Every object in the universe exerts a gravitational pull on every other object. Magnitude of gravitational force is ∝ to the product of the objects’ masses. Magnitude of graviational force is inversely ∝ to the square of the distance between them. Constant of proportionality is defined as G (Newton’s universal gravitational constant).
Fgrav = G (Mm/r²)
Key Equations
| c = fλ | f = w/2π | k = 2π/&lambda | E = Emaxcos(kx – wt) | B = Bmaxcos(kx – wt) | c = E/B | I = E²max/2µ0L | I = cB²max/2µ0 | p = s/c | s = E x B/µ0 | p = 2s/c | I = I0 cos²θ |
| θ1‘ = θ1 | n1sinθ1 = n2sinθ2 | λn = λ0/n | M = h’/h | p-1 + q-1 = F-1 | f = ½F |
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.
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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.
Geometric optics: ray aproximation = a wave moving through a medium trvels in a straight line parallel to its rays.
Reflection:
incident ray reflected ray
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_________\|/_________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 = fλmedum
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 nλmedium. Therefore, nλmedium 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
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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
Linearly polarized light = electric (E with vecto) field vibrates along specific axis within a plane perpindicular to propagation. UNpolarized light goes out. First polarized light along the polarizing axis. linearly polarized light transmitted to polarizing axis. Second polarizarer (nalyzer)
Linearly polarized light transmitted to parallel to polarizing axis, ampitude reduced by cosθ factor. Amplitude E0cosθ
Intensity proportional to amplitude2
I = I0cos2θ (Malus’ Law)
I0 = intensity of incident polarized light
I = intensity of transmitted polarized light.
θ = angle between incident electric field and polarizing axis.
Ipolarized = ½Iunpolarized
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.
Iunpolarized | I0 / I
I0 should be expressed in terms of Iunpolarized amd I in terms of I0
I0 = ½Iunpolarized
and I = I0cos²θ=½Iunpolarizedcos²θ
We want I = 1/10 Iunpolarized = ½Iunpolarizedcos½θ
Power measures howquickly work gets done. For example, a force of 100J of work in 20 seconds is being done at a rate of 100J/20s = 5J/s. This is the power (P), and from sample calculation above the units are obviously J/s. This is defined as the watt, W, which should not be confused with the W which defines work.
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 = KEfinal – KEinitial.Wtotal = ΔKE.
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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. ΔPEgrav=-Wby Fgrav=mgh. If the brick had fallen, it would have been -mgh.
PE = -mgh
Gravity is a conservative force.
Levi Meir Clancy
SID 703 254 175
Physics 6AH
Instructor: Dr. Wallny
Final Paper (03.07.06)
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.
Two vectors are equal if they have the same units, magnitude, and direction.
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.
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