#$&* course PHY 201 August 1 around 2:30 pm 029. `query 29*********************************************
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Given Solution: `a**This situation is strictly analogous to the one you encountered early in the course. As before acceleration is change in velocity / change in clock time. However now it's angular acceleration. We have angular acceleration = change in angular velocity / change in clock time. The average angular velocity is change in angular position / change in clock time. This question assumes you know the angle through which the object rotates, which is its change in angular position, as well as the change in clock time. So you can calculate the average angular velocity. If angular accel is uniform and initial angular velocity is zero then the final angular velocity is double the average angular velocity. In this case the change in angular velocity is equal to the final angular velocity, which is double the average angular velocity. From this information you can calculate angular acceleration. ** Principles of Physics and General College Physics Problem 7.46: Center of mass of system 1.00 kg at .50 m to left of 1.50 kg, which is in turn .25 m to left of 1.10 kg. Using the position of the 1.00 kg mass as the x = 0 position, the other two objects are respectively at x = .50 m and x = .75 m. The total moment of the three masses about the x = 0 position is 1.00 kg * (0 m) + 1.50 kg * (.50 m) + 1.10 kg * (.75 m) = 1.58 kg m. The total mass is 1.00 kg + 1.50 kg + 1.10 kg = 3.60 kg, so the center of mass is at position x_cm = 1.58 kg m / (3.60 kg) = .44 meters, placing it a bit to the left of the 1.50 kg mass. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qQuery problem 7.50 3 cubes sides L0, 2L0 and 3L0; center of mass. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The total of the lengths between the three cubes would be the addition of all three of the sides: L0 + 2L0 + 3L0 = 6L0. Now, you would divide how many cubes there are by this total number to get the center of mass: 6L0 / 3 = 2L0 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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) = 49 m1 L0 so the center of mass is at center of mass in y direction: 49 m1 L0 / (36 m1) = 1.36 L0. ** STUDENT QUESTION I don’t understand why in the Y direction the “equation” isn’t identical to that of the X…why is X 2L0 and Y just L0…since the cubes have to be uniform this doesn’t make sense to me. INSTRUCTOR RESPONSE The coordinates are for the center of mass. Each cube rests on the x axis. The first cube extends in the vertical direction from the x axis to y = L0, so its center of mass in the y direction is at 1/2 L0. The second cube extends in the vertical direction from the x axis to y = 2 L0, so its center of mass in the y direction is at L0. The third cube extends in the vertical direction from the x axis to y = 3 L0, so its center of mass in the y direction is at 3/2 L0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): How do you know that the mass of the second is 8 times that of the first, and the mass of the third 27 times that of the first, with just the given information????? Is this just approximation?? I guess it takes time to get custom to the situation.
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Given Solution: `a** 3 dimensions: the mass will be 2^3 = 8 times as great. It takes 8 1-unit cubes to make a 2-unit cube. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Why is it a multiple of 3???
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Given Solution: `a** The mass of the third cube is 3^3 = 27 times the mass of the first. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qHow far from the outside edge of the first cube is its center of mass? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The whole has a x-distance of L0, so the distance from the outside edge to its center of mass of half: 1/2 L0. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** In the x direction the distance is 1/2 L0 (the center of the first cube). In the y direction the distance is also 1/2 L0 (the center of the first cube). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok, so you want the distance in the x and y direction. ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qHow far from the outside edge of the first cube is the center of mass of the second cube? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The x-distance from the outside edge of the first cube to the center of mass of the second cube would be the length of the first cube L0 plus half of the second cube 2L0 / 2 = 1L0: L0 + L0 = 2L0. The y-distance from the outside edge would be 1L0. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** In the x direction the distance is L0 + L0 (the L0 across the first cube, another L0 to the center of the second), or 2 L0. In the y direction the distance is L0 (the center of the second cube). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qHow far from the outside edge of the first cube is the center of mass of the third cube? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The x-distance from the outside edge of the first cube to the center of mass of the third cube would be the distance of the first and second cubes L0 and 2L0, plus half of the third cube 3L0 / 2 = 1.5 L0 or 3/2 L0: L0 + 2L0 + 3/2 L0 = 9/2 L0. The y-distance from the center of mass of the third cube would be half of the distance of the whole third cube: 3L0 / 2 = 3/2 L0 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** In the x direction the distance is L0 + 2 L0 + 3/2 L0 (the L0 across the first cube, another 2 L0 across the second and half of 3L0 to the center of the third), or 9/2 L0. In the x direction the distance is 3/2 L0 (the center of the third cube). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qHow do you use these positions and the masses of the cubes to determine the position of the center of mass of the system? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The positions and the mass of each of the cubes can be combined to get the position of the center of mass of all the cubes, by using this equation: m1 * `ds + m2 * `ds + m3 * `ds = CM. This will be used for the x and y directions. x-direction = m1 * L0/2 + 8 m1 * (2 L0) + 27 M1 * (9/2 L0) = 276 m1 L0 / 2 = 138 m1 L0 Then, we need to find the total mass of all of the cubes, which is: m1 + 8 m1 + 27 m1 = 36 m1. Now, you can just divide the x-direction by the total mass to get the center of mass in the x-direction: 138 m1 L0 / 36 m1 = 3.83 L0. y-direction = m1 * L0/2 + 8 m1 * (L0) + 27 m1 * ( 3/2 L0) = 49 m1 L0 Then, again we need to divide the y-direction by the total mass to get the center of mass in the y-direction: 49 m1 L0 / 36 m1 = 1.36 L0. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** In the x direction you have moment m1 * L0/2 + 8 m1 * (2 L0) + 27 M1 * (9/2 L0) = 276 m1 L0 / 2 = 138 m1 L0. The total mass is m1 + 8 m1 + 27 m1 = 36 m1 so the center of mass is at 138 m1 L0 / (36 m1) = 3.83 L0. In the y direction the moment is m1 * L0/2 + 8 m1 * (L0) + 27 m1 * ( 3/2 L0) = 45 m1 L0 so the center of mass is at 45 m1 L0 / (36 m1) = 1.25 L0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK I think that the y-direction in the given solution is supposed to be 49 m1 L0. ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qUniv. 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Since water resistance is negligible the net force acting on the system is zero. Since the system is initially stationary the center of mass of the system is at rest; since zero net force acts on the system this will continue to be the case. Assuming that the center of mass of the canoe is at the center of the canoe, then when the woman is 1 m from the left end the center of mass of the system lies at distance c.m.1 = (1 m * 45 kg + 2.5 m * 60 kg) / (45 kg + 60 kg) = 195 kg m / (105 kg) = 1.85 m from the left end of the canoe. A similar analysis shows that when the woman is 1 m from the right end of the canoe, then since she is 4 m from the left end the center of mass lies at c.m.2 = (4 m * 45 kg + 2.5 m * 60 kg) / (45 kg + 60 kg) = 310 kg m / (105 kg) = 2.97 m. The center of mass therefore changes its position with respect to the left end of the canoe by about 1.1 meters toward the right end of the canoe. Since the center of mass itself doesn't move the canoe must move 1.1 meters toward the left end, i.e., backwards. Note that since the woman moves 3 m forward with respect to the canoe and the canoe moves 1.3 m backwards the woman actually moves 1.7 m forward. The sum -1.3 m * 60 kg + 1.7 m * 45 kg is zero, to within roundoff error. This is as it should be since this sum represents the sum of the changes in the centers of mass of the canoe and the woman, which is the net change in the position of center of mass. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* "