Determine the linear velocity of a point rotating at an angular velocity of 7 pi radians/sec.(c details)?
819170
7 pi radians/second = 3.5 revolutions per second.\nCircumference =  20 ft * pi.\n\nLinear velocity = 3.5 * 20pi  = 70pi ft/sec.
970150
We calculate in radians for convenience.\n\nFixed part of rotation from initial v is\n48 * 7 = 336\n\nVariable part from additional acceleration is\nd = 1/2 a t^2\nwhere acceleration a is\n(120 - 48) / 7 = 10.3 r/sec^2\ntherefore d is\nd = 1/2 * 10.3 * 7 * 7 \nd = 252\n\nTotal d is then\n252 + 336 = 588 radians\nand total revolutions is 588 / (2*pi) = 93.6
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!
911487
The equations of motion for the angular displacement of each disk are given by:\n\nDisk A: R1(t) = (1/2)*alpha1*t² + w1*t +Ri1\nDisk B: R2(t) = (1/2)*alpha2*t² + w2*t +Ri2\n\nFor part (a), we're looking for the time, T, for which R1(T) = R2(T)\n\n(1/2)*alpha1*T² + w1*T +Ri1 = (1/2)*alpha2*T² + w2*T +Ri2\n\nFrom the problem statement, we obtain the following information:\n\nRi1 = Ri2\nalpha1 = 0\nw1 = 9.5 rad/s\nalpha2 = 2.2 rad/s²\nw2 = 0\n\nSubstituting these values and simplifying, we obtain:\n\nw1*T= (1/2)*alpha2*T²\n\nw1= (1/2)*alpha2*T\n\nT = 2*w1 / (alpha2)\n\nT = 2(9.5 rad/s) / (2.2 rad/s²)\n\nT = 8.64 s\n\nTechnically, part (b) is impossible to say without knowing how the reference line is defined, but I'll give it a whirl. If the reference line is measured from the center of each disk to its outer edge, then yes.\n\nHowever, if the reference line is placed across the diameter of the disk, then no. In this case, the reference lines will be aligned when the angular displacement of Disk A is 180° or pi radians ahead of Disk B. To solve for this time, simply add pi to Disk B's equation of motion:\n\n(1/2)*alpha1*T² + w1*T +Ri1 = (1/2)*alpha2*T² + w2*T +Ri2 + pi\n\nSubstituting out known values again leads us to:\n\nw1*T= (1/2)*alpha2*T² + pi\n\nor:\n\n(1/2)*alpha2*T² - w1*T + pi = 0\n\n1.1*T² - 9.5*T + pi = 0\n\nUsing the quadratic formula, we find that \n\nT = .344 seconds or 8.292 seconds\n\nThus, if the reference line is directed across the entire disk, then they are momentarily aligned twice before having the same angular displacement.\n\n(This added after a while)\n{BTW, monohan47130 - no I didn't. Please read what I wrote. If the reference lines are defined along the radius only, then you are correct - there is only one point at which the disks have the same angular displacement. But, reference lines are typically made along the entire circle (diameter). This means that they will be aligned 3 times total in 8.64 seconds :) }
187232
LET:\nr = radius of pulley\ns = arc length = belt length that moved around the pulley.\nt = time of belt to move that length\nw = angular speed of pulley\nv = linear speed of belt\n\nGIVEN:\nr = 13.34 cm\ns = 52 cm\nt = 16 sec\n\nSOLUTION:\n\na) w = v/r = s/(rt) = 52 cm/[(13.34 cm)(16 sec)] \n = 0.24 rad/sec ANSWER\n\nb) v = s/t = 52 cm/16 sec = 3.25 cm/sec ANSWER\n\nI hope the solution is clear enough...
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.
598106
you can use the following proportion:\n\ndegree/180 = radians/pi
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.
940333
54 m/s / 52 rev/s = 54/52 m/rev\nC = 2 pi r\nr = 54 / (104 * pi) = 0.165 m (rounded down)
932723
This is a classic conservation of angular momentum question.\nWe will assume frictionless motion.\nThe angular momentum is always:\n\n1/2 r^2 m  for the disk\nplus\nd^2 m for the man\nthe sum multiplied by the angular speed\n\nsince the momentum is conserved:\n\n(1/2*4*25+85*4)*.15=\n(1/2*4*25+85*.7^2) speed\n\nspeed=.64 rev/s\n\nI'll let you convert it to radians per second\n\nj
812026
a) There are 2pi radians in a revolutuion\nThe angular speed in radians pers sec is simply 2pi xRPS\nwhere RPS is the number of revolutions per sec (1RPS = 60RPM) \n\n2400x2pi /60 sec = 251.33 radians per sec\n\nb) The circumference = pi Dia =  23.562 in = 23.562/12 =1.9635ft\n The speed of the saw teeth is 2400/60 = 40 RPS (x 19635 ft) =78.54 ft/sec\n\nJust a note about angular speed:\nThe angular speed is the answer to the question: how many times does it rotate in a given amount of time? This can be expressed in radians per sec, RPS or RPM etc.\nThe speed of the teeth (or of some other point on the blade) is the answer to the question: What distance do the teeth travel in a given amount of time, for a given RPS, RPM etc.?\nIf you take a disk, or circular blade, and draw a radial line on it and then give the disk a certain RPM then the points on your line will have the same angular speed, since different points on the radial line will all travel 360 degrees(or 2 pi radians) every revolution. But the points will be travelling at different speeds. Points at the edge of the disk will have the greatest speed since they travel a greater distance (i.e. a greater circumference) per revolution than points which are closer to the center of the disk. So the teeth of the saw will have a greater speed than points on the blade which are closer to the center, but they all have the same angular speed.
