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These laws only apply in inertial reference frames  our reference frame must be stationary or moving with a constant velocity.
The most important quantity related to an object's motion is it's momentum
$\vec{p}=m\vec{v}$
Momentum can only be changed by a net force acting on the object
$\frac{d\vec{p}}{dt}=\Sigma\vec{F}$
So for collisions in which no external force is acting the total momentum is conserved. This is really just another way of expressing Newton's 2nd Law.
The equation of conservation of momentum is a vector equation.
$m_{A}\vec{v}_{A}+m_{B}\vec{v}_{B}=m_{A}\vec{v'}_{A}+m_{B}\vec{v'}_{B}$
Therefore if we have a collision involving more than one dimension we need to consider conservation of each component of momentum, for example in 2 dimensions
$m_{A}v_{Ax}+m_{B}v_{Bx}=m_{A}v'_{Ax}+m_{B}v'_{Bx}$
$m_{A}v_{Ay}+m_{B}v_{By}=m_{A}v'_{Ay}+m_{B}v'_{By}$
If a collision is perfectly elastic we can add a 3rd equation
$\frac{1}{2}m_{A}^{2}v_{A}+\frac{1}{2}m_{B}v^{2}_{B}=\frac{1}{2}m_{A}v'^{2}_{A}+\frac{1}{2}m_{B}v'^{2}_{B}$
$v'_{B}=\frac{2m_{A}}{m_{A}+m_{B}}v_{A}$ $v'_{A}=\frac{m_{A}m_{B}}{m_{A}+m_{B}}v_{A}$
$m_{A}=m_{B}$
$m_{A}>m_{B}$
$m_{B}>m_{A}$
Conservation of momentum $v_{A}=v'_{Ax}+v'_{Bx}$ $0=v'_{Ay}+v'_{By}$  Conservation of energy $v_{A}^2=v'^{2}_{A}+v'^{2}_{B}$ 
$v'_{Bx}=v'_{B}\cos\theta$ $v'_{By}=v'_{B}\sin\theta$  $v'_{Ax}=v'_{A}\sin\theta$ $v'_{Ay}=v'_{A}\cos\theta$ 
Square the conservation of momentum equations
$v'^{2}_{Ay}=v'^{2}_{By}$
$v_{A}^22v_{A}v'_{Ax}+v'^{2}_{Ax}=v'^{2}_{Bx}$
Add these together
$v_{A}^2+v'^{2}_{A}2v_{A}v'_{Ax}=v'^{2}_{B}$
Use kinetic energy equation to eliminate $v'_{B}$
$v_{A}^2+v'^{2}_{A}2v_{A}v'_{Ax}=v_{A}^{2}v'^{2}_{A}$ → $v'^{2}_{A}=v_{A}v'_{Ax}$
$v'^{2}_{A}=v_{A}v'_{A}\sin\theta$ → $v'_{A}=v_{A}\sin\theta$
and conservation of energy equation gives us $v'_{B}=v_{A}\cos\theta$
We can use conservation of momentum for the collision and conservation of energy for the subsequent motion in a ballistic pendulum to find the velocity of a bullet fired in to a block from the height that the bullet and block rises to.
$mv=(M+m)v'$
$\frac{1}{2}mv^2\neq\frac{1}{2}(M+m)v'^{2}$
but
$\frac{1}{2}(M+m)v'^{2}=(M+m)gh$ → $\frac{1}{2}v'^{2}=gh$
so
$v=\frac{M+m}{m}\sqrt{2gh}$
In a perfectly inelastic collision the two object stick together after the collision so conservation of momentum now is written
$m_{A}\vec{v_{A}}+m_{B}\vec{v_{B}}=(m_{A}+m_{B})\vec{v}'$
In general the displacement vector for the center of mass of a system of particles $m_{i}$ can be written as
$\vec{r}_{CM}=\frac{\Sigma_{i}m_{i}\vec{r_{i}}}{\Sigma_{i}m_{i}}=\frac{\Sigma_{i}m_{i}\vec{r_{i}}}{M}$
Differentiating with respect to time gives
$M\frac{d\vec{r}_{CM}}{dt}=\Sigma_{i}m_{i}\frac{d\vec{r}}{dt}$ or $M\vec{v_{CM}}=\Sigma_{i}m_{i}\vec{v}_{i}$
and thus the total momentum of a system is
$\vec{P}=M\vec{v}_{CM}$
differentiating once more
$M\vec{a_{CM}}=\Sigma_{i}m_{i}\vec{a}_{i}$
$\Sigma_{i}m_{i}\vec{a}_{i}=\Sigma_{i}F_{i}=\Sigma \vec{F}_{ext}$
so we have obtained a new form of Newton's Second Law that works for a system of particles or an extended object.
