How many revolutions does the wheel turn through during this time interval?
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.
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
525309
1) Convert rpm to revolutions per second, by dividing rpm by seconds (so the minutes cancel).\n\n2)Convert revs per sec to radians per second, by multiplying rev/sec *2pi rad/rev (the revs cancel).\n\n3 Acceleration is in the form of distance/(time^2), so dividing radians per second by the time required to stop will give you the magnitude of acceleration over that time period. \n\nFor your example:\n\nrotational acceleration = \n            1000 rev/min *1 min/60 sec* 2Pi rad/rev *1/2 sec\n           =  52.4 rad/sec^2\n                 \n\nPart 2:"\n\nIf by distance you mean radial distance traveled by the tip of a spinning blade, then you know that \n\n1 revolution = 2*pi*r \n\nso to find the number of revolutions, multiply the distance by 1 rev/(2*pi*r)- the units of linear distance will cancel (see a pattern?).\n\n(distance) * 1rev/(2*pi*r) = number of revolutions
341657
OK, you are right, there are 2(pi) radians in a circle and spinning 10 times is 20 radians.\n\nYour angular velocity is the rate the mass (the clothespin?) spins within a given time. You give 10 revolutions in 5 seconds. or you can say 2 revolutions per second or 120 rpm (revs per minute). Multiply that by 2 to get radians per minute. \n\nAngular acceleration doesn't seem to be considered here since you say the mass has spun 10 times at a constant rate (or speed). So, in what you are asking, the length of the string (the radius of the circle of which the mass spins - your half a meter) Does not come into play here.
174325
Yes.  One ROTATION is one day (which varies for whatever planet you're on...for Earth it is 24 hours)\n\nOne REVOLUTION is one year for the planet (again, varies for each planet...earth is 364 1/4 days...and the reason for LEAP YEAR).
908729
(a) ω = ωo + αt\n0 = 150 + α * 132\nα = -150/132\n≈ -1.136 rev/s²\n\n(b) θ = ωot + ½αt²\n= 150 *132 - ½*1.136*132²\n≈ 9903 revolutions (nearest rev)\n\n(c) See (b)\n\n(d) 75 rpm = 75*2π rad/min\nα = rω²\n= 50 *(75*2π)²\n≈ 11103300 cm/min²\n= 111033 m/min²\n= 111033/3600 m/s²\n≈ 30.84 m/s²\n\n(e) ???? I do not understand what it is you are seeking.
1165249
This is the part of the wikipedia that I understood.  I can explain it to you if you want to send me a message.  I think it makes alot of sense.\nNow let’s see what happens to our accelerating particles when a torque is applied to the spinning wheel. Assume the axis of rotation created by the torque is through the center of the wheel at 90 degrees to the primary rotation of the wheel. Let’s look at a particle that is on this axis of rotation. Since the particle is on the axis of rotation there is no direct motion applied to the particle at the instant of the applied torque. But let’s look at what will need to happen at the next moment in time. The particle is now going to be forced to curve again. This time in the direction of the curve so as to accommodate the tilt of the wheel. Now we have a particle that is already moving and it wants to keep moving in a straight line. So the particle will exert a force on the wheel. If you look at a particle on the other side of the wheel you will see that the force of the second particle is in the opposite direction of the first particle. That pair of unmatched forced is what causes the precession torque that is 90 degrees to the applied torque
632041
The number of elements in the sample space are all the possibilities that could happen after spinning the wheel twice.\nImagine two empty boxes which you will fill out with the numbers the wheel gives. _  _\nIn the first one there are five possibilities,\nin the second one also five possibilities, therefore a total of 5*5=25.
858881
12.25cycles/20sec = 0.6125 cycles/sec, = 0.6125Hz, which is the frequency.  The period is the reciprocal of the frequency, which equals approx. 1.6327 seconds.\n\nI didn't count that first half cycle because it looks longer than the rest.  I started counting at the first positive zero-crossing at time 2.6, and counted 12.25 cycles from there to time 22.6\n\nThis is coming from someone who has made his living for the last few decades looking at and interpreting thousands of such displays.
725445
You can't without the radius of the wheel...m/s is a tangential speed.
