course phy 121 ^~x⟛Dhϭassignment #029
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23:13:49 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 --> To find the ave angular velocity you divide the change in angular position by the time interval. To find the angular position in radians you multiply the angle by pi rad/180 degrees. To find the angular acceleration you divide the ang velocity by the time interval.
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23:17:02 **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 --> I think I might have said to use th ave ang vel to find the accel, but I know that isn't right - I just forget all the time - you use the final ang velocity.
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23:19:24 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 --> [(1 kg * 0) + (1.5kg * .5m) + (1.10kg * .75m) ]/ 3.6kg = .4375m
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23:19:34 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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23:27:50 Query problem 7.50 3 cubes sides L0, 2L0 and 3L0; center of mass.
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RESPONSE --> Principles of Physics wasn't assigned this problem, but I will take a stab at it. m = mass, l = length [(1m * .5l) + (8m * 2l) + (27m * 4.5l)] / 36m = (.5 + 16 + 108) / 36 = 124.5 / 36 = 3.5 l The CM is 1/2 length into the the largest cube from the left hand side (the side closest to the middle and small cubes).
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23:29:16 What is the mass of the second cube as a multiple of the mass of the first?
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RESPONSE --> 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. The second cube is 8 times as great as the 1st cube in mass
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23:29:23 ** 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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23:29:40 What is the mass of the third cube as a multiple of the mass of the first?
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RESPONSE --> The mass of the third cube is 27 times as great
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23:29:44 ** The mass of the third cube is 3^3 = 27 times the mass of the first. **
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23:30:06 How far from the outside edge of the first cube is its center of mass?
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RESPONSE --> 1/2 the length of the first cube
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23:30:12 ** 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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23:31:24 How far from the outside edge of the first cube is the center of mass of the second cube?
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RESPONSE --> The CM of the 2nd cube is one length from the outside edge of the first cube
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23:31:55 ** 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 --> Oh! from the outside edge, not the inside edge
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23:32:15 How far from the outside edge of the first cube is the center of mass of the third cube?
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RESPONSE --> 4.5 lengths
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23:32:25 ** 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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23:33:41 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 --> You multiply the position of each cube by the mass of each cube and add all of the results. Then divide the answer by the total mass of the three cubes. I don't get why it works, but its fun
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23:33:46 ** 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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23:33:54 Univ. 8.94 (8.82 10th edition). 45 kg woman 60 kg canoe walk starting 1 m from left end to 1 m from right end, moving 3 meters closer to the right end. How far does the canoe move? Water resistance negligible.
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RESPONSE --> Not applicable
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23:33:57 ** Since water resistance is negligible the net force acting on the system is zero. Since the system is initially stationary the center of mass of the system is at rest; since zero net force acts on the system this will continue to be the case. Assuming that the center of mass of the canoe is at the center of the canoe, then when the woman is 1 m from the left end the center of mass of the system lies at distance c.m.1 = (1 m * 45 kg + 2.5 m * 60 kg) / (45 kg + 60 kg) = 195 kg m / (105 kg) = 1.85 m from the left end of the canoe. A similar analysis shows that when the woman is 1 m from the right end of the canoe, then since she is 4 m from the left end the center of mass lies at c.m.2 = (4 m * 45 kg + 2.5 m * 60 kg) / (45 kg + 60 kg) = 310 kg m / (105 kg) = 2.97 m. The center of mass therefore changes its position with respect to the left end of the canoe by about 1.1 meters toward the right end of the canoe. Since the center of mass itself doesn't move the canoe must move 1.1 meters toward the left end, i.e., backwards. Note that since the woman moves 3 m forward with respect to the canoe and the canoe moves 1.3 m backwards the woman actually moves 1.7 m forward. The sum -1.3 m * 60 kg + 1.7 m * 45 kg is zero, to within roundoff error. This is as it should be since this sum represents the sum of the changes in the centers of mass of the canoe and the woman, which is the net change in the position of center of mass. **
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