#$&* course PHY 121 7/30 12 032. Moment of inertia
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Given Solution: A mass of m = .05 kg at distance r = .30 meters from the axis of rotation has moment of inertia I = m r^2 = .05 kg * (.30 m)^2 = .0045 kg m^2. The moment of inertia of the new system will therefore be the sum .0713 kg m^2 + .0045 kg m^2 = .0758 kg m^2 of the moments of inertia of its components, the disk and the magnet. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q002. A uniform rod with mass 5 kg is 3 meters long. Masses of .5 kg are added at the ends and at .5 meter intervals along the rod. What is the moment of inertia of the resulting system about the center of the rod? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For the rod: 1/12 mL^2 = 1/12 (5 kg)(3m)^2 = 1/12 (5 kg)(9m^2) = 3.75 kg m^2 For the weights at the ends (1.5 meters from the center)(mr^2): .5 kg (1.5m)^2 = .5 kg (2.25 m^2) = 1.125 kg m^2 There are two weights, so 2 * 1.125 kg m^2 = 2.25 kg m^2 For the weights at .5 m in (1.0 meters from the center)(mr^2): .5 kg (1.0 m)^2 = .5 kg (1m^2) = .5 kg m^2 There are two weights, so 2 * .5 kg m^2 = 1 kg m^2 For the weights at .5 more meters in (.5 meters from the center)(mr^2): .5 kg (.5 m)^2 = .5 kg (.25 m^2) = .125 kg m^2 There are two weights, so 2 * .125 kg m^2 = .25 kg m^2 The weight at the center would be at 0 radius, so the moment inertia would be 0 kg m^2. Then you have to add this all up: 3.75 kg m s^2 + 2.25 kg m^2 + 1.00 kg m^2 + .25 kg m^2 + 0 kg m^2 = 7.25 kg m^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Inertia for the disk: .5mr^2 .5 (8 kg)(.4 m)^2 .5 (8 kg)(.16 m^2) .64 kg m^2 Inertia for the rod: 1/12 mL^2 length of rod is 2 * radius of disk: 2 * .4 m = .8 m 1/12 (10 kg)(.8 m)^2 1/12 (10 kg)(.64 m^2) .53 kg m^2 Total Inertia for the system: .64 kg m^2 + .53 kg m^2 = 1.17 kg m^2 `alpha = .8 m N / 1.17 kg m^2 `alpha = .684 radians/s^2 .684 radians/s^2 / `pi = .218 `pi radians/s^2 I put it in `pi terms because I know that I'm trying to find out how long it takes for the system to go through 1 rotation, which is 2 `pi. But now I am stuck..... confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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.. INSTRUCTOR'S FURTHER CLARIFICATION You have the acceleration and you know how far the system travels from rest. If the system was accelerating along a line these quantities would be a, `ds and v0. The third equation of uniformly accelerated motion (`ds = v0 `dt + 1/2 a `dt^2) would apply, with `dt as the unknown. Since v0 = 0 the equation becomes `ds = 1/2 a `dt^2. That equation isn't appropriate here because this is a rotating system. So instead of using `ds we use `dTheta, which is the symbol for angular displacement, and instead of a we use alpha, the symbol for angular acceleration. Our equation therefore translates to `dTheta = 1/2 alpha `dt^2. Now using alpha = .7 rad/s^2 and `dTheta = 2 pi radians, we solve for `dt. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Starting where I left off, I got: `ds = v0`dt + .5a`dt^2 `dt^2 = 2 `ds / a `dt^2 = 4 `pi radians/ .218 pi radians/s^2 `dt^2 = 18.34862385 s^2 `dt = +-4.28 s `dt = 4.28 s ------------------------------------------------ Self-critique rating: 3 If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: `q004. To a uniform rod of length 30 cm and mass 80 grams, initially at rest and constrained to rotate about an axis through its center, we add a 20-gram domino at one end and a 50-gram magnet on the other side of the axis, at a distance of 6 cm from the axis. Show that the rod will balance about its center. If the rod is subject to a torque of .03 m N for 5 seconds, what will be its angular velocity? What centripetal force will be required to keep the domino moving around its circular path? Answer the same for the magnet. What will be the kinetic energy of the domino? Answer the same for the magnet. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Moment of Inertia for Magnet: .05 kg (.06 m)^2 .05 kg (.0036 m^2) 1.8 * 10^-4 kg m^s Moment of Inertia for Domino: .02 kg (.15 m)^2 .02 kg (.0225 m^2) 4.5 * 10^-4 kg m^2 Moment of Inertia for the Rod 1/12 (.08 kg)(.3 m)^2 1/12 (.08 kg)(.09 m^2) 6 * 10^-4 kg m^2 Total Inertia for the System: (1.8 * 10^-4) + (4.5 * 10^-4) + (6 * 10^-4) = .00123 kg m^2 Acceleration = Torque/Inertia `alpha = .03 m N / .00123 kg m^2 `alpha = 24.4 radians/s^2 Angular velocity: 24.4 radians/s^2 * 6 s = 122 radians / s Circumference for magnet: 2 `pi*r 2 (3.14)(.06 m) = 3.77 m