$M\vec{a_{CM}}=\Sigma \vec{F}_{ext}$
The total momentum of a system is given by
$\vec{P}=M\vec{v}_{CM}$
but this does not necessarily imply that every part of the system has $\vec{v}=\vec{v}_{CM}$
However the motion of the mass around the center of mass in the reference frame of the center of mass needs to maintain the position of the center of mass. Rotation around the center of mass satisfies this requirement.
An object in a system will have an angular momentum around the origin (in the context of dealing with a rotating moving system you need to calculate angular momentum around the translating center of mass).
$\vec{L}=\vec{r}\times\vec{p}$
and to change this we need to apply a torque
$\vec{\tau}=\vec{r}\times\vec{F}$
$\Sigma\vec{\tau}=\frac{d\vec{L}}{dt}$
The cross product of two vectors $\vec{A}$ and $\vec{B}$ is another vector
$\vec{C}=\vec{A}\times\vec{B}$
The magnitude of the vector $\vec{C}$ is $AB\sin\theta$
The direction of $\vec{C}$ is perpendicular to the plane which contains both $\vec{A}$ and $\vec{B}$. Use of the right hand rule can help you determine this direction.
If we consider the vectors in unit vector notation, $\vec{A}=A_{x}\hat{i}+A_{y}\hat{j}+A_{z}\hat{k}$ and $\vec{B}=B_{x}\hat{i}+B_{y}\hat{j}+B_{z}\hat{k}$ the cross product can be expressed as the determinant of a matrix
$\vec{A}\times\vec{B}=\left \begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k}
A_{x} & A_{y} & A_{z}
B_{x} & B_{y} & B_{z} \end{array} \right = \hat{i}\left \begin{array}{cc}
A_{y} & A_{z}
B_{y} & B_{z} \end{array} \right

\hat{j}\left \begin{array}{cc}
A_{x} & A_{z}
B_{x} & B_{z} \end{array} \right
+
\hat{k}\left \begin{array}{cc}
A_{x} & A_{y}
B_{x} & B_{y} \end{array} \right$
$=(A_{y}B_{z}A_{z}B_{y})\hat{i}+(A_{z}B_{x}A_{x}B_{z})\hat{j}+(A_{x}B_{y}A_{y}B_{x})\hat{k} $
In a case such as that illustrated above we can calculate the magnitude of the torque using
$\tau=RF\sin\theta$
The units of torque are $\mathrm{Nm}$
In an object with a common angular acceleration which consists of many masses at various distances from the center of rotation, each of which may be subject to a torque we can say that
$\sum\limits_{i}\tau_{i}=(\sum\limits_{i} m_{i}R_{i}^{2})\alpha$
We call the quantity $\sum\limits_{i} m_{i}R_{i}^{2}$ the moment of inertia $I$ and write the rotational equivalent of Newton's Second Law as
$\sum \tau=I\alpha$
Usually a rotation axis that passes through the center of mass of an object will be one of the easiest to find the moment of inertia for, because the center of mass usually reflects the symmetry of the object. Moments of inertia on wikipedia.
If we know the moment of inertia of an object around an axis that passes through it's center of mass there is a theorem that can help us find the moment of inertia around a different axis parallel to the axis through the COM.
If the axis of rotation is a distance $h$ from the through the COM then
$I=I_{COM}+Mh^{2}$
where M is the total mass of the object.
