1. Vectors have magnitude and direction. Scalars have only magnitude.
2. we are allowed to move vectors around as long as magnitude and direction stays the same – they “stay the same vector”
3. We add and substract vectors according to the parallelogram or triangle construction.
4. We construct a COORDINATE SYSTEM (1,2,3-dimensional) using an ORIGIN and UNIT vectors ( X, Y, Z). We can decompose VECTORS in COMPONENTS using these unit vectors.
5. Using a COORDINATE SYSTEM, we can “address” a point in space by giving its COORDINATES. For the 2-dim case, we have encountered CARTESIAN and POLAR coordinates.
6. We can represent the POSITION of a point in space by a POSITION “VECTOR” whose COMPONENTS are the CARTESIAN COORDINATES.
7. DISPLACEMENT denotes a vector which points from one point to another. DISPLACEMENT is INDEPENDENT of the choice of coordinate system, as are vectors in general.

A straight line curve has constant velocity, meaning it lacks acceleration. There are several important formulas:

• Sample Problem

  If you throw up a basketball, how long does it take to hit the ground? Gravity is 9.8m/s².

  x_i = +1m
  x_f = 0m
  v_i = +5m/s
  a= = -9.8 m/s² ≈ -10m/s²

  ½gt² – v_it + (x_f – x_i) = 0

  t_{1 or 2} = (-b ± √(b² – 4ac) )/(2a) = v_i &#177 √(v_i² – 2g(x_f – x_i) )/g

  DET: v_i² – 4½g(x_f – v_i) = 25m½/s½ – 20m/s½ (0 -1) = 45m½s½

  t_{1 or 2} = (5m/s ± √(45m/s) )/10m/s² = 1.17 s or -0.17 s

  v_f – v_i = -g(v_i ± √(v_i² – i₀) )/g

  v_f – v_i = -v_i &#177 -√(v_i² – 2g(x_f – x_i))

  v_f = √(v_i² – 2g(x_f – x_i) ⇒ v_f² – v_i² = -2g (x_f – x_i)

Important Note: if you throw up a ball with a certain v_i, and you thrown down a ball with the same v_i, they will both have the same velocity when they hit the ground.

Physics of Breakdancing

In breakdancing, a headspin is a basic power move (a move requiring tremendous strength to complete). Headspins are performed by engaging in a handstand, then lowering the body such that the head touches the ground. The breakdancer then lifts their hands and pushes off, causing rapid spinning on his or her head. The breakdancer stretches their legs as in the splits. By contracting or extending his or her legs, a breakdancer can make a headspin go faster or slower, respectively. This relies upon angular momentum and moment of inertia. Momentum measures how likely an object will change in position along a straight path at constant speed. Angular momentum measures how likely an object will spin. It correlates the mass and speed of the object to the radius of the circular path. Angular momentum is constant unless torque is applied. It can be used to measure torque applied to an object over time based upon differences between initial and final angular momentum. In the case of a breakdancer, there is a friction force decelerating the headspin.

The formula for angular momentum (L) is:

L = mass x velocity x radius

The formula for linear momentum (p) is:

P = mass x velocity

The radius is the distance from the point around which the object spins. In the case of a breakdancer with legs fully extended, this would be the distance from between his or her crotch and toes. For an object with constant mass rotating in a constant path, the angular momentum is the product of the object’s moment of inertia and its angular velocity vector. I is the moment of inertia of the object and ω is the angular velocity.

L = I x ω

Sample Problem

If somebody is 60kg and 1.8m tall, then their legs are approximately .9m long. This means that the breakdancer will have a diameter of 1.8 m from the tip of one toe to another. This problem will describe the motion of a breakdancer performing a headspin for 1.5 seconds. The coefficient of kinetic friction between leather and steel is .6, so this problem will assume the coefficient of kinetic friction between a scalp and a linoleum floor is .4.

Friction causes a decelerating force of:

F = 60 x 9.8 x .4 = 235.2N

Therefore, it decelerates a breakdancer by:

235.2N = 60 x a
a = 3.92m/s2

In order to remain in motion for 1.5 seconds, the breakdancer must therefore begin with a velocity of:

0 = vxi + 3.92 x 1.5
vxi = 5.88m/s

The angular momentum will be:

L = m x r x v = 60 x .9 x 5.88 = 323.52 kgm2/s

If the person bends their knees to cut split the length of their legs, then their radius is cut in half. As a result, their velocity doubles to become 11.76 initially. Friction will require the breakdancer to eventually apply more fore to maintain high speed. However, by pulling in his or her legs the breakdancer will be able to move faster with much less force. This is due to angular momentum.

So, by decreasing radius, linear momentum increases to maintain a constant angular momentum. Since the person does not get significantly fatter while breakdancing, their velocity must increase to accommodate a constant angular momentum.

