If you decided to graph m on the y-axis, rather than m⁵ , this is wrong because m is not proportional to z: m is not equal to z times just a constant; it is m⁵ that is equal to z times a constant (where that constant is 2/(3 k)). So it is m⁵ that is proportional to z, hence graphing m⁵ vs z must yield a straight line.

 - what you need to graph on your x axis:  z      

- the expression for the slope you will get IF you do graph this:         slope =2/(3 k)

Remember the structure of a straight line: y = slope . x + y-intercept. So the slope is the constant expression that multiplies x, where by x we mean whatever is graphed on the x axis.

- the expression for k in terms of the slope:  k = 2/(3 slope)


 Now highlight with your mouse the lines above to check your answer.



2) You are told to measure the value of some constant quantity named c and you know that:

A³  = 5 c d

where A and d are variables that you can measure. In the experiment, you can control the value of d, and A varies as a result. Write down:

- what you need to graph on your y axis:        A³


- what you need to graph on your x axis:     d  

If you decided to graph A on the x-axis, this is wrong because A is not proportional to d: A is not equal to d, just multiplied by a constant; it is A³ that is equal to d times a constant (where that constant is 5 c). So it is A³ that is proportional to d, hence graphing A³ vs d must yield a straight line.



- the expression for the slope you will get IF you do graph this:         slope = 5c

Remember the structure of a straight line: y = slope . x + y-intercept. So the slope is the constant expression that multiplies x, where by x we mean whatever is graphed on the x axis.

- the expression for c in terms of the slope:  c = slope / 5


 Now highlight with your mouse the lines above to check your answer.



3) You are told to measure the value of some constant quantity named v and you know that:

b³  = v² j⁵

where b and j are variables that you can measure. In the experiment, you can control the value of j, and b varies as a result. Write down:

- what you need to graph on your y axis:        b³
- what you need to graph on your x axis:        j⁵
Indeed b³ and j⁵ are proportional to one another since the one is equal to the other times a constant. So this graph must give a straight line.
- the expression for the slope you will get IF you do graph this:         slope = v²

- the expression for v in terms of the slope:  v = square root of the slope

Now highlight with your mouse the lines above to check your answer.