A Wheel of radius 1.0m is rotating with a constant angular speed of 2.0 rads/s.?
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
498394
You have a bevel gear on the end of the drive shaft (the input shaft to the differential). This gear right-angle drives a large "planetary" or ring gear that rotates about the axle but is not attached to it directly, but is attached via a yoke that carries two intermediate bevel gears, and these gears mesh with the gears that drive the two wheels. (You really need to look at the ref.) When the two wheels want to rotate at the same speed, the two bevel gears do not need to rotate and the transmission acts like a simple nondifferential mechanism. However any differential motion of the wheels is accommodated by rotation of the intermediate gears.\nP.S. For some reason I had to go to another page (at the ref. website) and then back to get the animations to work right.
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
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.
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
52386
The easiest answer I can give you is assuming you mean that our rotational velocity (radians per second) goes to 0 instantaniously. Being that the world is rotating 360 degrees (2pi radians) every 24 hours (86400 seconds), we are turning 2pi radians per 86400 seconds or .0000727 radians per second. Assuming you are close to the equator, you will be launched (well perspectively you will continue moving at normal speed as the Earth abruptly stops) forward at approximately 1037.6 miles per hour, being that the earth rotates 15 degrees per hour at 69.7 miles per degree.
487670
None. It takes no power to maintain that speed (on level ground, no friction). The object has kinetic energy (0.5 * m * v^2). Power was required to put it in motion and power can be recovered by decelerating it. If you do the math, the energy is 3551.86 ft-lb, equivalent to 1 horsepower for 6.458 seconds. Answer 1 neglected the fact that weight must be converted to mass by dividing by g, and energy is proportional to v^2.
791900
Ok.  From the hint start by figuring out the ratio of wheel speeds\nCircumference of A =2*pi*10\nCircumference of C = 2*pi*25\nso C/A=2.5  ie A must go around 2.5 times to make C go around 1 time.\nThere are 2*pi radians to 1 revolution so 100 rev/min =200*pi rad/min.  A must go 2.5 times as fast as this or 2.5*200*pi=500 rad/min\nAccelleration is the derivative of velocity which means that we must integrate the angular acceleration to obtain the speed at any given point of time.\nIntegrating 1.6rad/s^2 with respect to time and setting velocity equal to zero at time equals zero we obtain angular velocity= 1.6*time elapsed (rad/sec).  Note that this speed is in "per seconds" and we need it to be in "per minutes" so we must multiply by 60 for the units conversion.\nThis gives us the angular speed in rad/min for any chosen passage of time.  We shall set this equal to the goal speed of 500*pi rad/min as mentioned above. So\n1.6*time elapsed*60 =500*pi\nSolving for time elapsed we obtain a time of 1.63 minutes.
909202
a= r * w^2 = .12 * 2^2 = .12 *4 = .48 m/s^2
701707
The Average Linear Speed is {total distance / total time taken} = 65/ 5.7 = 11.4 m/s. \n\nDivide this value by the radius of the wheel to get the ‘The Average Angular Speed’. \n\nThe Average Angular Speed is = 11.4 / 0.02 = 570 radian per second.\n\nSince the motion is uniform,\n\n The Average Angular Speed = {Initial Angular Speed + Final Angular Speed} / 2   \n\nBut the Initial Angular Speed is zero.\n\nTherefore, the Final Angular Speed   is twice the Average Angular Speed\n= 2 x 570 \n=1140 radian/ second. (The answer for the second question)\n\nSince the Initial Angular Speed is zero,\n\nThe Angular acceleration is, “Final Angular Speed/ total time”.\n= 1140/5.7 = 200 [radian / second] / second. The answer for the first question)\n\n\nThe number of revolutions is found by “total distance / (pi x diameter of the wheel)” = 65 / pi 0.04 = 517.25 revolutions.
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!
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
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.
609168
Excellent question!\n\nBecause film is composed of 25 frames per second the much-faster revolution of the holes in the wheel rim gets out of phase.
833966
The question is easy enough if you think about it properly.  What you want is the speed at which the centrifugal force (given by mv^2/r) equals the gravitational force given by the bank, and that will be mg sin(2.5 degrees).  Note that m cancels out in this, and you are given r, so v is the only unknown and you can solve for it with a scientific calculator.
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.
939877
E=kinetic energy of wheel\nr= radius of wheel\nw=angular speed of wheel\nm=mass of wheel\nT=required torque\nt=stopping time \nW=necessary work to stop the wheel = E\n\nE=m r^2 w^2\nW=T * (w+0)/2 * t\nE=W\nT=2mw r^2/t\nT=10.77 N.m