Power = work/time = W/t

The faster work gets done, the greater the power. If there is constant v and force is parallel to path, then P = Fv

Energy is the ability to do work. Moving objects inherently have energy because they can crash into something and exert a force over a distance. Kinetic energy is this energy of motion.

Fd=.5mv²

W=.5mv²

KE = .5mv²

This assumes the initial speed of the object is 0. If it were not, then W = KE_final – KE_initial.W^total = ΔKE.

Â

Potential energy is energy of an object due to position. There is gravitational, electrical, and elastic potential energy. Potential energy is placed on position in a gravitational field. ΔPE_grav=-W_{by Fgrav}=mgh. If the brick had fallen, it would have been -mgh.

PE = -mgh

Gravity is a conservative force.

Linearly polarized light transmitted to parallel to polarizing axis, ampitude reduced by cosθ factor. Amplitude E₀cosθ

Intensity proportional to amplitude²

I = I⁰cos²Î¸ (Malus’ Law)

I⁰ = intensity of incident polarized light
I = intensity of transmitted polarized light.
θ = angle between incident electric field and polarizing axis.

I^polarized = ½I^unpolarized

Example: what is the angle between the axes of the two polarizers if the original unpolarized beam intensity is reduced by 90% after passing through both sheets.

I^unpolarized | I⁰ / I

I⁰ should be expressed in terms of I^unpolarized amd I in terms of I⁰

I⁰ = ½I^unpolarized

and I = I⁰cos²θ=½I^unpolarizedcos²θ

We want I = 1/10 I^unpolarized = ½I^unpolarizedcos½θ

Reflection:

incident ray           reflected ray
   \           |             /
    \          |normal  /
     \         |          /
      \        |         /
       \       |        /
        \      |       /
         \     |      /
          \    |     /
           \   |    /
            \  |   /
             \ |  /
_________\|/_________surface

θ₁ = angle between incidden ray and normal ro tusrface (angle of incidence).
θ₁^‘ = angle between reflected ray and normal to surface (angle of reflection).

Index of refraction: n = indext of refraction = speed of light in vacuum/speed of light in medium. c/v≥1

n(air) ≅ 1
n(water) ≅ 1.33

v = c/n = fλ_medum

As light passes from one medium to another, the frequency remains constant because it still emits pulses at the same rate. The value of c also remains constant. Therefore, the ratio c/f is a constant. However, c/f is also nλ_medium. Therefore, nλ_medium should also remain constant. n₁λ₁ = n₂λ₂. This tells how wavelength changes from medium 1 to medium 2.

Refraction: the apparent bending of light at an interface beteen two media (eg, air and water). n₁sinθ₁ = n₂sinθ₂ (Snell’s law of refraction)

n₁ = refractive index of meidum with incident ray
n₂ = refractive index of medium with refracted ray
θ₁ = angle between incident ray and normal to surface
θ₂ = angle between refracted ray and normal to surface

Two Cases:

1. n₂>n₁ (eg, air to water)
n₁sinθ₁ = n₂sinθ₂
sinθ₂<sinθ₁
θ₂<θ₁ (light bends toward normal)

incident

\θ1 airn1
——————————————————————-
θ2 | watern2

glass and air for case 2

2. n₂<n₁ (eg, glass to air)
θ₂>θ₁ (light bends away from normal)

Total internal reflection can occur only in Case 2 when n₂ < n₁. It only happens under certain conditions.

What is the critical angle of incidence, θ₂, at which a refracted light ray travels along the boundary between the media. We use Snell’s Law: n₁sinθ₁ = n₂sinθ₂

θ₁ = θ₂ = 90

Total internal reflection

3 x 10⁸ m/s in air
Speed of light in any medium is speed of light in vaccum times refractive index of that medium.

Flat mirrors. Simplest case: point source (or object) O forms image I.

                                  |
__________________________________|____________________________________Oobject
                                  |
                                  |
                                  |
                                  |     /
                                  |    /
                                  |   /
                                  |  /
                                  | /    θ1
                                  |/__________________
                                  |\
                                  | \    θ1
                                  |  \
                                  |   \
                                  |    \
                                  |     \
                                  |
                                  |
                                  |

p=distance between object and mirror. (object distance)
q=distance between image and mirror (image distance)

Virtual=image formed at point from which reflected light rays appear to emanate, cannot be displayed on a screen.

Multiple reflections. When mulitple mirrors are present, the image formed ab one mirror acts as the object for reflection from another mirror.

Spherical mirrors. Concave=object O and center of curvature C are on he same side. 2. convex=o and c are on opposite sides.

F_grav = G (Mm/r²)

F_net = ma

• F_net is the sum of all forces acting on the object.
• If F_net = 0, then a = 0.
• A force of 1 kgºm/s² is 1 netwon (N).