TORQUES - ANGULAR FORM OF NEWTON'S 2ND LAW

This webpage contains the following sections:

- Introduction

- Steps required: overview and comparison to the linear case (this contains a summary of differences from the linear case at the end of the section; make sure you read it).

- Difficult steps: detailed explanations and examples. This section focuses on 2 skills:
          - how to choose the pivot point
          - how to find the expression for a torque. This subsection is in turn divided into the 3 skills required:
                             - figure out whether the torque is positive or negative (for your choices of + direction and pivot point).
                              - write down the magnitude of the force that creates the torque
                              - figure out the lever arm (for your choice of pivot point).

- More practice/diagnostic example (starts about 3 quarters of the way down). If you already have a fair amount of practice with these problems, you may want to start there. Do what is asked there, and you should be able to tell whether you are lacking some critical skills, and where to do on this page to acquire them.

INTRODUCTION

The law which involves torques is the angular version of Newton's 2nd law. Recall that Newton's 2nd law in its usual, linear form tells you that the linear acceleration of an object is due to all the forces acting on this object: the object accelerates if there is a net force acting on it, and the larger its mass the less it accelerates:

 Σ F→ = m . a→

The angular version of Newton's 2nd law tells you that the angular acceleration of an object is due to all the torques acting on this object: it accelerates in its rotation, in its spin, if there is a net torque acting on it, and the larger its moment of inertial, the less it accelerates in its rotation:

 Σ τ→ = I . α→

STEPS REQUIRED:

OVERVIEW AND COMPARISON TO THE LINEAR CASE

Recall that when you apply the linear version of Newton's 2nd law, you need to:

1) Write down Σ F→ = m . a→

2) Decide what "system" (what object, etc...) you want to apply it to.
Example:

You choose the cat.

3) Draw a free body diagram: i.e. draw a diagram representing all the forces that act on this "system".
Example:

4) Decide which directions you are going to call "positive", by drawing the axes you choose near the free body diagram.

In our example:

5) Write down Newton's 2nd law in terms of components, for your specific object:

- for as many directions as there are in the problem ( Σ F_x = m . a_x  , and / or Σ F_y = m . a_y )

- Use the free body diagram and the axes from 4) to replace Σ F_x  and / or Σ F_y by expressions that involve the particular forces that act on your specific object.

Example: 

Σ F_y = m . a_y
m g = m . a_y

The corresponding steps for the angular version of Newton's 2nd law are:

1) Write down  Σ τ→ = I . α→

2) Decide what "system" (what object, etc...) you want to apply it to.

Example:

You choose the boy.

3) Draw a free body diagram: i.e. draw a diagram representing all the forces that act on this "system".
Now you need to draw the arrows representing the forces with their tail where the force acts. For weight, that's at the center of mass of the object (i.e. at the center of the object if its density is the same throughout the object, i.e. uniform).

Example:

4a) Decide which direction you are going to call "positive", by drawing a counterclockwise (or clockwise) arrow near the free body diagram.
There is no right or wrong decision, this is arbitrary. However, it makes calculations simpler if the direction that you choose as the positive is the direction of the angular acceleration of your object (if it has one).

Example:

4b) Decide which point you are going to choose as the pivot point.

Example:

5) Use:
- the free body diagram
- the direction from 4a)
- and the pivot point from 4b)
in order to replace Σ τ in  Σ τ= I α by an expression that involves the particular forces that act on your specific object.

This means that you need to express each one of the torques in Σ τ by an expression that contains the magnitude of the corresponding force.

Example:
                                                                                     Σ τ  = I . α

+ mg . (d/2) cos θ = I α

SUMMARY OF DIFFERENCES BETWEEN THE ANGULAR AND THE LINEAR CASE:
- The forces on the free body diagram must be drawn with the tail of the arrow at the point where the force acts.
- The choice of a positive direction refers to either clockwise or counter-clockwise, rather than to choosing axes.
- In addition to choosing a positive direction, you now need to choose a pivot point as well.
- You need to find an expression for each of the (non-0) torques, instead of for each of the (non-0) components of the forces.


