a centrifuge rotor is spinning  at 3000 rpm. what is the angular velocity in radians per second?
364675
3000 rev per minute = 6000 pi radians/min = 6000/60 pi radians per second\n\n= 100 pi radians per second
807395
http://eta.physics.uoguelph.ca/tutorials/torque/Q.torque.inertia.html\n\ngives 2I=mr^2  so I = 163*0.306*0.306/2\n\nhttp://buphy.bu.edu/~duffy/py105/notes/AngularMo.html\n\ngives 2 * kinetic energy = Iw^2\n\nso KE at 1500rpm = I*4/2*pi^2*(1500/60)^2 joules - everything is frictionless so this is the work done in accelerating the flywheel to 1500 rpm\n\nyou can work out the KE at 1181 rpm and find the difference that was dissipated as heat.\n\nbest of luck - Mike
793054
The moment of inertia of the disk is 0.5*m*r^2 (ref.). (In kg-m-s units, m=.017 and r=.06.) The average angular acceleration A=23/0.86 rad/s^2. The torque T is the product of moment of inertia and angular acceleration. T=0.5*m*r^2*23/0.86 N-m.
473508
The revolutions per minute that an optical disk drive spins is dependent on which one you buy. They are given in the form of (number)x. X is 150 MegaBytes/second. So the speed is the number times 150 MB/sec. RPM specifications must be acquired from the manufacturers.
512039
I hope this is what you were looking for: \n\nLet A = your acceleration(RPS^2)\nLet D = your deceleration(RPS^2) (use positive value)\nLet RPS = your target velocity in RPS\nLet RPM = velocity in RPM\nLet N = the number of revolutions the motor needs to turn\nLet t (sec) = time needed\nSqrt = square root. Make sure you pay attention to the brackets.\n\nThen RPS = ( -t + Sqrt(t^2 + 2N(A+D)/AD))x AD/(A+D) \n\nAnd RPM = RPS/60\n\nI believe this formula is correct since I checked that both sides of the equation have the units of velocity.\nHowever, I have noted that many people on this site give out incorrect info. It is always a good idea to get a second source confirmation. Perhaps a servo motor manufacturer or distributor could provide the info and give helpful tips for a specific application. Also physics or math faculty at most colleges and universities are pretty co-operative when it comes to requests like this. If you fax the details they'll usually respond. Best wishes!
37325
first thing you need to do is calculate the moment of inertia of the wheel.  this can be done by treating the poles and the main wheel separately.  the moment of inertia of a circular disc (the wheel) is (1/2)MR^2, and of the rods (also a circular disc!) is (1/2)mr^2.  add them together to get the total moment of inertia.  \n\nyou'll need to use conservation of energy.  specifically, when the wheel drops a height h, the potential energy lost is -mgh.  this energy gets split into two parts, the rotational energy of the wheel and the translational energy (since it's moving downwards).  you can then write an equation setting the potential energy loss equal to the sum of the rotational and kinetic energies:\n\n-E_p = E_rot + E_trans\n-mgh = (1/2)Iw^2 + (1/2)(m + M)v^2\n\nhere I used (1/2)Iw^2 for the rotational kinetic energy and (1/2)mv^2 for the translational kinetic energy, where I substituted (m + M) for the total mass of your object.\n\nthe key point is now to realize that because the wheel unwinds along a string that is wrapped around the pole, the angular velocity w = v/r where r is the pole's radius and v is the translational velocity of the pole.  so you can substitute that into your expression.  it's not clear from your question whether you want the angular acceleration or the translational acceleration.  so you can either substitute the rw in for v, or v/r in for w.  \n\nin either case you then have a formula for the velocity as a function of the height of the wheel, which you can then solve for the velocity, and then take the derivative to get the acceleration.\n\ngood luck!
890959
I think I know what is confusing you. Total angular momentum of your system M=M1+M2=I1*w1+I2*w2, while I1=I2, w2=-w1, Thus M=I*0=0. Angular momentum is a vector, because w (angular speed) is a vector. How could we describe the difference of the system with w=0 and w not equal to zero? If w=0, then kinetic energy of the system E=0. When they are rotating, kinetic energy of the system E=E1+E2=I1*w1^2/2+I2*w2^2/2, which is not 0, as both w’s are squared. That means you have to do some work to obtain a certain angular speed w no matter what direction disks would spin.
