What is angular velocity? ALSO EXPLAIN OMEGA , RADIANS AND FORMULA FOR THETA?
363428
it is the angle swept by the ray in unit time\nomega:angular velocity\n180 degrees=pi radians\ntheta = omega.time +1/2(angular acceleration).(time)^2
751049
First find Vo:  Vo^2 = Rg/sin(28)  Where R = range & 8 = theta\n\nUsing your numbers, Vo = 41.523 m/s\n\nImpulse = change in momentum ----> Ft = mVo---> F = mVo/t\n\nSo, F = (.0431kg*41.523m/s)/.00438sec = 408.59 N\n\nVoila!
831615
R being half way on the circle then:\n\nb)\nave velocity on x axis is v*Integral(sin(a)) (delta a) /(45deg) from 0deg to 45deg, on y the integral is v*Integral(cos(a)) (delta a) /(45deg) from 0deg to 45deg, the x integral evaluates to -0.292, divided by pi/4 is -0.373, the y integral is 0.7071, divided by pi/4 is 0.9\n\nv=w*r, which is 1.25m*4*pi/s=15.7m/s\n\nso v_ave PQ = 0.9*15.7m/s (j) - 0.373*15.7m/s (i)=\n14.13 (j) - 5.86 (i),, the magnitude is correct, i've checked\n\na) over PR this is very simple since integrals both evaluate to 1, so you have +/- 2/pi in both cases, sign is +for y - for x\nthe ave velocity is 10m/s (j) -10m/s (i)\n\nc) at P, it is -15.7m/s (i) +0 (j)\n\n\nNote: the magnitude of ave velocity can be any number <= 15.7m/s. consider the case for full revolution. the ave velocity vector is zero vector\n\nPS ok im not modifying this anymore
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
871488
Torque=F*D\nT=18*.35\nT=6.3\nWe know that T=a*i     [a=angular acc.\n                                       i=moment of inertia]\n         T=a*mr^2       [m=mass\n                                  r=radias]\n         6.3=a*2.65*(.35)^2\n         a=19.4 radian/sec^2\n\nProblem solned. Have a nice day.
756404
a = rω² → ω = √(a/r) = √(9.8/1100) = .09438 rad/sec\n\n*86400sec/day*1rev/2π rad = 1297.928 rev/day
674213
The angular momentum of a particle of mass m with respect to a chosen origin is given by L = mvr sin θ or more formally by the vector product L = r x p The direction is given by the right hand rule which would give L the direction out of the diagram. For an orbit, angular momentum is conserved, and this leads to one of Kepler's laws. For a circular orbit, L becomes L = mvr.
701875
We're given r = 10 m, t = 5.0 s, and that \n\nθ = αt^2, where α = 0.3 rad/s^2\n\n(a) Rotational velocity, ω = θ/t = αt^2/t = αt = (0.3)(5.0) = 1.5 rad/sec\n\n(b) Tranlational velocity, v = ωr = (1.5)(10) = 15 m/s\n\n(c) Tangential acceleration, a = ω^2r = (1.5)^2(10) = 22.5 m/s^2\n\n(d) Radial acceleration, α = ω/t = 0.3 rad/s^2
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?
846820
The formulae for linear acceleration and angular acceleration are very similar:\n\nF=ma\nT = I a  (the T and a here are usually written Tau & alpha in greek)\n\nYour 'torque' of 60N can't be right though, a torque value would be in N-m. Assuming its 60N-m then\n\na = T/I = 60 /12 = 5 rad/s^2
849646
-\na=(2, 1, 1)\nb=(1, root3-1, -root3 -1)\n\ncos(theta)= a.b/|a||b|\n= (2 +sqrt(3)-1 -sqrt(3)-1 )/|a||b| = 0\nso theta = pi/2 (90 degrees)
945234
[Edit -- two things.  #1 Oops, I made a mistake.  When I worked this, I used a mass of 4 kg instead of 0.4.  Now I'll go back through my answer and make the correction.\n\n[#2  There are two different methods here.  Look at both of them.  The <second> way is <much better> for this problem.  I think you'll see that.]\n\nSnow White has an answer, but she doesn't say how she got it.\n\nLet's start by getting the net force F = F1 + F2 = -0.6Ni + Nj with magnitude sqrt(1.36)N = 1.166 N and acceleration F/m = 2.915476 m/s^2.\n\nNow we have two ways to proceed.  This is linear motion with constant acceleration, and we know the direction ... tan theta = -1 / 0.6 = -5/3, so the angle theta in the second quadrant is 59 degrees north of west.\n\nAfter 1.6 seconds, the velocity is acceleration x time = 2.915476 x 1.6 = 4.66476 m/s in the direction indicated above.\n\nThe displacement s after 1.6 seconds is s = vt/2 or s = 1/2 at^2 = 4.66476 x 1.6 / 2 = 3.7318 m.\n\nTo get the position, we could use sines and cosines, but notice that tan theta above is -5/3, so the hypoteneuse of that triangle is sqrt(25+9) = sqrt(34).  Using that geometry, the position after 1.6 seconds is\n\nx = -3/sqrt(34) x 3.7318 = -1.92 m\n\ny = 5/sqrt(34) x 3.7318 = 3.2 m\n\nThat's your answer: (-1.92, 3.2) with velocity 4.66476 m/s and direction 59 degrees north of west.\n\n\nThere's another way to do it that may be better, especially when you're using vectors.  Do it by components, and keep the components separate.\n\nThe net force is F = F1 + F2 = -0.6Ni + Nj.\n\nThe constant acceleration a = F/m = -1.5i + 2.5j m/s^2\n\nThe velocity after 1.6 seconds is v = at = -2.4i + 4j m/s\n\nThe position after 1.6 seconds is s = vt/2 = -1.92i + 3.2j\n\nand this is the same answer we got by doing it the other way.  (You can, of course, get all the magnitudes and angles.)\n\nLooks to me that it's easier doing it by components.\n\nHope this is the same answer Snow White got.  This way, at least, you have two ways of doing it.
262624
Angular velocity is the distance a rotating object, like a wheel, accellerates over a certain period of time.  You must switch rotational speed into rad/s, however, in order for it to work.
217864
J=Iw, w=v/r (v = tangental velocity) I is diffrent for each object, or group of objects.  so J=Iv/r I=Jr/v\n\nI for uniform objects does = sigma MR^2.
986763
To convert from cartesian to polar coordinates, the formulas are:\n\nr = √(x² + y²)\nθ = arctan(y/x)\n\nso r = √((-2)² + (3)²) = √(4 + 9) = √13\n\nand θ = arctan(y/x) = arctan(3/-2) = -56.3°\n\nBut you can also add or subtract 360° as many times to theta as you want and you still get the same angle.\n\nSo two possible representations would be:\n\n(√13, -56.3°) and\n(√13, 303.7°)
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)
852750
T=I*w’, T=16 torque, w’= angular acceleration, I=1.2/1000; In time t (SECONDS) its angular speed w=w’*t;\nOr w’=w/t or w’=T/I; thus w/t=T/I or I*w=T*t is angular momentum, & w=T*t/I