Steps 4b) and 5) are those that you are going to find new and most difficult, even if you are comfortable with applying the usual, linear version of Newton's 2nd law. So let's discuss them in detail.

DIFFICULT STEPS:

DETAILED EXPLANATIONS AND EXAMPLES

How to choose the pivot point (i.e. step 4b above)

There is no correct nor incorrect pivot point in the sense that the angular form of Newton's 2nd law would apply to the former and not to the latter. The law is true for all points, that is for any point that we may choose as the pivot point. This is the angular equivalent of the fact that Newton's (linear) 2nd law is true in all directions, i.e. whatever we choose to pick as x or y.
However, just like for Newton's linear 2nd law we did not choose directions randomly, but chose those that we knew would make solving the problem easier. The same goes for the angular version of Newton's 2nd law.
Recall that in the linear case, if the acceleration of the object was non-0, we usually chose one of the directions to be the direction of the linear acceleration. This was because it made the acceleration 0 in the direction normal (perpendicular) to that one (we'd pick x in the direction of the acceleration so that ay would be 0). Something similar goes in the angular case. As we will see in greater detail below when we discuss lever arms, what a torque turns out to be depends both on what force gives rise to it, and on where that force acts. It so happens that when a force acts directly on the pivot point (or the axis) considered, the corresponding torque is 0. This gives us a neat way of getting rid of some of the torques. To avoid having a torque appear in an equation, we just choose to apply Newton's (angular) 2nd law using as pivot point the point where the corresponding force acts.

Example: 
The forces on your object of interest are:

You'd rather your equation does not involve the tension, hence not the torque it creates, since the torque would contain T (you might feel that way, say, because you are not asked to find T and it is not given to you; for instance you are asked to find f and given W and N). Then pick as pivot point the point where the tension acts:

But if instead you want the torques due to N and f  to not appear in the equation (say, you are asked for the tension T and given W), then choose the point where N and f act as the pivot point:

And if it is W and its torque that you don't want to see appear, choose the point where W acts (i.e. the center of mass) to be the pivot point.

So when:
- you are not interested in finding a certain torque or the force it is due to, and:
   - you know it is going to be the most difficult one to find (usually, because you don't know what the corresponding force is),
   - or there are several forces acting on the same point and you are not interested in finding any of them or their torques,
it is usually a good idea to pick the point where it/they act to be the pivot point.
Note that you can apply Newton's (angular) 2nd law about several pivot points in the same problem, generating a different equation every time. In the same way that in the linear case, you can generate as many different equations as there are dimensions in the problem (i.e. in 2 dimensions, one equation for x and another for y), in the angular case, you can generate as many different, useful equations as there are points on which forces act. That doesn't mean you will need them all to solve the problem, but depending on the situation you might, or you might need several of them.

How to find the correct expression for the sum of the torques Σ τ (i.e. step 5 above):

Recall that the idea is to find the sum of the torques Σ τ in order to replace it in Σ τ= I α. 
So you need to find the correct expression for each (non-0) torque, and add them all up.

The definition of a torque is that it is the cross-product:   r→x  F→. There are several methods to find the expression that this leads to. The one I am going to present here involves finding a distance called the lever arm (c) below).

Whatever method you use to find it, the expression for the torque has the following structure:

In this expression, the expression (d/2) cos θ is the lever arm, which is itself a distance. In the example given, it is not the same distance that appears explicitly, i.e. (d/2) because it must be given in terms of a distance the problem tells you about, i.e. d here.

In order to find this expression you need to:

a) figure out whether the torque is positive or negative with your choice of a positive direction in 4a. That is, decide whether the expression has a + or a - sign in front.

