course phy 121 }UҖR۪sassignment #032
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19:55:19 `q001. Note that this assignment contains 3 questions. The moment of inertia of a concentrated mass m lying at a distance r from the axis of rotation is m r^2. Moments of inertia are additive--that is, if an object with a moment of inertia about some axis is added to another object with its moment of inertia about the same axis, the moment of inertia of the system about that axis is found by simply adding the moments of inertia of the two objects. Suppose that a uniform steel disk has moment of inertia .0713 kg m^2 about an axis through its center and perpendicular to its plane. If a magnet with mass 50 grams is attached to the disk at a point 30 cm from the axis, what will be the moment of inertia of the new system?
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RESPONSE --> Initial I = .0713 kg m^2 m magnet = 50g = .05kg r magnet = 30 cm = .3m I = mr^2 I = .05kg * (.3m)^2 I = .05kg * .09m^2 = .0045 kg m^2 I of system = sum of I I of system = .0713 kg m^2 + .0045 kg M^2 I of system = .0758 kg m^2 confidence assessment: 2
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19:55:35 A mass of m = .05 kg at distance r = .30 meters from the axis of rotation has moment of inertia I = m r^2 = .05 kg * (.30 m)^2 = .0045 kg m^2. The moment of inertia of the new system will therefore be the sum .0713 kg m^2 + .0045 kg m^2 = .0758 kg m^2 of the moments of inertia of its components, the disk and the magnet.
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RESPONSE --> self critique assessment: 3
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20:06:06 `q002. A uniform rod with mass 5 kg is 3 meters long. Masses of .5 kg are added at the ends and at .5 meter intervals along the rod. What is the moment of inertia of the resulting system about the center of the rod?
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RESPONSE --> m = 5kg length = 3m I of rod about center = 1/12 m L^2 I of rod = 1/12 * 5 kg * (3m)^2 = 3.75 kg m^2 I of each mass = mr^2 I 1 = .5kg (1.5m)^2 = 1.125 kg m^2 I 2 = .5kg (1m)^2 = .5 kg m^2 I 3 = .5kg (.5m)^2 = .125 kg m^2 I 4 = .5kg (0m)^2 = 0 I 5 = .125 kg m^2 I 6 =.5 I 7 = 1.25 sum of all moments of inertia = 7.25 kg m^2 confidence assessment: 2
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20:07:09 The rod itself, being rotated about its center, has moment of inertia 1/12 M L^2 = 1/12 * 5 kg * (3 m)^2 = 3.75 kg m^2. The added masses are at distances 1.5 meters (the two masses masses on the ends), 1.0 meters (the two masses .5 m from the ends), .5 meters (the two masses 1 m from the ends) and 0 meters (the mass at the middle of the rod) from the center of the rod, which is the axis of rotation. At 1.5 m from the center a .5 kg mass will have moment of inertia m r^2 = .5 kg * (1.5 m)^2 = 1.125 kg m^2; there are two such masses and their total moment of inertia is 2.25 kg m^2. The two masses lying at 1 m from the center each have moment inertia m r^2 = .5 kg * (1 m)^2 = .5 kg m^2, so the total of the two masses is double is, or 1 kg m^2. {}The two masses lying at .5 m from the center each have moment of inertia m r^2 = .5 kg ( .5 m)^2 = .125 kg m^2, so their total is double this, or .25 kg m^2. The mass lying at the center has r = 0 so m r^2 = 0; it therefore makes no contribution to the moment of inertia. The total moment of inertia of the added masses is therefore 2.25 kg m^2 + 1 kg m^2 + .25 kg m^2 = 3.5 kg m^2. Adding this to the he moment of inertia of the rod itself, total moment of inertia is 3.75 kg m^2 + 3.5 kg m^2 = 7.25 kg m^2. We note that the added masses, even including the one at the center which doesn't contribute to the moment of inertia, total only 3.5 kg, which is less than the mass of the rod; however these masses contribute as much to the moment of inertia of the system as the more massive uniform rod.
