course Phy 121 ??V?g???€??????}?Student Name:
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23:10:53 `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 --> Steel disk: I = 0.0713 kg*m^2 Magnet: m = .05 kg r = .30 m I = m*r^2 = 0.05 kg * (0.3m)^2 = 0.0045 kg*m^2 In order to find the total moment of inertia for the new system, I simply added the two moments of inertia for the disk and the magnet. I total = 0.0713kg*m^2 + 0.0045kg*m^2 = 0.0758 kg*m^2
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23:11:54 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 understand this problem.
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23:13:28 `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 = 5 kg L = 3 m 6 Distances (every 0.5 meter from center): 0.5m, 1.0m, 1.5m m2 = 0.5 kg Inertia around center: I = 1/12 (m)(L^2) = 1/12(5kg)(3m)^2 = 3.75 kg*m^2 Now we find the moments of inertia for each distance at which 0.5kg is added: I = m*r^2 = (0.5kg)(0.5m)^2 = 0.125 kg*m^2 I = (0.5kg)(1.0m)^2 = 0.5 kg*m^2 I = (0.5kg)(1.5m)^2 = 1.125 kg*m^2 Since there are six distances from the center on both sides of the rod we mulitply 2 by the total of moments of inertia from the one side we have calculated: I = 2(0.125kg*m^2 + 0.5kg*m^2 + 1.125kg*m^2) = 3.5 kg*m^2 Now we add this moment of inertia to the moment of inertia for the rod spinning around the center to get the total moment of inertia: I total = 3.5kg*m^2 + 3.75kg*m^2 = 7.25 kg*m^2
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23:15:55 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 understand this process. The mass at the center of the rod does not contribute to the moment of inertia.
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23:17:59 `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 = 8kg r = 0.4 m I = 1/2(m*r^2) = 1/2(8kg)(0.4m)^2 = 0.64 kg*m^2 Uniform Rod: m = 10kg r = 0.4m * 2 = 0.8m I = 1/12(m*r^2) = 1/12(10kg)(0.8m)^2 = 0.533 kg*m^2 Total moment of inertia = I disk + I rod I = 0.64kg*m^2 + 0.533kg*m^2 = 1.173 kg*m^2 `tau = 0.8 N*m To find the acceleration of the system, we divide the given torque by the moment of inertia for the whole system: `alpha = `tau / I = 0.8 N*m / 1.173 kg*m^2 = 0.682 rad/s^2 Now to find the time interval to complete one rotation, we use one of the equations of uniformly accelerated motion: `ds = (vo)(`dt) + 1/2(a)(`dt)^2 The initial velocity is zero, therefore the equation changes to, `ds = 1/2(a)(`dt^2) For one rotation, the distance in radians is represented as 2 pi*radian. The acceleration is what we just found. 2 pi*rad = 1/2(0.682 rad/s^2)(`dt^2) `dt^2 = (2 pi*rad)/(0.341 rad/s^2) = 5.865 pi*s^2 = 18.43 s^2 `dt = 4.29 s
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23:18:43 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 understand this concept. Time cannot have a negative value.
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????U????????assignment #032 032. `query 32 Physics I 07-21-2006
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07:28:43 Query experiment to be viewed. What part or parts of the system experiences a potential energy decrease? What part or parts of the system experience(s) a kinetic energy increase?
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RESPONSE --> PE decrease: The object on the string will lose potential energy because it is falling. KE increase: The axel, the weight, and the wheel will gain kinetic energy.
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07:29:39 ** The mass on the string descends and loses PE. The wheel and the descending mass both increase in KE, as do the other less massive parts of the system (e.g., the string) and slower-moving parts (e.g., the axel, which rotates at the same rate as the wheel but which due to its much smaller radius does not move nearly as fast as most of the wheel). **
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RESPONSE --> I understand how this experiment works. I forgot to take into account that the rest of the system will experience a KE increase as well.
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07:31:25 What part or parts of the system experience(s) an increase in angular kinetic energy?
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RESPONSE --> Angular KE increase: (All objects that are turning) styrofoam disk, axle, bolts
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07:31:36 ** The wheel, the bolts, the axle, and anything else that's rotating experiences an increase in angular KE. **
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RESPONSE --> I understand this concept.
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07:32:50 What part or parts of the system experience(s) an increasing translational kinetic energy?
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RESPONSE --> Translational KE increase: (Any object that is falling) weight on the string
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07:33:11 ** Only the descending mass experiences an increase in translational KE. **
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RESPONSE --> I understand this concept.
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07:36:18 Does any of the bolts attached to the Styrofoam wheel gain more kinetic energy than some other bolt? Explain.
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RESPONSE --> The bolts that are closer to the outer edge of the styrofoam disk will experience more KE increase because they are turning at a greater angular velocity. Since KE and velocity are directly proportional, if velocity increases, KE increases as well.
