what is the name of the angular velocity of a star?
929160
I assume you're referring to the apparent motion of a star, not its spin rate. The apparent angular rate of a star on the celestial sphere is called proper motion (see ref.).
731964
There is no force that causes the planets to rotate. Most of the rotation comes about from the conservation of angular momentum. Angular momentum is given by L=m*w*r2 where m is the mass, w is the angular velocity in radians per second, and r is the radius of the circular motion. Due to conservation of angular momentum, if the radius of the orbit decreases, then its angular velocity must increase (as the mass is constant).\n\nAll planetary and stellar systems are born from the collapse of dense interstellar clouds. The clouds may originally be very large (even thousands of light years across). Consider a portion of the cloud the collapses from a size of a light year or so to the size of the solar system. That is a huge change in the size of the system. So, the very slight rotation that the cloud has in the beginning is increased dramatically when the collapse takes place. In fact, this is one of the barriers in star formation: there is excess angular momentum and there has to be a way of losing angular momentum before you can form a star.\n\nAnyway, the bottom line is that stars like the Sun spin from the original angular momentum that was there in the solar nebula from which it formed. Not only that, all orbital motion of the planets (including the spin) is due to this orginal angular momentum.\n\nYou are saying that original angular momentum of the cloud causes orbital motions and rotations of the planets(mostly). But in the case of orbital motions we have gravitational force that gives us some restrictions of movement(Kepler laws,for example).\n\nWhat I am saying is that there will be no planets if there was no initial angular momentum in the primordial solar nebula. If a nebula with absolutely no rotation collapses, then there will only be a central non-rotating star and there will not be any planets. Planets form out of a protostellar disk, which itself forms only because of the initial angular momentum of the cloud. The dynamics of a rotating body is of course controlled by forces like gravity. Kepler's laws are a direct consequence of gravity. \n\nAll the planets in our solar system spin.
26747
ok.  call the downward translational velocity of the whole thing v.  this means that the tangential velocity at the edge of the poles (at the radius r) is equal to v.  the reason for this is that the strings are wound AROUND THE POLES, so this winding and the condition that the string has to unwrap for the thing to move, sets the angular velocity w = v/r.\n\nnow, if you look at the main wheel part, the angular velocity has to be equal to the angular velocity of the poles, since they are one solid piece.  however, as you note the radius of the wheel R > r.  so, the fact that v / r = w = V / R implies that V > v.  in other words, the motion of the outer edges of the wheel is faster than the motion of the outer edges of the rods.\n\ni think you are getting confused by saying that V is the translational velocity of the wheel.  it's not.  v is the translational velocity of the wheel, and of the rods.  the fact that the string is wound *around the rods* and not around the wheel sets the angular velocity of the whole apparatus equal to v/r.  V, the tangential velocity of the outer edges of the wheel, is much faster than the translational velocity.\n\nthat clear?
884559
There are a number of angular equvalents in physics.  Here are some:\n\nDistance, x (meter) is angle, b (radian)\nForce, F (Newton) is torque, T (Newton-meter)\nVelocity, v(meter/second) is angular velocity, w (radian/second)\nAcceleration, a (meter/second^2) is angular acceleration, z (radian/second^2)\n\nSo far these are all straight forward.  \n\nTo make Newton's law (F=m*a) work in angular terms, we need to define an angular mass.  It is defined by considering all the differential masses in an object and how they under go angular acceleration when subjected to a torque (an angular force).  Without too much trouble, you can show that this angular mass is:\n\nS r^2 dm (S means integration)\n\nThis mass property is called moment of inertia and is represented by I.  It just takes into account the fact that mass located farther from the axis is harder to accelerate.\n\nWith that defined, all the other properties fall into place.\n\nMass m (kg)  becomes moment of inertia I (kg-m^2) \nNewton's Law F=ma becomes T=Iz\nKinetic energy E = (m*v^2)/2 becomes E = (I*w^2)/2\nMomentum P = m*v, is angular momentum H = I * w
140474
Use the formula Ac = v^2/r = r w^2, where w = angular speed.\n\nSince you already have the angular speed, you can use the second form of the equation above.\n\nAc= r w^2 =(1.0 m) (2.0 rad/s)^2
232256
Same direction as the Sun and the Moon, to the west.\n\n---Addition----\nChristina has a good point. If you were close to one of the poles, the stars would be just turning around and you could not tell if it is west or east direction.
70579
Interesting. I vaguely remember a "great attractor" from chaos theory, but what I find when I look it up now is that it refers to a cluster of galaxies, "in the direction of Centaurus and Hydra" where nobody is sure why they're grouped like that.\n\nHere's a direct quote from another source though which I think directly addresses what you're asking about: The galaxy rotational problem refers to the phenomenon of the descrepency between the observed rotational motion of the matter of the galactic disc from the predictions of Newtonian dynamics. (See links below and search for "12.25.I".)\n\nI think in layman's terms what they're boggling about is the "bar" shape that occurs in most galaxies, our own apparently included. A recent issue of Scientific American had a very good theory on how that occurs and why.\n\nIf I'm getting the gist of it right, they're not sure why clusters of stars occur in a semi-pinwheel pattern, since based on observed mass there simply isn't enough to cause such a structure to form or remain stable.  The theory in SciAm had something to do with a coincidence of travelling wave patterns rather than any deep dark mysteries of physics, filiments, or anything else.\n\nSee http://www.sciam.com/, October 2005 issue, p. 43, "Ripples in a Galactic Pond." Very nice article and they explained it more clearly than anything I could put together.