There is a + sign in front if the corresponding force tends to make the object spin in the direction that you chose to be positive, and negative otherwise. Be careful however: what the torque tends to do depends not only on the force, but also on the pivot point that you chose in 4b.
The sign of the torque depends on 3 things: the direction of the force, the pivot point you chose, the direction you chose as positive.
Concretely, look at the diagram focusing your attention on only the force and the pivot point, ignoring everything else. Ask yourself in which direction that force would make your object spin about that pivot point. Once you "see" that direction, check whether that is the same direction as the one you chose as the positive one. If it is, this torque is negative (with your choice of what constitutes positive!).


Example:

With the pivot point and choice of positive direction indicated:
- the torque due to T is 0.
- the torque due to W is negative.
- the torque due to N is positive.
- the torque due to f is positive.


Now take a piece of paper and write down whether the torques due to T, W, N and f are positive, negative  or 0 in the following case:

Now check your answers by highlighting the lines below with your mouse:

- the torque due to T is negative.
- the torque due to W is positive.
- the torque due to N is 0.
- the torque due to f is 0.


Now write down whether the torques due to T, W, N and f are positive, negative or 0 in the following case (careful: the choice of positive direction has been changed too, for variety):

Now check your answers by highlighting the lines below with your mouse:

- the torque due to T is positive.
- the torque due to W is 0.
- the torque due to N is negative.
- the torque due to f is negative.



b) write down the magnitude of the force that gives rise to the torque. This step's easy. In the expression taken above as an example, + mg (d/2) cos θ , this is just mg (or W for weight if you prefer).

c) figure out what the lever arm is for this torque. In + mg (d/2) cos θ, that would be (d/2) cos θ, but figuring this out is usually the toughest part.

In order to be able to do this, the first thing you need to do, before you even begin trying to figure out the lever arms, is to write down on your diagram expressions for the lengths that you can express in terms of the one given to you, and the angle given to you.

Example:

You are given:

The first thing you must do is write down on your diagram what is added below in red:

What is the lever arm?

The lever arm is the shortest distance between the direction of the force (i.e. a line going through the force) and the pivot point.
So be aware that the lever arm depends on both the direction of the force whose torque you are after, and the pivot point that you chose.
Concretely, look at the diagram focusing your attention on only the force and the pivot point, ignoring everything else (yep, so far it's like what you did to find the sign of the torque).
Now imagine a straight line going through the force. Then imagine the shortest line segment between that line and the pivot point. Once you "see" it, identify what its length must be in terms of the distances given in the problem.

Example:

You want to find lever arms for the non-0 torques in the following situation:

To find the lever arm for the torque due to N, these are the elements that you must focus on, and this is the lever arm that you should imagine:

This implies that for the torque due to N the lever arm is equal to (L/2) cos θ:

If you have trouble understanding why, picture the following:

and look at the first rule on this link.

Ok so that was for the lever arm associated with N. Now ponder the diagram below and write down on a piece of paper what the lever arm is for f:

Now check your answer by highlighting the line below with your mouse:

(L/2) sin θ

If you still can't see it, ponder this diagram:

How about W? What is the lever arm for W? 

Write it down on a piece of paper.
Now check your answer by highlighting the lines below with your mouse:

It is 0. The shortest distance between the direction of W and the pivot point is 0, since W acts directly on the pivot point (usually chosen by you). This is why the torque due to a force that acts on the pivot point/axis is 0.

And what about T? What is the lever arm associated with it?

Write it down on a piece of paper.
Now check your answer by highlighting the lines below with your mouse:

It is 0 as well! Indeed, the line that goes through T goes through the pivot point, so the shortest distance between that line and the pivot point is once again 0. It is not only torques from forces that act on the pivot point that are 0: the torque due to any force whose direction goes through the pivot point is also 0!

Remember this if you have trouble choosing a pivot point that works for your problem using the simplest rule given above!