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RESPONSE --> self critique assessment: 3
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20:29:57 `q003. A uniform disk of mass 8 kg and radius .4 meters rotates about an axis through its center and perpendicular to its plane. A uniform rod with mass 10 kg, whose length is equal to the diameter of the disk, is attached to the disk with its center coinciding with the center of the disk. The system is subjected to a torque of .8 m N. What will be its acceleration and how to long will it take the system to complete its first rotation, assuming it starts from rest?
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RESPONSE --> uniform disk: m = 8 kg r = .4m I = 1/2 m r^2 I disk = 1/2 (8kg) (.4m)^2 = .64kg m^2 rod about center: m = 10kg L = .8m I rod = 1/12 m L^2 I rod = 1/12 (10kg) (.8m)^2 = .53 kg m^2 I sum = .64 + .53 = 1.17 kg m^2 Tau = .8 mN alpha = tau / I sum alpha = .8mN / 1.17 kg m^2 = .684 rad/s^2 I cannot figure out how to find the Time for one complete revolution. I feel like I should know it, but I can't come up with it. The only thing I can think to do is plug the numbers into vf ^2 = v0 ^2 + 2a * ds and solve for vf and then put that into vf = v0 +a * dt and solve for dt. But I don't know if that would work. confidence assessment: 1
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20:41:24 The moment of inertia of the disk is 1/2 M R^2 = 1/2 * 8 kg * (.4 m)^2 = .64 kg m^2. The rod will be rotating about its center so its moment of inertia will be 1/12 M L^2 = 1/12 * 10 kg * (.8 m)^2 = .53 kg m^2 (approx). ( Note that the rod, despite its greater mass and length equal to the diameter of the disk, has less moment of inertia. This can happen because the mass of the disk is concentrated more near the rim than near the center (there is more mass in the outermost cm of the disk than in the innermost cm), while the mass of the rod is concentrated the same from cm to cm. ). The total moment of inertia of the system is thus .64 kg m^2 + .53 kg m^2 = 1.17 kg m^2. The acceleration of the system when subject to a .8 m N torque will therefore be `alpha = `tau / I = .8 m N / (1.17 kg m^2) = .7 rad/s^2, approx.. To find the time required to complete one revolution from rest we note that the initial angular velocity is 0, the angular displacement is 1 revolution or 2 `pi radians, and the angular acceleration is .7 rad/s^2. By analogy with `ds = v0 `dt + 1/2 a `dt^2, which for v0=0 is `ds = 1/2 a `dt^2, we write in terms of the angular quantities `d`theta = 1/2 `alpha `dt^2 so that `dt = +- `sqrt( 2 `d`theta / `alpha ) = +- `sqrt( 2 * 2 `pi rad / (.7 rad/s^2)) = +-`sqrt( 12.56 rad / (.7 rad/s^2) ) = +-4.2 sec. We choose the positive value of `dt, obtaining `dt = +4.2 sec..
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RESPONSE --> I was on the right track, I just didn't trust myself. Of course I chose the ""wrong"" formula, in that my way would have taken longer. But I believe I would have come up with the right answer. self critique assessment: 3
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~ƺˣāբT湻 assignment #033 033. Rotational KE and angular momentum Physics II 11-25-2008
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22:04:16 `q001. Note that this assignment contains 11 questions. A rotating object has kinetic energy, since a rotating object has mass in motion. However we cannot easily use 1/2 m v^2 to calculate this kinetic energy because different parts of a typical object are rotating at different velocities. For example a rigid uniform rod rotated about one of its ends is moving faster near its far end than near its axis of rotation; it has a different speed at every distance from its axis of rotation. However as long as the rod remains rigid the entire rod moves at the same angular velocity. It turns out that the kinetic energy of a rotating object can be found if instead of 1/2 m v^2 we replace m by the moment of inertia I and v by the angular velocity `omega. Thus we have KE = 1/2 I `omega^2. What is the kinetic energy of a uniform sphere of radius 2.5 meters (that's a pretty big sphere) and mass 40,000 kg when its angular velocity is 12 rad/sec (that's almost two revolutions per second)?
