course phy201 ?r???????o?{???]}assignment #029
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13:29:57 Query class notes #28. Explain how we can calculate the average angular velocity and the angular acceleration of an object which rotates from rest through a given angle in a given time interval, assuming constant angular acceleration.
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RESPONSE --> If we know the angle and the radius of the arc we also know the portion of the circumference the object has moved through which is the change in position. Divding this by the change in time gives us average angula velocity. Since the object started from rest, we know that the final angular velocity will be twice the average and will also equal the change in velocity. Knowing the change in velocity we can divide by the change in time to get angular acceleration.
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13:30:40 **This situation is strictly analogous to the one you encountered early in the course. As before acceleration is change in velocity / change in clock time. However now it's angular acceleration. We have angular acceleration = change in angular velocity / change in clock time. The average angular velocity is change in angular position / change in clock time. This question assumes you know the angle through which the object rotates, which is its change in angular position, as well as the change in clock time. So you can calculate the average angular velocity. If angular accel is uniform and initial angular velocity is zero then the final angular velocity is double the average angular velocity. In this case the change in angular velocity is equal to the final angular velocity, which is double the average angular velocity. From this information you can calculate angular acceleration. **
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RESPONSE --> got it
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13:35:23 Principles of Physics and General College Physics Problem 7.46: Center of mass of system 1.00 kg at .50 m to left of 1.50 kg, which is in turn .25 m to left of 1.10 kg.
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RESPONSE --> (1kg*0 + 1.5Kg*.5m + 1.1Kg*.75m) / (1Kg+1.5Kg+1.1Kg) = 1.575/3.6 = CG at .43m
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13:35:53 Using the position of the 1.00 kg mass as the x = 0 position, the other two objects are respectively at x = .50 m and x = .75 m. The total moment of the three masses about the x = 0 position is 1.00 kg * (0 m) + 1.50 kg * (.50 m) + 1.10 kg * (.75 m) = 1.58 kg m. The total mass is 1.00 kg + 1.50 kg + 1.10 kg = 3.60 kg, so the center of mass is at position x_cm = 1.58 kg m / (3.60 kg) = .44 meters, placing it a bit to the left of the 1.50 kg mass.
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RESPONSE --> got it
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13:43:12 Query problem 7.50 3 cubes sides L0, 2L0 and 3L0; center of mass.
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RESPONSE --> (l^3*.5) + (2l^3*1.5) + (3l^3*4) / l^3 + 2l^3 + 3l^3 = 15.5l^3 / 4l^3 = 3.875 CM
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13:44:43 ** The mass of the second will be 2^3 = 8 times as great as the first. It takes 8 1-unit cubes to make a 2-unit cube. The mass of the third will be 3^3 = 27 times as great as the first. It takes 27 1-unit cubes to make a 3-unit cube. In the x direction the distance from left edge to center of first cube is 1/2 L0 (the center of the first cube). In the y direction the distance is from lower edge to center of the first cube is 1/2 L0 (the center of the first cube). In the x direction the distance from left edge to center of the second cube is L0 + L0 (the L0 across the first cube, another L0 to the center of the second), or 2 L0. In the y direction the distance from lower edge to center of the second cube is L0 (the center of the second cube). In the x direction the distance from left edge to center of the third cube is L0 + 2 L0 + 3/2 L0 (the L0 across the first cube, another 2 L0 across the second and half of 3L0 to the center of the third), or 9/2 L0. In the x direction the distance from lower edge to center of the first cube is 3/2 L0 (the center of the third cube). Moments about left edge and lower edge of first cube: If m1 is the mass of the first cube then in the x direction you have total moment m1 * L0/2 + 8 m1 * (2 L0) + 27 M1 * (9/2 L0) = 276 m1 L0 / 2 = 138 m1 L0. The total mass is m1 + 8 m1 + 27 m1 = 36 m1 so the center of mass is at center of mass in x direction: 138 m1 L0 / (36 m1) = 3.83 L0. In the y direction the moment is m1 * L0/2 + 8 m1 * (L0) + 27 m1 * ( 3/2 L0) = 45 m1 L0 so the center of mass is at center of mass in y direction: 45 m1 L0 / (36 m1) = 1.25 L0. **
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RESPONSE --> I got the right answer in the X direction but negelected the Y direction. Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
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13:47:47 What is the mass of the second cube as a multiple of the mass of the first?
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RESPONSE --> the first cube has a mass of one l^3 The second has a mass of (2l)^3 or eight l^3 So multiple is 8
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13:47:57 ** 3 dimensions: the mass will be 2^3 = 8 times as great. It takes 8 1-unit cubes to make a 2-unit cube. **
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RESPONSE --> ok
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13:48:55 What is the mass of the third cube as a multiple of the mass of the first?
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RESPONSE --> The third has a mass of (3l)^3 or 27 l^3 so the multiplier is 27
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13:49:01 ** The mass of the third cube is 3^3 = 27 times the mass of the first. **
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RESPONSE --> ok
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13:49:24 How far from the outside edge of the first cube is its center of mass?
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RESPONSE --> .5l
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13:49:35 ** In the x direction the distance is 1/2 L0 (the center of the first cube). In the y direction the distance is also 1/2 L0 (the center of the first cube). **
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RESPONSE --> ok
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13:50:12 How far from the outside edge of the first cube is the center of mass of the second cube?
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RESPONSE --> 1.5l
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13:52:02 ** In the x direction the distance is L0 + L0 (the L0 across the first cube, another L0 to the center of the second), or 2 L0. In the x direction the distance is L0 (the center of the second cube). **
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RESPONSE --> I didn't read the problem carefully enough. I thought it was center to center not edge to center.
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13:52:32 How far from the outside edge of the first cube is the center of mass of the third cube?
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RESPONSE --> 4.5l
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13:52:47 ** In the x direction the distance is L0 + 2 L0 + 3/2 L0 (the L0 across the first cube, another 2 L0 across the second and half of 3L0 to the center of the third), or 9/2 L0. In the x direction the distance is 3/2 L0 (the center of the third cube). **
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RESPONSE --> ok
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13:54:16 How do you use these positions and the masses of the cubes to determine the position of the center of mass of the system?
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RESPONSE --> calculate the mass of each object and multiply it tile it's distance from a common refernce point in each axis then divide this product by the otal of all the masses.
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13:54:32 ** In the x direction you have moment m1 * L0/2 + 8 m1 * (2 L0) + 27 M1 * (9/2 L0) = 276 m1 L0 / 2 = 138 m1 L0. The total mass is m1 + 8 m1 + 27 m1 = 36 m1 so the center of mass is at 138 m1 L0 / (36 m1) = 3.83 L0. In the y direction the moment is m1 * L0/2 + 8 m1 * (L0) + 27 m1 * ( 3/2 L0) = 45 m1 L0 so the center of mass is at 45 m1 L0 / (36 m1) = 1.25 L0. **
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RESPONSE --> ok
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