MORE PRACTICE / DIAGNOSTIC EXAMPLE

In the previous section, we chopped the difficult steps up into small bits in order to explain them in detail. In this section, we work out a full example, without going into explanations as to why the answers are what they are - that's in the previous section!
You need to do this along, so grab a nice piece of paper and your favorite pen ;-)

You are given a problem that you know is going to involve you applying the angular form of Newton's 2nd law. The physical situation is the following. A firefighter stands on a uniform ladder of length L, a third of the way from the top of the ladder. The ladder is standing on a frictionless floor, at an angle θ to the floor. It is leaning against a wall. The friction coefficient between the wall and the ladder is non-0. Everything is going well at the moment of interest: neither firefighter nor ladder are accelerating! They give you this drawing to go with:

Do step 1: write down the relevant law in its most general form.
Now check your answer by highlighting the line below with your mouse:

Σ τ→ = I . α→



Do step 2: decide what object you want to apply this law to.
Now check your answer by highlighting the lines below with your mouse:

Considerable experience with homework problems makes it clear to you that you want to apply the law to the ladder.

Do step 3: draw the free body diagram (don't peek!).
Now check your answer by scrolling down:

Note that we have indicated the weight of the firefighter on the free body diagram. This is a sloppy habit that most physicists have taken, in fact this force is a normal force: the normal force exerted by the firefighter's feet on the ladder. This normal force so happens to be equal in magnitude and direction to the firefighter's weight when (s)he isn't accelerating, hence the sloppy habit.


Step 4a, i.e. choosing a positive direction, I am going to impose, so you can end up with the same equation as I have below. I am choosing counter-clockwise.
Draw it on your diagram!

Do step 4b, i.e. choose a pivot point, such that you can make the greatest number of torques equal to 0.
Write down your answer. 
Check it by highlighting the lines below with your mouse:
For that you must choose the point of contact between the ladder and the wall as the pivot point. Two forces are exerted there (the normal force due too the wall and the friction force due to the wall). No other point has more forces applied on it, and no point has more forces having directions going through it, either.

Do step 5. First, add onto your diagram everything else that you need to do this (again, don't peek!).
Now check your answer by scrolling down. Your diagram should look like what is below:

Now write down the angular version of Newton's second law, for these choices of positive direction and pivot point.
After you have written all of it down, check your answer by highlighting the lines below with your mouse:

- N_G . L cos θ  +    M_L g  . (L/2) cos θ  +   M_F g  . (L/3) cos θ  = 0

We have set the right hand side to 0 because the ladder has no angular acceleration (as usual in "statics" problems).

Compare what you wrote down to this answer! 

- If you got some of the signs wrong, and you can't immediately see why, study the relevant part in the previous section. That's the one that starts with:
"a) figure out whether the torque is positive or negative with your choice of a positive direction in 4a. That is, decide whether the expression has a + or a - sign in front."), under the colorful expression.
- If you got wrong some of the expressions involving either L or cos and sin, and you can't immediately see why, study the relevant part in the previous section. That's the one that starts with:
"c) figure out what the lever arm is for this torque. In + mg (d/2) cos θ, that would be (d/2) cos θ, but figuring this out is usually the toughest part."

More practice:

Now write down the angular version of Newton's second law, for the choices of positive direction and pivot point below:

After you have written all of it down, check your answer by highlighting the line below with your mouse:

      M_L g  . (L/2) cos θ +   M_F g  . (2L/3) cos θ   -   N_w . L sin θ -    f . L cos θ  = 0


Now write down the angular version of Newton's second law, for the choices of positive direction and pivot point below:

After you have written all of it down, check your answer by highlighting the line below with your mouse:

N_G  . (L/2) cos θ    + M_F g  . (L/6) cos θ   - N_w . (L/2) sin θ    - f  . (L/2) cos θ    = 0



Now write down the angular version of Newton's second law, for the choices of positive direction and pivot point below:

After you have written all of it down, check your answer by highlighting the line below with your mouse:

- N_G  . (2L/3) cos θ     + M_L g  . (L/6) cos θ     +  N_w  . (L/3) sin θ  +  f  . (L/3) cos θ  = 0

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.