The sum of the torques on the pulley will be given by
$\Sigma \tau =(T_{2}T_{1})R$
and as we saw that the moment of inertia for solid cylinder is $I=\frac{1}{2}MR^2$ we can find the angular acceleration of the pulley
$\large \alpha=\frac{\Sigma \tau}{I}=\frac{2(T_{2}T_{1})}{MR}$
This can be related to the tangential acceleration of a point on the edge of the pulley by multiplying by R as $a=\alpha R$
$\large a=\frac{\Sigma \tau}{I}=\frac{2(T_{2}T_{1})}{M}$
As we did before when we neglected rotation we should write Newton's Second Law for the two weights
$m_{2}a=m_{2}gT_{2}$ → $T_{2}=m_{2}gm_{2}a$
$m_{1}a=T_{1}m_{1}g$ → $ T_{1}=m_{1}g+m_{1}a$
$T_{2}T_{1}=m_{2}gm_{2}am_{1}gm_{1}a$
$\frac{1}{2}Ma=m_{2}gm_{2}am_{1}gm_{1}a$
$\large a=g\frac{m_{2}m_{1}}{\frac{1}{2}M+m_{1}+m_{2}}$
For any object which is symmetric around it's axis of rotation
$\vec{L}=I\vec{\omega}$
Otherwise the angular momentum along the rotation axis will be of magnitude $L_{\omega}=I\omega$, but there will be other angular momentum components.
For any problem I will ask you you can use $L=I\omega$
In the absence of a net external torque
$\Sigma \tau=\frac{dL}{dt}=0$
and angular momentum is conserved.
$L=I\omega=\mathrm{constant}$
Suppose I with two weights in my hands can be approximated by an 80 kg cylinder of radius 15 cm. My moment of inertia if I am spinning around an axis going down my center will be
$I=\frac{1}{2}MR^{2}=0.9\,\mathrm{kgm^2s^{1}}$
With my arms (which we approximate as 3.5 kg and a length of 0.75 m from my shoulder) extended holding 2.3 kg weights my moment of inertia will be considerably higher. When I am holding my arms and weights out I should remove their mass from the cylinder
$I=\frac{1}{2}68.4\times0.15^2=0.77\,\mathrm{kgm^2s^{1}}$
The moment of inertia of the two weights around the axis when my arms are extended is
$I=2\times2.3\times(0.9)^{2}=3.726\,\mathrm{kgm^2s^{1}}$
The moment of inertia of an arm if it is rotated around it's center of mass is
$I=\frac{1}{12}\times3.5\times(0.75)^2=0.16\,\mathrm{kgm^2s^{1}}$
But if it rotates around a point $0.375+0.15\,\mathrm{m}$ from it's center of mass then its moment of inertia is
$I=\frac{1}{2}\times3.5\times(0.75/2)^2+3.5\times(0.375+0.15)^2$
$=0.16+0.96=1.12\,\mathrm{kgm^2s^{1}}$
So the total moment of inertia due to the extended arms, weights and the cylinder is
$I=0.77+3.726+2.24=6.736\,\mathrm{kgm^2s^{1}}$
From the previous calculation we have
With arms in $I=0.9\,\mathrm{kgm^2s^{1}}$
With arms out $I=6.736\,\mathrm{kgm^2s^{1}}$
If I am rotating with angular velocity $\omega$ with my arms out and I bring them in then my angular velocity afterward is $\omega'$ and from conservation of momentum
$L=I\omega=I'\omega'$
$\frac{\omega'}{\omega}=\frac{I}{I'}=\frac{6.916}{0.9}=7.5$
Kinetic energy is defined as
$K=\frac{1}{2}mv^{2}$
but if we want to treat an extended object rotating around a moving center of mass this
is
$K=\frac{1}{2}mv_{CM}^{2}+\frac{1}{2}I\omega_{CM}^{2}$
To change kinetic energy a force must do work on the object,
$W=\int_{a}^{b}\vec{F}\cdot d\vec{l}$
if only conservative forces act the potential energy must change by an equal and opposite amount, which is a statement of the principle of conservation of mechanical energy.
For rotation
$W=\int \vec{F}\cdot\,d\vec{l}=\int F_{\perp}R\,d\theta=\int_{\theta_{1}}^{\theta_{2}}\tau\,d\theta$
For an object to roll without slipping there is a frictional requirement, it is actually not possible for an object to roll on a frictionless surface, it would instead slide. Note that the point in contact with the ground during rolling is actually motionless, so the relevant coefficient of friction is the static coefficient of friction, whereas for sliding the kinetic coefficient of friction is relevant. There is no work done by this force as it is directed perpendicular to the displacement.
The point around which the wheel turns which is also it's center of mass is seen to execute translational motion with velocity $\vec{v}$
If we instead consider the motion of a point on the edge of the wheel we can see it comes to rest, in the reference frame of the ground when it is contact with the ground. We can however also look at it from the perspective of the wheel, in which case the motion of the point of the edge of the wheel is purely rotational, and as the wheel sees the ground moving to the left with velocity the same magnitude as it moves forward in the grounds reference frame the rotational velocity of the wheel must be $\omega=\frac{v}{r}$ or $v=\omega r$.