819170
7 pi radians/second = 3.5 revolutions per second.\nCircumference =  20 ft * pi.\n\nLinear velocity = 3.5 * 20pi  = 70pi ft/sec.
1036388
Torque is given by the product of moment of Inertia and angular acceleration. \n  \nR - Radius = 13/2 inch =0.1651m\nMoment of inertia of the disc is ½ M R^2. = ½ x 15 x 0.1651 x 0.1651 \n=0.204435 kg m^2.\n\nAssuming that the disc is stopped in t seconds by the applied torque, the angular acceleration is 1800 x 2 x pi/ 60t radian per second^2 =188.49556/ t (radian per second^2)\n\nTorque = 0.204435 kg m^2 x 188.49556/ radian per second^2\n..............= 38.535/t Nm.
810633
Well, convert 720 rpm into seconds\nYou get 12 rps, right?\nWell, anglular velocity is change in angle divided by change in time\nSo you got 12 revolutions per second, right?\nThe change in time is obviously 1 second\nWell, recall how many radians in a revolution?\n2*pi radians in a circle = 1 revolution\nSo you have 12 revolutions, so whats 12*2*pi?\n24 * pi!!\nSo the answer is 24 pi rad/sec\n\nThe anglular velocity is not dependant on the radius, actual velcity depends on that\nSo in case you want the actual velocity, the equation is:\nangular velocity * radius\nSo in your case the radius is 6 (diameter/2)\nSo you have a pretty decent angular velocity (144 m/s)
797003
ω = rotational speed = v/r = 3.6/.2 = 18 r/s\nt = √(2y/a) = √(2*2.1/9.8) = .654 sec\n\nθ = ωt = 18*.654 = 11.783 rad
789343
Rotational velociy is given by Radians / sec. = Rotations/sec*2PI since one rotation=2PI radians.(PI=3.14159...)\n\nThe linear (tangentional, translational) Velocity is given by the formula:\nV=R* w  (m/sec)\n\nWhere R is Radius and w=Rotational velocity in Radians / sec.\nthe Centrifugal force is \nF=M * w^2 *  R  Therefore the acceleration is \na=w^2 * R and is directed toward the center.  (Vectorial quantity)\n(note:w^2 means w squared)\nfor more info: http://en.wikipedia.org/wiki/Centrifugal_force
922381
If the initial angular velocity is 3.0 and the final is 5.2, then the average would be 4.1 rev/sec.  The inverse of this is .244 seconds per revolution.  1-1/2 revolutions would take 1.5 * .244 and this would be .366 seconds.
799248
let RA and RC be the angular speed of wheel A and C respectively\nthen linear speed of belt B = RA*rA = RC*rC \nhence RA= RC * rC/rA = RC 25/10 = 2.5 RC\n\nWhen C reach 100 rev/min\nthen A reach 2.5 100 = 250 rev/min\nRA = RAo + a*t = 0 + 1.6 t rad / sec\nRA = 1.6 t /(2*pi) = 1.6 t /6.28 =0.2548 t rev/sec =15.288 t rev/min\n15.288 t = 250\nt = 250/15.288 = 16.35  seconds
943735
a) T = dL / dt = 16\nL = 16 t \n\nI don't think you can do any better than that since they don't ask for the angular momentum at a particular moment in time, and it certainly is not constant.\n\nb) L = Iw\nw = L / I = 16 t / I = 16(33 * 10^-3) / (1.2 * 10^-3) = 440 rad/s
710256
You got the CG just right.  Your method might work with enough concentration., but I would look at making the angular momentum come to zero:\nL = Iω = (2.7*.325²  + 1.3*.675²) * 26rpm*2π/60 = 2.389 kg m²/s\n\nTo stop the movement the angular impulse must equal the - of the angular momentum:\n\nT * t = L→ T = L/t = 2.389/5.8 = .4118 Nm (ccw)
828009
One revolution = 2pi radians.  One minute = 60 seconds.  So 1 rev/min = 2pi/60 rad/s.  Multiply 11.2 rev/min by 2pi/60 to get 1.17286 rad/s.