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RESPONSE --> KE = 1/2 I omega^2 uniform sphere r = 2.5 m m = 40,000kg omega = 12 rad/s I sphere = 2/5 m r^2 I sphere = 2/5 (40,000kg) ( 2.5m)^2 I sphere = 100,000 kg m^2 KE = 1/2 (100,000kg m^2)(12 rad/s)^2 KE = 7,200,000 Joules confidence assessment: 2
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22:04:27 The KE is 1/2 I `omega^2. We first need to find I; then we can use the given angular velocity to easily find the KE. For this sphere we have I = 2/5 M R^2 = 2/5 * 40,000 kg * (2.5 m)^2 = 100,000 kg m^2. The kinetic energy of the sphere is thus KE = 1/2 I `omega^2 = 1/2 * 100,000 kg m^2 * (12 rad/s)^2 = 7.2 * 10^6 Joules.
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RESPONSE --> self critique assessment: 3
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22:09:16 `q002. By carefully measuring the energy required to accelerate it from rest to an angular velocity of 500 rad/s, we find that the KE of a certain uniform disk is 45,000 Joules. What is the moment of inertia of that disk?
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RESPONSE --> ang vel (omega) = 500 rad/s KE = 45,000J KE = 1/2 I * omega^2 45,000J = .5 * I * (500rad/s)^2 45,000J = .5 * I * 250,000 I = .36 confidence assessment: 2
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22:09:34 We know that KE = 1/2 I `omega^2, and we know the KE and we know `omega. Solving this equation for I we obtain I = 2 * KE / `omega^2. So for this disk I = 2 * (45,000 J) / (500 rad/s)^2 = 90,000 J / ( 250,000 rad^2 / s^2) = .36 kg m^2. [ Note that if we know the mass or the radius of the disk we can find the other, since I = 1/2 M R^2 = .36 kg m^2. ]
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RESPONSE --> self critique assessment: 3
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22:19:54 `q003. A 3-kg mass is tied to a thin cord wound around the thin axle of a disk of radius 20 cm and mass 60 kg. The weight descends 200 meters down a long elevator shaft, turning the axle and accelerating the disk. If all the potential energy lost by the weight is transferred into the KE of the disk, then what will be the angular velocity of the disk at the end of the weight's descent?
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RESPONSE --> m weight = 3kg r disk = 20 cm = .2m m disk = 60 kg ds weight = 200m PE transferred to KE disk ? ang velocity KE disk = 1/2 I * omega^2 PE = m*g*h PE = 3kg * 9.8 m/s^2 * 200m PE = 5880J KE disk = 5880J I disk = 1/2 mr^2 = 1/2 (60kg)(.2m)^2 = 1.2kg m^2 5880J = .5 (1.2kg m^2)* omega^2 5800J = .6 omega^2 omega^2 = 9800 omega = 98.99 rad/s confidence assessment: 2
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22:24:07 The 3-kg mass has a weight of 3 kg * 9.8 m/s^2 = 29.4 Newtons. As it descends 200 meters its PE decreases by `dPE = 29.4 N * 200 m = 5880 Joules. The disk, by assumption, will gain this much KE (note that in reality the disk will not gain quite this much KE because of frictional losses and also because the descending weight will have some KE, as will the shaft of the disk; however the frictional loss won't be much if the system has good bearings, the weight won't be traveling very fast if the axle is indeed thin, and a thin axle won't have much moment of inertia, so we can as a first approximation ignore these effects). The KE of the disk is KE = 1/2 I `omega^2, so if we can find I our knowledge of KE will permit us to find `omega = +-`sqrt( 2 KE / I ). We know the radius and mass of the disk, so we easily find that I = 1/2 M R^2 = 1/2 * 60 kg * (.2 m)^2 = 1.2 kg m^2. Thus the angular velocity will be +- `sqrt( 2 * 58800 J / (1.2 kg m^2) ) = +- 310 rad/s (approx).