791539
Wow. Your classmate just asked this question. But here's your answer anyways...\n\nThere are a few things to keep in mind here: all of the energy in the initial system is rotational, and the initial energy is equal to the energy lost to friction. Also note that the energy is lost to friction at a constant rate.\n\n(a) The initial energy of the system is:\n\nE = (1/2) * I * w²\n\nWhere I is the moment of inertia and w is the angular velocity. The power transferred from the system due to friction is equal to the time derivative of the energy loss:\n\nP = dE/dt = I * w * a\n\nwhere a is the angular decceleration.\n\nSince the power is transferred at a constant rate and removes all of the energy from the system, we can take this as:\n\nP = -E/t (negative sign denotes power loss)\n\nCombining the power equations yields:\n\nI * w * a = -E/t\n\nSolving for a:\n\na = -E / (I * w * t)\n\nFinally, substitute the expression for E to obtain:\n\na = -I * w² / (I * w * t)\n\na = -w/(t)\n\nSubstituting the values yields:\n\na = -(150 rev/min * 2*pi rad/rev * 1/60 min/sec)/(2.2 hr * 3600 sec/hr)\n\na = -(15.71 rad/s)/(7920 s)\n\na = -.001983 rad/s²\n\n(b) The rotational behavior of the fly wheel can be modeled as simply:\n\nR(t) = -1/2*a*t² + w*t + Ri\n\nwhere R(t) is measured in radians and Ri is the initial angular displacement. Assuming that Ri = 0 (we'll measure the number of rotations from its initial position), then the flywheel will revolve:\n\nR(t) = -1/2*a*t² + w*t (radians)\n\nR(t) = (-1/2*a*t² + w*t)/(2 * pi) (revolutions)\n\nSubtituting the values:\n\nR(2.2 hr) = R(7920 s) = (-.5*.001983*(7920 s)² + (15.71 rad/s)(7920 s))/(2*pi)\n\nR = 9900 revolutions\n\n(c) The tangential component of the linear acceleration of a point on a rotating body is given as:\n\nA-tangent = a * r\n\nwhere A = linear acceleration. The angular acceleration is constant, so for all w (including w=75 rev/min), the linear acceleration will also be constant. Thus:\n\nA-tangent = (-.001983 rad/s²)(.5 m)\n\nA-tangent = -9.915 x 10^(-4) m/s²\n\nA-tangent = -0.9915 mm/s²\n\n(d) The magnitude of the net linear acceleration on the particle is composed of a tangential and a normal component. The equation is of the form:\n\nAnet = [ (A-tangent)² + (A-normal)² ]^(1/2)\n\nOr:\n\nAnet = [ (a*r)² + (v-tangent²/r)² ]^(1/2)\n\nNote that the tangential velocity is given by:\n\nv-tangent = w * r\n\nThus, the magnitude of net linear acceleration is:\n\nAnet = [ (a * r)² + (w² * r)² ]^(1/2)\n\nSubstituting the known values yields (note that for this case, the angular velocity of 75 rev/min is equal to 7.854 rad/s):\n\nAnet = [ (-.001983 * .5)² + (7.85² * .5)² ]^(1/2)\n\nAnet = 30.81 m/s²\n\nFINAL ANSWERS:\n\n(a) a = -.001983 rad/s²\n(b) R = 9900 revolutions\n(c) A-tangent = -0.9915 mm/s²\n(d) Anet = 30.81 m/s²
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.
959353
All of the power from the motor goes to friction because they just let the wheel coast and burn up all the energy.  But, I'm sure that wasn't the intended answer for the problem\n\nYou need to make a few preliminary calculations:\nFirst you need the Inertia of the flywheel which you can find given the mass and radius of the wheel.\n\nThen you need to calculate the angular acceleration required to bring an object at rest up to 5.88 rad/s in 0.531 sec.  Using this, you can get the torque of the motor - friction.\n\nFinally, calculate the angular acceleration required to bring an object spinning at 5.88 rad/s down to rest in 26.8 revolutions.  Using this, you can get the torque of the friction by itself.\n\nAdd the second result from the first to get the torque of the motor.  Take the friction torque and divide by the motor torque times 100 to get percentage of friction.
764197
(a)  How long does it take the propeller to go 1/3 of a revolution?  Figure that out, and you'll have your answer.\n(b)  Similarly, how long does it take to go 2/3 of a revolution?
748663
The answer hippoterr... gave you was OK till the last step. It's always is wise to check the units and see if they make sense. His result is in sec/rev.\n\n2 x Pi x11.5 / 30 = 2.4 He left out the units in the math part, but if you include them you see that the result is in sec/rev.\n\nUnit conversion can be confusing. A better way - you know you want it to be something per sec so start with:\n\n30 meter/sec. Then to convert to rev/sec multiply by 1, in a special form.\n\nIn the form 1 rev/(2 x Pi x11.5 meters). This = 1, right? The top and bottom are equivalent. So back to \n\n30 meter/sec * 1 rev/(2 x Pi x11.5 meters) = .415 rev/sec\n\nYou understand why weight/mass wasn't needed? It's the acceleration that results from the rotation,  and  a = F/m.
745809
1 revolution  =2*3.1416 (pi) radian.\n\nSo, a revolution = pi radias i.e; 3.1416 radian
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