698608
Have you been given the standard formulas?\nx = a cos (nt + c),  frequency = n, period = 2*pi/n, phase = c (Not sure of that -- it may be -c.  Look it up.) This gives you the answers to (d), (e) and (f) immediately.  [No, I'm not doing it for you: You can substitute values I'm sure.]\n\nx is the displacement, so just sub t = 2 in this equation.  If you need touse your calculator (you shouldn't!), make sure it's set in radians.\n\nVelocity is dx/dt.  You can do that I hope, and get\n\nv = - 18*pi*sin(3*pi*t + pi/3)  Notice I omit units.  I prefer to use equations relating to numbers only (though I know units are sometimes useful especially in a Physics context) and then state the correct units in presenting the answer -- which you are going to do!\nSub t = 2 to get v in m/s\n\nAcceleration is dv/dt, so differentiate and then sub t = 2\n\nAll the best
288679
One full rotation of the tip is:\n\n2 pi r \n2 pi 9.8cm\n\nin 25 min the tip will make 0.42 rotation (25/60)\n\nso 25/60 * 2 pi 9.8 =  25.7
338257
a) vcenter = (m1*v1+m2+m2)/(m1+m2)\n\nb) v1 rel = v1 - v center\nv2 rel = v2 - v center\n\nremember velocity is a vector so you must keep track of magnitude and direction
791525
questions are very easy. You must try them by yourself. I will only give hints-\n1> Instantaneous accl  = dv/dt\nAvg accl = (v2-v1)/(t2-t1)\n\n2> draw the vectors, take the components separately along x and y axes.\nremember accl component along x axis increases velocity along x axis, and accl component along y axis increases velocity along y axis.\n3> use the basic kinematics equations like v^2=u^2+2as etc
900310
A wheel mounted on a wall is rotating. Every second, its velocity advances 4 radians/second, which is equivalent to 229.183 degrees/second. That is, if it is moving at 0 degrees/second at time 0, it will be moving at 229.183 degrees/second after 1 second, and it will be moving at 458.366 degrees/second after 2 seconds. This is actually the answer to part (a).\n\n(a) The angular speed at t=2 seconds is 458.366 degrees/second, or 8 radians/second.\n\nWe assume that the wheel is rotating counter-clockwise. That is,  if you look at it, a point at the top of the wheel will move to your left.\n\nAt t=0, we focus our eyes on a point on the rim of the wheel. That point is located 57.3 degrees above the horizontal. That is, if the wheel was a clock face, the point would be located around 1 o'clock. 1 o'clock is located at 60 degrees above the horizontal, so the point is actually located at where the hour hand would point at 1:05, which is just passed 1 o'clock. (keep in mind that 3 o'clock is located at 0 degrees and 12 o'clock is located at 90 degrees)\n\nNow, at t=2 seconds, we already found that the wheel is rotating at 8 radians per second. Since the radius of the wheel is 2 meters, if the wheel moves 2*pi radians, the wheel moves 2*pi*r meters. Thus, there are r=2 meters per every radian. Thus,\nthe point at the edge of the wheel is traveling around the wheel at (8 radians per second)*(2 meters per radian) = 16 meters per second. Similarly, since the wheel is accelerating at 4.0 radians/second/second, the point at the edge of the wheel is accelerating along the circle that is the edge of the wheel at at (4.0 rad/s/s)*(2 meters/rad) = 8.0 meters/s/s. \n\nHowever, there is also an acceleration normal to the edge of the circle pointing toward the center of the circle. That acceleration is thus a "centripetal" acceleration, and is given by v*v/r, where "v" is the tangential speed of the point. We know the tangential speed is 16 meters/second, so the centripetal acceleration is 128 m/s/s. \n\nThese two accelerations (tangential and normal/centripetal) are perpendicular to each other, so the TOTAL acceleration is the length of the hypotenuse of the right-triangle formed with each of them on a leg. That is, the total acceleration is the square root of ( (128 m/s/s)^2 + (8 m/s/s)^2 ), which is 128.25 m/s/s.\n\nThus, the answer to (b) is:\n\n(b) The tangential speed is 16 meters/second. The tangential acceleration is 8 meters/s/s. The centripetal acceleration is 128 m/s/s. The TOTAL acceleration is 128.25 m/s/s.\n\nFinally, you want to know the angle where P is after 2 seconds. Since everything started at rest, you know that (you can derive these with a little calculus; keep in mind that acceleration is change in velocity over time and velocity is change in position over time):\n\nacceleration = 4 rad/s\n\nvelocity = acceleration*t = (4 rad/s)*t = 8 rad/s\n\nDISPLACEMENT = 0.5*acceleration*t^2 = 0.5*(4 rad/s)*t^2 = 8 radians\n\nSo the angle of point P has moved 8 radians, which is equal to 458.366 degrees. Since the point P has started at 57.3 degrees, then you just add 57.3 degrees to this number and you get 515.666 degrees. Now, there are only 360 degrees in a circle, so this means that the point has completed one revolution and has gone 155.666 degrees into the next circle. Thus, your answer to (c) is:\n\n(c) The angular position of the point P is 155.666 degrees.\n\nI hope that helps.
746380
Torque = change in angular momentum over change in time\n\nT=Iw/t      T=mrrw/t     T=1.1(.09)(93.6)/1.7  T=5.45 N/m