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RESPONSE --> You've got a typo I believe. At the beginning the Joules is calculated to be 5,880, but at the end you use 58,800 in your calculation of angular velocity. From my calculations angular velocity therefore ends up being 98.99 rad/sec instead of 310 rad/sec. self critique assessment: 3
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22:29:01 `q004. A rotating object also has angular momentum L = I * `omega. If two rotating objects are brought together and by one means or another joined, they will exert equal and opposite torques on one another and will therefore end up with an angular momentum equal to the total of their angular momenta before collision. What is the angular momentum of a disk whose moment of inertia is .002 kg m^2 rotating on a turntable whose moment of inertia is .001 kg m^2 at 4 rad/s?
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RESPONSE --> ang momentum = I * omega I determine the ang momentum for each object and then add them together to get the new ang momentum. ang mom disk = .002 kg m^2 * 4 rad/s = .008 ang mom ttable = .001kg m^2 * 4 rad/s = .004 .008 + .004 = .012 kg m/s is that the correct units for angular momentum? confidence assessment: 1
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22:31:44 The total moment of inertia of the system is .002 kg m^2 + .001 kg m^2 = .003 kg m^2. The angular momentum of the system is therefore L = I * `omega = .003 kg m^2 * (4 rad/s) = .012 kg m^2 / s.
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RESPONSE --> self critique assessment: 3
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22:41:35 `q005. If a stick with mass .5 kg and length 30 cm is dropped on the disk of the preceding example in such a way that its center coincides with the axis of rotation, then what will be the angular velocity of the system after frictional torques bring the stick and the disk to the same angular velocity?
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RESPONSE --> stick m = .5kg stick L = 30 cm = .3m I stick = 1/12 m L^2 I stick = 1/12 (.5kg)(.3m)^2 I sitck = .00375 kg m^2 ang mom of stick = .00375 * 4 rad/s = .015 kg m^2/s previous system ang mom = .012 kg m^2/s plus ang mom of stick = .027 kg m^2/s confidence assessment: 2
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22:48:10 Since the stick and the disk exert equal and opposite torques on one another, the angular momentum of the system will be conserved. Since we know enough to find the moment of inertia of the new system, we will be able to easily find its angular velocity. The moment of inertia of the stick is 1/12 * .5 kg * (.3 m)^2 = .00375 kg m^2, so the moment of inertia of the system after everything settles down will be the sum of the original .003 kg m^2 and the stick's .00375 kg m^2, or .00675 kg m^2. If we designate this moment of inertia by I ' = .00675 kg m^2 and the new angular velocity by `omega ', we have L = I ' `omega ' so `omega ' = L / I ', where L is the .012 kg m^2 total angular momentum of the original system. Thus the new angular velocity is `omega ' = L / I ' = .012 kg m^2 / s / (.00675 kg m^2) = 1.8 rad/s, approx.. Thus when the stick was added, increasing the moment of inertia from .003 kg m^2 / s to .00675 kg m^2 / s (slightly more than doubling I), the angular velocity decreased proportionately from 4 rad/s to 1.8 rad/s (slightly less than half the original angular velocity).
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RESPONSE --> I misread the question. And unfortunately I don't quite understand how to do this. I follow the explanation, but I don't understand the ""why"" Why is the original system's ang mom divided by the moment of inertia of the new system to get the omega of the new system? self critique assessment: 1
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22:58:43 `q006. An ice skater whose moment of inertia is approximately 1.2 kg m^2 holds two 5 kg weights at arm's length, a distance of 60 cm from the axis of rotation, as she spins about a vertical axis at 6 rad/s (almost 1 revolution / sec ). What is her total angular momentum and her total angular kinetic energy?