541140
Well if you look at the spectrum of a star you'll see a smooth continuum as well as the spectral lines.  Those lines are either absorbtion lines or emission lines of such things as Hydrogen, Helium, Calcium, Iron and a whole heap of others and we know the wavelength of those lines very well so by comparing what we measure them to be with what we have determined in the lab we can calculate how fast the star is moving towards or away us (radial velocity).\n\nThen once we know the radial velocity we just measure where the peak in the continuum is, correct for the radial velocity to find what the wavelength where the emission actually peaks and then since the star is very close to a blackbody we can calculate the temperature (a blackbody of a given temperature will peak at a given wavelength).
1145113
This should help you:  http://www.le.ac.uk/physics/faulkes/web/stars/o_st_overview.html
847656
The mass is moving parallel to the x-axis with a velocity of magnitude, say v. then its velocity vector is V(vector) = vi,i is the unit vector along x-axis. Its momentum vector is p(vector) = m*V(vector).\n\nAt any instant let the mass be at A(x,y) so that OA(vector) = r(vector) [ O is the origin].\nthen by definition, the angular momentum,\n\nJ(vector) = r(vector) X p(vector) [ X implies cross product)\nor, J(vector) = r(vector) X m*v(vector)\nor, J(vector) = m( xi + yj) X vi\nor, J(vector) = mvy ( -k) [ since iXi=0 & jXi= -k]\n\nsince m,v,y are constants,[y is const because the particle is moving parallel to x-axis so its ordinate will remain const.] the angular momentum J will remain constant & its magnitude at a pt (x,y)is mvy ; direction negative z- axis [v= the velocity of the mass] .(ans)
215475
The Earth is rotating around an axis (called its rotational axis). Some objects rotate about a horizontal axis, like a rolling log. Some objects, such as a skater, rotate about a vertical axis. The Earth's axis is tipped over about 23.5° from vertical.\n\nHow do we define up and down in space? What would "vertical" mean? For the Earth, we can think of vertical as straight up and down with respect to the plane in which the Earth orbits the Sun (called the ecliptic).\n\nEarth's rotational axis points in the same direction relative to the stars, so that the North Pole points towards the star Polaris. Think of the Earth as a spinning top, tipped over to one side. Over very long time periods (thousands of years) the direction of Earth's axis slowly changes due to precession.\n\nThe Earth rotates around once in 24 hours - that's a rate of 1000 miles per hour!. The time it takes for the Earth to rotate completely around once is what we call a day. It's Earth's rotation that gives us night and day.\n\nThe combined effect of the Earth's tilt and its orbital motion result in the seasons
157772
Except for stars nearest the Pole Star, all stars move in one single direction all the time --- east to west. This isn't because the stars are moving but instead because Earth is rotating.\n\nIf you're seeing stars moving 'rapidly' and 'sporadically' you're either not observing stars, or you should tell me where you're gettin' that good stuff :)
161310
Motion towards or away from the observer. This motion in effect changes the wavelength of the light emitted by the star, and it's wavelength that determines the color of light.\n\nA picture is worth a thousand words, so check the animated graphics at this website:\nhttp://www.colorado.edu/physics/2000/applets/doppler.html\n\nNOTE..!  The graphics at the website are for sound waves, but light waves behave the same way.
334497
Astronomers, for the longest time, had been puzzled over some objects in the sky.  On photographic plates, there were stars, planetary nebulae, and some mysterious objects.  They wouldn't resolve themselves into singular stars.  And some of them were shaped like spirals.  This gave the mystery objects the name 'spiral nebulae'.  In the 1920s, Edwin Hubbell was examining one of those spiral nebulae called the 'Andromeda Nebula'.  He managed to find something that astronomers here had known for a while: a Cepheid variable.  This is a star that pulses very slowly.  These stars have very predictable pulsations.  The longer the pulsation period, the more bright the star is.  It's called the 'period-luminosity relationship'.  In any case, when he found a Cepheid variable, he was able to know what the absolute magnitude of the star should be.  So he was able to measure the apparent magnitude, and compare it to the predicted magnitude, and he got a distance of 2 million light-years (he used the distance modulus equation).  So he knew that this 'nebula' COULDN'T be in our galaxy, and that he had found a new 'island in the sky'.
535952
Assuming a big ring that rotates, we need to fix the radius and the velocity out on the edge.  The acceleration toward the middle (which causes the feeling of gravity) is given by\n\na = v^2/r\n\nTo match gravity on earth, assume 9.80 m/s^2\n\nYou can make the radius what ever you want.  Assume 100 meters.\n\nThen 9.80 = v^2/100.  So, v^2 = 980.  v = 31.30495.\n\nSo, how fast is it spinning?\n\nThe circumference is 2*pi*r = 2*100*3.14159 = 200*pi.\n\nThe time it takes for it to do one rotation is:\n\n200*pi (meters)\n--------------------\n31.30495 (meters/sec)\n\n= or 20.07088 seconds.\n\nThe angular velocity is 360 deg/20.07088 seconds = 17.9364 deg/sec\n\nor, 2pi radians/20.07088 = .3130495 rads/sec.\n\nNotice, this is exactly what you would have gotten by taking v/r and getting the angular velocity.  I don't know if you are studying that part of rotational dynamics yet.\n\nSo, the bottom line is, as long as v^2/r = 9.80, you can then get the rotational velocity in radians from v/r.
753124
The centripetal acceleration of circular motion a = w*2*r, where w is the angular velocity.  Since you know a = 16m/sec^2, and r = 1m, then w=√[a/r]  The tangential speed = r*w.  In these formulas w is in radians/sec. The frequency is in cycles (rotations) per second, one rotation = 2π radians.  The period is one over the frequency.
660999
THe orbital angular momentum is determined by the number L (usually a small l, but I want to be clear here). For L=0, you have an s orbital, for L=1 a p orbital. In general, the orbital angular momentum is given by L*hbar.