angular acceleration problem?
373075
Acceleration= 10 rad/s2\n\nAngular vel w= 10 x t rad/s\n\nAngular displacement d= 10 x t^2 / 2  :(t squared)\n\nThey say that angular displacement is 150 rad in 5 seconds, this means in an interval between t and (t+5).\nd= 10 t^2 /2 \nd2-d1= 10/2 (t+5)^2 - 10/2 t^2 =150 rad\nt^2 + 10t + 25 - t^2 = 30\nt= 0.5 sec\n\nSo the wheel totates for 0.5 seconds before they start to count.  \n\nHope its clear enough.
889368
The first step is to set up two simultaneous equations for the acceleration upward and the acceleration downward\n\nFor upward:\nmg+ma=771.7\nfor downward\nmg-ma=318.8\n\nadd the two equations together:\n2mg=771.7+318.8\n\nm=(771.7+318.8)/(2*9.8)\n=55.64\n\nThen solve for a in either equation and check with the other\n\na=771.7/55.64 - 9.8\n=4.1\n\ncheck\na=9.8-318.8/55.64 \na=4.1\n\nj
718152
weight equals mass times acceleration due to gravity
728641
acceleration can be found using\na=(m1-m2)g/(m1+m2) = 3*9.8/5 = 5.88 m/s^2
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
952712
Since the angular acceleration is constant, you can take the average of 48 and 120 = 83 rads/sec.\n\nThus 7*83=581 revs.
750653
The rotation rate is such that the centripetal acceleration equals the component of the downward acceleration of gravity that acts normal to the cylinder's wall at 68 deg. Thus, solve the following for w (rotation rate):\nw^2*r = g*sin(68).\nSince you asked what equation to use and did not ask for a worked-out answer, that's it. I hope it helps.
824167
The radius is actually 2/3 of the altitude.  Try that
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.
913766
The angular acceleration is 10/6 = 1.66667 rad/sec^2\n\na) I = Ta/(alpha)a = 36/1.6667 = 21.6 kg-m^2\n\nSince it loses 10 r/s in 60 sec, the deceleration is 10/60 = -1/6 r/s^2\nb) Td = I*(alpha)d = 21.6*(-1/6) = -3.6 N-m\n\nTheta (acc) = .5(alpha)a t^2 = .5*(10/6)*10^2 = 83.333 rad\nTheta (decel) = .5(alpha)d t^2 = .5*(1/6)*60^2 = 300 rad\n\nc) Theta total = 383.33 rad = 61 rev
998322
Lets make an enormous assumption to fill in the gaps.\nYou have a thing on a string of length L and spin it round at a certain rate of rotation. Keeping the rate of roataion what would the tangential velocity of the thing be if the string was shortened to L/2?\nIn this case the tangential velocity (= the radius x the angular velocity) will also halve.
753766
First we determine the total moment of inertia of the system (with everyone on the rim).\n\nPeople: I = m r^2 = (65*150) (96^2) = 8.986 * 10^7\nSpace Station: I = 5.08 * 10^8 (it says 5.08 108 in your question, so I'm gonna assume that this is what you meant)\nSum = 5.679 * 10^8 kg*m^2\n\nIf the apparent acceleration = 1g\nThen, using a = v^2 / r , we can get the linear velocity of the edge\n\n9.8 = v^2 / 96\nv = 30.672 m/s\n\nTo get the angular velocity, use w = v / r \n(w is angular vel.)\n\nw = 30.672 / 96 = 0.3195 radians/sec\n\nSince angular momentum is conserved, we can equate the following:\nI * w (before meeting) = I * w (during meeting)\n\nObtain I (during meeting)\n\nPeople: I = m * r^2 = (65*50) * (96^2) = 2.995 * 10^7\nSpace Station: I = remains the same\nSum = 5.380 * 10^8 kg*m^2\n\nConservation of angular momentum:\nBefore = During\n(5.679*10^8) * (0.3195) = (5.380*10^8 ) * w\nw = 0.337 radians/sec\n\nFind the linear velocity:\nv = w * r = 0.337 * 96 = 32.377 m/s\n\nAcceleration to maintain circular motion:\na = v^2 / r\na = 32.377^2 / 96 = 10.92 m/s^2\n\n(that's 1.11g, if you're wondering)\n\nTada!
1054447
I had this happen once (thanks to my two year old playing with the computer).  The way I fixed it was to right click on the desktop and choose Graphics Options.  Then chose Rotation and set it back to Normal.  I am using Windows XP Home Edition.  I hope this helps!
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
27065
I've been studying this stuff recently, and I'm pretty sure this is how it goes:\n\nK is now (-2,1) and D is now (-3,-2).\n\nFor future reference on determining rotations, simply turn the graph counter-clockwise one time for 90 degrees, upside down for 180 degrees, and for 270 degrees, just turn the graph clockwise once to see what the new cooridnates are.
805281
OK, first we need a formula for the motion, which, if it's simple harmonic motion is\nf(x) = Asin(kx)\nwhere A is 1.8*10^-4\nand k = 2000*2*pi\nSo f(x) = 1.8*10^-4*sin(4000*pi*x)\n\nTo compute the maximum force, we need the accelleration f''(x)\nf'(x) = 1.8*10^-4*(cos(4000*pi*x))*4000*pi, or \n1.8*4*pi*10^-1*cos(4000*pi*x)\n\nf''(x) = 1.8*4*pi/10*(-sin(4000*pi*x))*4000*pi\nor\n1.8*16*pi^2*100*(-sin(4000*pi*x))\nthe MAXIMUM acceleration is then 1.8*16*pi^2*100.\nThis is because you can find a value for x where -sin(4000*pi*x) is 1.\n\nUsing F=ma, the maximum force is 50 *1.8*16*pi^2*100 KG m/s^2, or approximately 1.42*10^6 KG m/s^2
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