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RESPONSE --> I original = 1.2 kg m^2 I weights = 10kg (.6m)^2 = 3.6 kg m^2 I total = 4.8 kg m^2 ang mom = I * omega ang mom = 4.8 kg m^2 * 6 rad/s = 28.8 kg m^2/s KE = 1/2 I* omega^2 KE = .5 (4.8 kg m^2) (6 rad/s)^2 = 86.4 J confidence assessment: 0
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22:59:03 The moment of inertia of each of the two weights is m r^2 = 5 kg * (.6 m)^2 = 1.8 kg m^2, so the total moment of inertia of both weights is 3.6 kg m^2 and the moment of inertia of the system consisting of the skater and the weights is 1.2 kg m^2 + 3.6 kg m^2 = 4.8 kg m^2. The angular momentum of the system is therefore 4.8 kg m^2 * 6 rad/s = 28.8 kg m^2 / s. The total angular kinetic energy is KE = 1/2 I `omega^2 = 1/2 * 4.8 kg m^2 * (6 rad/s)^2 = 86.4 Joules.
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RESPONSE --> self critique assessment:
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22:59:11 The moment of inertia of each of the two weights is m r^2 = 5 kg * (.6 m)^2 = 1.8 kg m^2, so the total moment of inertia of both weights is 3.6 kg m^2 and the moment of inertia of the system consisting of the skater and the weights is 1.2 kg m^2 + 3.6 kg m^2 = 4.8 kg m^2. The angular momentum of the system is therefore 4.8 kg m^2 * 6 rad/s = 28.8 kg m^2 / s. The total angular kinetic energy is KE = 1/2 I `omega^2 = 1/2 * 4.8 kg m^2 * (6 rad/s)^2 = 86.4 Joules.
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RESPONSE --> self critique assessment: 3
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23:05:20 `q007. The skater in the preceding example pulls the 5 kg weights close in toward her stomach, decreasing the distance of each from the axis of rotation to 10 cm. What now is her moment of inertia, angular velocity and angular KE?
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RESPONSE --> I skater = 1.2 kg m^2 I weights = mr^2 = 10kg * (.1m)^2 = .1 I total = 1.2 + .1 = 1.3 kg m^2 ang mom = 28.8 (remains the same) = I * omega 18.8 = 1.3 kg m^2 * omega omega = 22.15 rad/s ang KE = 1/2 I * omega^2 = 1/2 (1.3 kg m^2) (22.15 rad/s)^2 ang KE = 319 J confidence assessment: 2
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23:05:40 Her angular momentum must be conserved, so L = angular momentum remains at 28.8 kg m^2 / s. The moment of inertia for each of the two 5 kg masses is now only m r^2 = 5 kg * (.1 m)^2 = .05 kg m^2 and her total moment of inertia is thus now 1.2 kg m^2 + 2 (.05 kg m^2) = 1.3 kg m^2. If we let I ' and `omega ' stand for the new moment of inertia and angular velocity, we have L = I ' * `omega ', so `omega ' = L / I ' = 28.8 kg m^2 / s / ( 1.3 kg m^2) = 22 rad/s, approx.. Moment of inertia decreased from 4.8 kg m^2 to 1.3 kg m^2 so the angular velocity increased by the same proportion from 6 rad/s to about 22 rad/s. Her new kinetic energy is therefore KE ' = 1/2 I ' * ( `omega ' )^2 = 1/2 * 1.3 kg m^2 * (22 rad/s)^2 = 315 Joules, approx.. [ Note that to increase KE a net force was required. This force was exerted by the skater's arms as she pulled the weights inward against the centrifugal forces that tend to pull the weights outward. ]
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RESPONSE --> self critique assessment: 3
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23:17:18 `q008. When a torque `tau acts through an angular displacement `d`theta, it does work. Suppose that a net torque of 3 m N acts for 10 seconds on a disk, initially at rest, whose moment of inertia is .05 kg m^2. What angular velocity will the disk attain, through how many radians will it rotate during the 10 seconds, and what will be its kinetic energy at the end of the 10 seconds?
