course Phy 201 ɖ{assignment #032
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13:55:30 `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 --> We know that to find the sum we have to add the moments of inertias; however, we only know one, the other we have to find. The first we know is .0713 kg m^2 The new one we have to use the formula to find: mr^2 =.05 kg * (.30 m)^2 = .0045 kg m^2 Now we add the two to get .0758 kg m^2 confidence assessment: 2
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13:55:37 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 --> I undestand. self critique assessment: 3
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13:58:08 `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 --> I have no idea how to do this really. I know I am supposed to use the moment of inertia formula, but I am not sure how to fill in the length after the masses are added and I am not even sure which equations to use. confidence assessment: 0
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15:04:30 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 --> I should have first known to use the formula for a rod with uniformly distributed mass and then filled in the variables appropriately. Then, one should find the inertias for all the masses at each corresponding distance and add them together for the sum of the moment inertias. This should have been easy, but for some reason I was trying to make it more complicated than what it was: simply fill in the mass and the corresponding distance from the center to the equation and add the sum of the moments of inertias. self critique assessment: 2
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19:56:39 `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 --> In order to use the equation torque divided by inertia, we must first find the moment of inertia for both the rod and the disk. So, for the disk: .5MR^2 =1/2 * 8 kg * (.4 m)^2 =.64 kg m^2 For the rod: 1/12 M L^2 =1/12 * 10 kg * (.8 m)^2 =.533333 kg*m^2 In orderto calculate 'alpha, we have to add the moments of the inertias to find the sum of them: .64 + .53 =1.17 kg m^2 Now we have found I and are given 'tau, so we can fill the variables into the equation: 'alpha='tau/I =.8 m N / (1.17 kg m^2) = about .7 rad/s^2 So now we know the acceleration, the torque, the inertia, and the initial velocity (we are told it starts from rest). We also know the displacement is one revolution, 360 degrees, or 2 pi radians. From here, I am not sure which equation to use... confidence assessment: 2
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19:57:45 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 see now that I should have merely used the equation of acceleration for 'ds and then solve it for 'dt. That would have been simple enough to do--I merely overcomplicated things! self critique assessment: 2