Disco Ball -- How fast do the dots move? - find rotating disco ball light in ottawa
A 1 m radius disco ball in the middle of a room located at 10 square meters. The ball rotates once every ten seconds. A light is 1 meter from the ball and shine on the ball.
Find the equation of the velocity of points of light traveling through the wall parallel to the boundary between light and the ball.
Here is a picture: http://i278.photobucket.com/albums/kk114 ...
(The things we believe when in a bar and drink too much, lol)
Wednesday, February 17, 2010
Find Rotating Disco Ball Light In Ottawa Disco Ball -- How Fast Do The Dots Move?
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2 comments:
x = distance between the point of the position of the source.
Θ = angle of rotation of the plane element in the ball
θ = 0 if the dwelling unit is located directly opposite the light.
http://i675.photobucket.com/albums/vv116 ...
x = 2 - cosθ - (5 - sinθ) / tan (α + 2θ)
2sinα where = sin (α + θ)
x = [2 - cosθ] - [(5 - sinθ) * (2 - cosθ - sinθtan2θ) / (sinθ + (2-cosθ) * tan2θ)]
The speed of the point x can be found, knowing that by differentiating
θ_dot = 2π/10 rad / sec
Reception
v = [2π/10] * [45 - 30cosθ - 8sin ³ θ θ] / [1 - 8cosθsin ³ ³ 15cos θ + θ - 16cos ⁴]
Cosθ ultimately the power of 4, if not to show the team the source.
We would not expect the speed ofPoint remain constant, if the ball is made of flat panels.
* EDIT *
Oh, that is 0.1 m? I have not seen that. Then the equations:
5V / π = NUM / DEN
Num = [10 - 2R (1 + r) sin θ + 5r ² ³ (4 + 3R - 3 (1 + r) cosθ)]
DEN = [(r - 2 (1 + r) cosθ) sinθ] ²
r = 0.1
5V / π = [10 - 2R (1 + r) sin θ + 5r ² ³ (4 + 3R - 3 (1 + r) cosθ)] / [(r - 2 (1 + r) cosθ) sinθ] ²
The plots of x and θ with respect to V are shown below.
http://i675.photobucket.com/albums/vv116 ...
http://i675.photobucket.com/albums/vv116 ...
The average velocity of the point through the wall is 8,22 m / s
For purposes of this problem, the ball is a flat mirror object (at this point in time) at an angle of 45 degrees against the light, the light makes a right angle, and crashed. It runs at a certain speed Omega.
The angle of the mirror relative to the beam is as follows:
thetaincident = pi / 4 + nt
The light is at an angle to the expression:
thetareflected = thetaincident = 2 * pi / 2 + 2 omega t
The angle with the wall:
thetafinal = thetaincident - pi / 2 = 2 NT
Tan (thetafinal) = distance to the wall of the place / distance to the wall
So, (the position of the points of a point on the wall next to the mirror is measured):
y = distance from the mirror on the wall * tan (2 omega t)
dy / dt = distance between the mirror on the wall2 * omega ^ 2 sec (2 omega t)
dy / dt (@ t = 0, distancetowall time of interest) = 2 * omega
But the mirror itself also moves parallel to the wall. The speed is:
= Vmirror rball * omega * sqrt (2) / 2
One must add that dt dy / for the velocity:
Overdrive = omega * (rball * sqrt (2) / 2 + 2 * distancetowall)
distancetowall = L / 2 - rball * sqrt (2) / 2,
where L is the length of the side of the square room
* Overdrive Omega = (L - rball * sqrt (2) / 2)
= (2 pi / 10 seconds) * (10m - (0.1 m) (, 707))
= 6.22 m / s
Oh, of course, is not the speed constant. I was looking for speed at the time that lies at the maximum speed.
In addition, for what it's worth, I had a ray of light in the paragraphTo say Hallel (light source that so far). I looked carefully drawing Remo and I realized that I looked up I, near D's.
Moreover, D is the radius of the ball 1 / 10 of a meter, not 1 meter. Remo, why do you always write a zero before the decimal point, which is not lost.
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