Hoop
$\large K=\frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}=\frac{1}{2}mv^{2}+\frac{1}{2}mr^{2}\frac{v^{2}}{r^2}=mv^{2}$
$mgh=mv^{2}$
$v=\sqrt{gh}$
Disk
$\large K=\frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}=\frac{1}{2}mv^{2}+\frac{1}{2}\frac{1}{2}mr^{2}\frac{v^{2}}{r^2}=\frac{3}{4}mv^{2}$
$mgh=\frac{3}{4}mv^{2}$
$v=\sqrt{\frac{4}{3}gh}$
Recall that for a sliding object (without friction) $v=\sqrt{2gh}$
Solid sphere $I=\frac{2}{5}mr^{2}$
Hollow sphere $I=\frac{2}{3}mr^{2}$
Translation
$\Sigma F =ma$
$v= v_{0}+at$
$x= x_{0}+v_{0}t+\frac{1}{2}at^2$
$v^{2}=v_{0}^2+2a(xx_{0})$
Rotation
$\Sigma \tau=I\alpha$
$\omega= \omega_{0}+\alpha t$
$\theta= \theta_{0}+\omega_{0}t+\frac{1}{2}\alpha t^2$
$\omega^{2}=\omega_{0}^2+2\alpha(\theta\theta_{0})$
For an object to have no translational acceleration
$\Sigma \vec{F}=0$
and for this to be true the sum of each component of the forces must be zero.
An object may have no net force on it, but be able to turn, which means for an object to be in equilibrium
$\Sigma \tau = 0 $
Where the torques can be calculated around any axis. Some axes are more sensible than others, a good idea is to choose an axis at place where a force is acting (as the torque due to that force around that axis will be zero.)
$\theta=\tan^{1}\frac{30}{60}=26.57^{o}$
Vertical forces, up is positive $T\sin\theta+Fh_{y}=m_{sign}g+m_{rod}g$
Horizontal forces, right is positive $T\cos\theta+Fh_{x}=0$
Torques $0.6T\sin\theta=0.8m_{sign}g+0.4m_{rod}g$
$T=\frac{0.8\times3\times9.8+0.4\times1\times9.8}{0.6\times\sin26.57^{o}}=102.24\mathrm{N}$
$Fh_{y}=4.981102.24\sin26.57^{o}=6.53\mathrm{N}$
So force points down.
$Fh_{x}=102.24\cos26.57^{o}=91.4\mathrm{N}$
Force points to the right
The forces that act on the ladder of length $l$ are the weight of the ladder that acts down at it's center of mass, $m\vec{g}$, a normal force exerted by the wall, $\vec{f}_{NW}$, an normal force exerted by the ground $\vec{f}_{NG}$, and a force due to the friction of the ground, $\vec{F}_{Fr}$. The friction between the ladder and the wall is typically neglected.
For equilibrium the sum of the forces in both the horizontal and vertical directions must be zero and also the sum of the torques around an appropriate axis must be equal to zero.
Balance of the forces in the horizontal direction implies that
$F_{NW}=F_{Fr}$
and in the vertical direction
$F_{NG}=m_{l}g$
If we calculate the torques around the point at which the ladder touches the ground then they will be
$F_{Nw}l\sin\thetam_{l}g\frac{l}{2} \sin(90^{o}\theta)=0$
$F_{Nw}l\sin\thetam_{l}g\frac{l}{2}\cos\theta=0$
This can be written in terms of the frictional force by using the equation we got from the horizontal forces.
$F_{fr}l\sin\thetam_{l}g\frac{l}{2}\cos\theta=0$
The maximum force that can be provided by friction is $\mu F_{NG}$ and so the maximum safe angle is when this frictional force is provided
$\mu F_{NG}l\sin\theta_{max}m_{l}g\frac{l}{2}\cos\theta_{max}=0$
$\mu mgl\sin\theta_{max}m_{l}g\frac{l}{2}\cos\theta_{max}=0$
$\tan\theta_{max}=\frac{1}{2\mu}$
Of course if a person is standing on the ladder, their weight would need to be taken in to account both in the balance of the vertical forces and the torques.