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RESPONSE --> dW = tau * d omega tau = 3 mN dt = 10 s omega initial = 0 I = .05 kg m^2 omega final = ? # radians = ? KE final = ? alpha = tau / I alpha = 3 mN / .05 kg m^2 alpha = 60 m/s^2 d omega = alpha * dt d omega = 60 m/s^2 * 10 s = 600 rad/s # radians = 600 rad/s * 10 sec = 6000 rads KE = 1/5 I omega^2 KE = 1/2 * .05 kg m^2 * (600 rad/s)^2 KE = 9000 J Or should I calculate it using dW = T * d omega dW = 3 mN * 600 rad/s = 900 J They should be the same shouldn't they? confidence assessment: 2
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23:19:47 A net torque of 3 m N acting on the disk whose moment of inertia is.05 kg m^2 will result in angular acceleration `alpha = `tau / I = 3 m N / (.05 kg m^2) = 60 rad/sec^2. In 10 seconds this angular acceleration will result in a change in angular velocity `d`omega = 60 rad/s^2 * 10 s = 600 rad/s. Since the torque and moment of inertia are uniform the acceleration will be uniform and the average angular velocity will therefore be `omegaAve = (0 + 600 rad/s) / 2 = 300 rad/s. With this average angular velocity for 10 seconds the disk will rotate through angular displacement `d`omega = 300 rad/s * 10 sec = 3000 rad. Its kinetic energy at its final 600 rad/s angular velocity will be KE = 1/2 I `omega^2 = 1/2 * .05 kg m^2 * (600 rad/s)^2 = 9000 Joules.
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RESPONSE --> I thought briefly about finding omega ave, but I'm tired and by the time I got to that part I forgot to do it. I still wonder why dW = tau * d omega doesn't come up with the same answer as KE = 1/2 I * omega ^2? self critique assessment: 2
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23:21:29 `q010. Show that this 9000 Joule energy is equal to the product of the torque and the angular displacement.
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RESPONSE --> DUH!!! It isn't d omega it's d theta!!! dW = tau * d theta dW = 3mN * 3000 rad = 9000 J Never mind my previous question. confidence assessment: 3
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23:21:37 The angular displacement is 3000 rad and the torque is 3 m N. Their product is 9000 N m = 9000 Joules. Note that the m N of torque is now expressed as the N m = Joules of work. This is because a radian multiplied by a meter of radius gives a meter of displacement, and work is equal to the product of Newtons and meters of displacement.
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RESPONSE --> self critique assessment: 3
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23:22:51 `q011. How does the previous example illustrate the fact that the work done by a net torque is equal to the product of the torque and the angular displacement?
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RESPONSE --> Because KE = dW confidence assessment: 3
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23:22:58 From the net torque, moment of inertia and time interval we found that the KE increased from 0 to 9000 Joules. We know that the KE increase of a system is equal to the net work done on the system, so 9000 Joules of net work must have been done on the system. Multiplying the angular displacement by the torque gave us 9000 Joules, equal to the KE increase, so at least in this case the work done was the product of the angular displacement and the net torque. It isn't difficult to prove that this is always the case for any system, and that in general the work `dW done by a net torque `tauNet acting through an angular displacement `d`theta is `dW = `tauNet * `d`theta. UNIVERSITY STUDENT COMMENT (relevant only to students who know calculus): Speaking in terms of calculus... 'dW=int('tau with respect to 'theta) from 'theta_1 to 'theta_2 = ('tau*'theta_2)-('tau*'theta_1)='tau*(theta_2-'theta_1)='tau*'d'theta Amazing! INSTRUCTOR RESPONSE: Very good. That will of course work if tau is known as a function of angular position theta (e.g., consider a cam accelerated by a falling mass, in the same manner as the disk with bolts except that the rim of the cam is not at constant distance from the axis of rotation). The shape of the cam may be described in terms of polar coordinates, where the coordinate r is given in terms of the angle theta from the polar axis of the cam.
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RESPONSE --> self critique assessment: 3
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