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07:37:12 ** The bolts toward the outside of the wheel are moving at a greater velocity relative to some fixed point, so their kinetic energy is greater since k = 1/2 m v^2 **
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RESPONSE --> I understand this concept.
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07:38:05 What is the moment of inertia of the Styrofoam wheel and its bolts?
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RESPONSE --> The moment of inertia for the styrofoam wheel is 1/2m*r^2 and the moment of inertia for each of the bolts is just m*r^2. Therefore the moment of inertia for the system is the sum of the moments of inertia for the wheel and the bolts.
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07:38:38 ** The moment of inertia for the center of its mass=its radias times angular velocity. Moment of inertia of a bolt is m r^2, where m is the mass and r is the distance from the center of mass. The moment of inertia of the styrofoam wheel is .5 M R^2, where M is its mass and R its radius. The wheel with its bolts has a moment of inertia which is equal to the sum of all these components. **
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RESPONSE --> I understand this concept.
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07:39:08 How do we determine the angular kinetic energy of of wheel by measuring the motion of the falling mass?
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RESPONSE --> Angular KE of wheel: The velocity of the falling mass is equal to the velocity of the center of the wheel. This velocity is divided by the radius of the wheel to find the angular velocity of the wheel. Once we know the angular velocity of the wheel we square it and multiply it by 1/2*mass to get its angular kinetic energy.
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07:39:29 ** STUDENT ANSWER AND INSTRUCTOR CRITIQUE: The mass falls at a constant acceleration, so the wheel also turns this fast. INSTRUCTOR CRITIQUE: We don't use the acceleration to find the angular KE, we use the velocity. The acceleration, if known, can be used to find the velocity. However in this case what we are really interested in is the final velocity of the falling mass, which is equal to the velocity of the part of the wheel around which it is wound. If we divide the velocity of this part of the wheel by the its radius we get the angular velocity of the wheel. **
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RESPONSE --> I understand this problem.
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07:40:39 Principles of Physics and General College Physics problem 8.43: Energy to bring centrifuge motor with moment of inertia 3.75 * 10^-2 kg m^2 to 8250 rpm from rest.
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RESPONSE --> I = 3.75*10^-2 kg*m^2 `omega = (8250 rpm)[(pi/30 rad/s)/rpm] = 864 rad/s The kinetic energy of the centrifuge is found by the equation, KE = 1/2(I)(`omega^2), which is similar to the equation we already know KE = 1/2(m)(v^2). KE = 1/2(3.75*10^-2 kg*m^2)(864 rad/s)^2 = 13997 Joules
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07:41:15 The KE of a rotating object is KE = .5 I omega^2, where I is the moment of inertia and omega the angular velocity. Since I is given in standard units of kg m^2, the angular velocity should be expressed in the standard units rad / sec. Since 8250 rpm = (8250 rpm) * (pi / 30 rad/sec) / rpm = 860 rad/sec, approx.. The initial KE is 0, and from the given information the final KE is KE_f = .5 I omega_f ^ 2 = .5 * 3.75 * 10^-2 kg m^2 * (860 rad/sec)^2 = 250 pi^2 kg m^2 / sec^2 = 14000 Joules.
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RESPONSE --> I understand this problem.
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07:42:58 Query gen problem 8.58 Estimate KE of Earth around Sun (6*10^24 kg, 6400 km rad, 1.5 * 10^8 km orb rad) and about its axis. What is the angular kinetic energy of the Erath due to its rotation about the Sun? What is the angular kinetic energy of the Earth due to its rotation about its axis?
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RESPONSE --> I was not required to answer this text problem.
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07:43:14 ** The circumference of the orbit is 2pi*r = 9.42*10^8 km. We divide the circumference by the time required to move through that distance to get the speed of Earth in its orbit about the Sun: 9.42 * 10^8 km / (365days * 24 hrs / day * 3600 s / hr) =29.87 km/s or 29870 m/s. Dividing the speed by the radius we obtain the angular velocity: omega = (29.87 km/s)/ (1.5*10^8 km) = 1.99*10^-7 rad/s. From this we get the angular KE: KE = 1/2 mv^2 = 1/2 * 6*10^24 kg * (29870 m/s)^2 = 2.676*10^33 J. Alternatively, and more elegantly, we can directly find the angular velocity, dividing the 2 pi radian angular displacement of a complete orbit by the time required for the orbit. We get omega = 2 pi rad / (365days * 24 hrs / day * 3600 s / hr) = 1.99 * 10^-7 rad/s. The moment of inertia of Earth in its orbit is M R^2 = 6 * 10^24 kg * (1.5 * 10^11 m)^2 = 1.35 * 10^47 kg m^2. The angular KE of the orbit is therefore KE = .5 * I * omega^2 = .5 * (1.35 * 10^47 kg m^2) * (1.99 * 10^-7 rad/s)^2 = 2.7 * 10^33 J. The two solutions agree, up to roundoff errors. The angular KE of earth about its axis is found from its angular velocity about its axis and its moment of inertia about its axis. The moment of inertia is I=2/5 M r^2=6*10^24kg * ( 6.4 * 10^6 m)^2 = 9.83*10^37kg m^2. The angular velocity of the Earth about its axis is 1 revolution / 24 hr = 2 pi rad / (24 hr * 3600 s / hr) = 7.2 * 10^-5 rad/s, very approximately. So the angular KE of Earth about its axis is about KE = .5 I omega^2 = .5 * 9.8 * 10^37 kg m^2 * (7.2 * 10^-5 rad/s)^2 = 2.5 * 10^29 Joules. **
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RESPONSE --> Ok. NOTE: The circumference of the orbit is 2pi*r = 9.42*10^8 km. Dividing the speed by the radius we obtain the angular velocity (`omega). The angular KE of earth about its axis is found from its angular velocity about its axis and its moment of inertia about its axis. The moment of inertia is I=2/5(m*r^2)
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07:44:31 Query problem 8.60 uniform disk at 2.4 rev/sec; nonrotating rod of equal mass, length equal diameter, dropped concentric with disk. Resulting angular velocity?
