Simple Neutonian physics is what makes up the rigorous, repetitive world of simple mechanics. As such we need to know the basic formula of kinematics. Those who have taken calculus know the basic kinematics formula then yields two other formulas: velocity and acceleration. We can then use these formulas to solve for almost any simple action if we neglect to include certain forces. Its ok though start getting used to being slowly brought up to the real world of physics. Lets begin by looking at the basic kinematic formula of motion.
X = ½ at² + vt + xX = distance traveled
a = acceleration
t = time
v = starting velocity
x = starting position
The first thing that needs to be taken into consideration is the fact that the starting and ending position all depends on the coordinate system that you set up. Second, we must also keep in mind the units we are using for each variable. I stick with the standard kms(kilogram, meter, second) scale. This has the following units for each variable...
X = metera = meter per second squared
v = meter per second
t = second
Taking the derivative of the position formula above yields the resultant velocity.
V = at + v
When i was taking this class, and to this day, I basically make a table to understand my knowns
and my unknowns. The table puts everything i know at the beginning and everything i know at the end.
| x0 = | xf= |
| v0 = | vf = |
| a0 = | af = |
| t0 = | tf = |
Now that we have some formulas to work with, the next thing we need to understand are the forces acting on a body.
To do this we will now look into a free body force diagram(FBFD). In order to do this, we must isolate
the body in question and then calculate the forces on it. To understand what goes onto a FBFD we must understand Newton's second law which states
The sum of the forces acting on an object equals its mass times its acceleration.
F = ma
All bodies have one force acting
on them, this is gravity, and it always acts in the down direction. Below are three
examples masses and their FBFD.
The group 'A' is just a simple mass only being effected by gravity. In practically it is a mass in free space. The FBFD associatd with it is drawn with the mass being represented by a point mass which is a dot in the center of the FBFD and the forces acting on it represented by arrows and their forces. Since gravity is the only acceleration, Mg is the only force acting on it.
Figure 'B' is the same mass but this time it is resting on some sort of object. The FBFD looks almost identical to the mass in free space except we have added an additional force, N. This is the normal force and is force created from Newton's third law, When one body exerts a force on another, the second exerts a force on the first. These two forces are always equal in magnitude and opposite in direction. Basically the table exerts a a force on the mass to keep it from falling.
Figure 'C' is the same as figure 'B' except for the fact we have now put the block on an inclined plane. The dotted line is for reference. It is easier to consider the summing of the forces on an inclined plane in the parallel and perpendicular directions to the plane.
These examples are of basic stationary blocks. Most of the time things are not stationary.
When adding additional forces to the FBFD, remember that the only thing that goes on it
are forces. A block may slide down a hill, but if there isn't a force pushing or pulling
it then there isn't an arrow in the direction of movement. When you solve the formula
for sum of the forces in the x and y directions, you will get an acceleration and a movement.
Let's also look at an aspect involved with motion called friction, ƒs for static friction and
ƒk for kinetic friction. They are listed like this on a FBFD and point in the
direction opposite to translation.
ƒs,k = µs,kN
The µs,k are the coefficients of friction and are dependant on the two bodies
that are in contact with each other.
Now that we got this down lets look at some other aspects of Newtonian mechanics.
Let's say you have a rotating object like a rock being twirled in a sling above your head. The rotational acceleration associated with the rock is v2/r. This means that your FBFD associated with the rock will have mg pointing down and Tension pointing toward the center of rotation. This will then equal mv2/r. You can now solve the force associated with the tension in the string and see how fast something would have to spin to hold up a mass.
Let's now look at momentum, p.
F = dp/dt.
p = mv
Impulse, J = dp = Fdt, is the change in momentum, like when
a bat strike a baseball. The baseball travels towards the bat at a momentum, then the bat strikes
the ball with a force equal to the change in momentum divided by the time that the ball is in contact
with the bat.
A useful rule to remember with respect to momentum is the conservation of momentum.
The total momentum of a system at the beginning is equal to the total momentum
of the system at the end.
The next thing i will explaining is rotational qualities. This is simple because it relates
to non rotational mechanics. Remember the table at the beginning, well here it is with rotational
figures. You can also replace the values in the kinematics formulas the same way i replace
them in the table. Distance transversed for this easy exchange must always be in the
form of radians. This also yields some easy calculations, gotta love radians. The following
formulas are translational interpretations of rotational quantities.
x = r * θ
v = dr/dt * θ = ω * θ
a = d2r/dt2 * ω = dω/dt * θ = α * θ
| θ0 = | θf = |
| ω0 = | ωf = |
| α0 = | αf = |
| t0 = | tf = |
Electromagnetism: E-M
This is the hardest and most complicated physics class that you will attempt to learn in your first two years of college. I have only taken it once and got a flimsy grasp of it, but will try and explain it the best that I can. To give some basic ideas of E-M, rubbing a ballon on your head induces a charge into the balloon that reacts with the carge of your hair and causes it to adhere to it. Another example is rubbing your feet on the ground and touching a door knob to release a charge.The first thing that must be realized and, in my opion, memorized, is the charge of an electrone since this is the basis for E-M.
e = 1.60 X 10-19Coloumbs
Charge is in the units of Coulumbs and is always found in quantities of integer values of e.
q = ne
Electric charges radiated a field known as an E-field. It is defined as the force per unit charge.
E = F/q0
Since all calculations that are done at this level are based on forces, the question that should arise is what is the force associated with charge. Te following will demonstarte how to determine forces associated with 2 point charges
F = kq1q2/r2
k in this formula is the Coloumb constant and no to be confused with the Boltzman constant.
k = 1/(4πε0) = 8.99 X 109Nm2/C2
'r' is the distance between the two charges, q1, q2.
This yields the result of an E-field with the following form with respect to how far away form the charge you are.
E = kq/r2
Let's now take a look at my favorite law in E-M, Guass' Law. First we have to do some priliminary explinations.
EFlux = ∫E•dA]
dA represents the area of the Guassian surface that you created. The idea of the guassian to to choose a shape that will yield the E-field exits the surface at 90° so that EFlux simplifies to E * dA since E•dA = E * dA Cos[theta]. By choosing the surface to allow the E-field to exit at 90°, that yields Cos[90] which equals 1.
Now for Guass' Law
ε0 * EFlux = q
Just to prove that this works and what i said about E = kq/r2 is true lets assume we are going to find the E-Field of a point charge using Guass' Law. We known a point charge is spherical in shape, as such we will choose a spherical guassian surface to surround the charge.
ε0 * ∫E•dA = q
The E-field will be constant ove rthe entire surface of the Guassian surface since we have choosen a symmetrical surface, thus it can be pulled out of the integral.
z0 * E * Integral[dA] = q
The Integral of dA is merely A, thus
ε0 * E * A = q
For a sphere, A = 4πε2 so...
ε0 * E * 4πε2 = q
And finally, solving for E
E = 1/(4πε0) * q/r2 = kq/r2
This is the primary way to find the charge of the E-field. It is easy and fast compared to the way they did it before Coloumb.
Some valuable simplifications that must be remembered for certain circumstances are as follows:
For a dipole situation, relationship between 2 charges seperated by a distance, 'd', there is a moment, p.
p = qd
For a uniform line of charge let's consider a rod of length L and the linear charge density, lamda.
λ = q/L
Finally, we will look at the situation of a surface like a sphere of charge.
σ = q/A
The next section of E-M gives an introduction to circuits and electricity. This will be explained in my Basic Electrical Circuit section
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