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RESPONSE --> This problem involves a uniform disk and a uniform rod. For the moments of inertia we know the following: Disk: I = 2/5(m*r^2) Rod: I = 1/12(m*r^2) We do not know the moments of inertia or the masses of the two, so the only way to find the disk's final angular velocity is to find the ratio of it to the total velocity through the ratio of the moments of inertia. I total = I disk + I rod = (2/5m*r^2) + (1/12m*r^2) = 29/60 m*r^2 I total / I rod = (29/60 m*r^2)/(1/12m*r^2) = 29/5 m*r^2 Therefore the rod's angular velocity is 5/29 of the total velocity and the disk's angular velocity is (29-5=24) 24/29 of the total velocity. Disk `omegaF = (24/29)(2.4 rev/s) = 1.986 revolutions per second.
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07:44:38 ** The moment of inertia of the disk is I = 2/5 M R^2; the moment of inertia of the rod about its center is 1/12 M L^2. The axis of rotation of each is the center of the disk so L = R. The masses are equal, so we find that the moments of inertia can be expressed as 2/5 M R^2 and 1/12 M R^2. The combined moment of inertia is therefore 2/5 M R^2 + 1/12 M R^2 = 29/60 M R^2, and the ratio of the combined moment of inertia to the moment of the disk is ratio = (29/60 M R^2) / (2/5 M R^2) = 29/60 / (2/5) = 29/60 * 5/2 = 145 / 120 = 29 / 24. Since angular momentum I * omega is conserved an increase in moment of inertia I results in a proportional decrease in angular velocity omega so we end up with final angular velocity = 24 / 29 * initial angular velocity = 24 / 29 * 2.4 rev / sec = 2 rev/sec, approximately.
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RESPONSE --> I understand this problem.
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07:45:48 Univ. 10.64 (10.56 10th edition). disks 2.5 cm and .8 kg, 5.0 cm and 1.6 kg, welded, common central axis. String around smaller, 1.5 kg block suspended. Accel of block? Then same bu wrapped around larger.
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RESPONSE --> I am not required to answer this question.
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07:45:51 ** The moment of inertia of each disk is .5 M R^2; the block lies at perpendicular distance from the axis which is equal to the radius of the disk to which it is attached. So the moment of inertia of the system, with block suspended from the smaller disk, is I = .5 (.8 kg) * ( .025 m)^2 + .5 * 1.6 kg * (.05 m)^2 + (1.5 kg * .025 m)^2= .0032 kg m^2 approx. The 1.5 kg block suspended from the first disk results in torque tau = F * x = .025 m * 1.5 kg * 9.8 m/s^2 = .37 m N approx. The resulting angular acceleration is alpha = tau / I = .37 m N / (.0032 kg m^2) = 115 rad/s^2 approx. The acceleration of the block is the same as the acceleration of a point on the rim of the wheel, which is a = alpha * r = 115 rad/s^2 * .025 m = 2.9 m/s^2 approx. The moment of inertia of the system, with block suspended from the larger disk, is I = .5 (.8 kg) * ( .025 m)^2 + .5 * 1.6 kg * (.05 m)^2 + (1.5 kg * .05 m)^2= .006 kg m^2 approx. The 1.5 kg block suspended from the first disk results in torque tau = F * x = .05 m * 1.5 kg * 9.8 m/s^2 = .74 m N approx. The resulting angular acceleration is alpha = tau / I = .74 m N / (.006 kg m^2) = 120 rad/s^2 approx. The acceleration of the block is the same as the acceleration of a point on the rim of the wheel, which is a = alpha * r = 120 rad/s^2 * .05 m = 6 m/s^2 approx. **
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RESPONSE -->
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