course Phy 241 ?E??y??H???i??|?yassignment #029
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14:31:00 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 --> average angular velocity refers to the travel in a circular path, therefore we can use the same principals as regular velocity, only relative to radians to account for the circular motion. The average angular velocity can be found by knowing the radius of the circular path and find the corresponding lenght from the angular degrees it passes through. How many degrees the object passes along can be divided by 180 degrees to find the value in radians. Then the value of radians can be divided by time to get a rate. From rest we can assume that v0 is 0 and we can use a `ds=.5a`dt^2 to obtain an acceleration.
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14:32:29 **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 --> ok yes, if acceleratin is uniform and we start at 0, the final velocity is double the average velocity that we calculated from knowing the angle of movement of the object
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14:32:34 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 --> n/a
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14:32:37 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 --> n/a
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14:35:25 Query problem 7.50 3 cubes sides L0, 2L0 and 3L0; center of mass.
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RESPONSE --> LO(0) + 2(LO) + 3 (LO) = 5LO total mass is 6LO so (5/6)LO is the center of mass
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14:41:11 ** 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 y 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 --> ok yes since these are cubes of multiple sizes then we must use the distance from the centers relative to the lowest unit to calculate this. I am a little confused of what the LO means and how we get 1 LO and then 2 LO but get 3/2 LO for the next block
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14:41:31 What is the mass of the second cube as a multiple of the mass of the first?
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RESPONSE --> it is 8 times the mass
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14:41:39 ** 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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14:41:47 What is the mass of the third cube as a multiple of the mass of the first?
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RESPONSE --> it is 27 x as massive
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14:41:51 ** 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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14:42:01 How far from the outside edge of the first cube is its center of mass?
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RESPONSE --> 1 LO
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14:42:27 ** 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 so LO is the lenght of the entire cube, so .5 to the center...?
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14:42:56 How far from the outside edge of the first cube is the center of mass of the second cube?
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RESPONSE --> 1LO
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14:46:24 ** 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 am nt sure why this takes into account the first LO all the way across the first one, and how we only add one more LO to the next one
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14:46:41 How far from the outside edge of the first cube is the center of mass of the third cube?
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RESPONSE --> 3/2
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14:48:03 ** 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 --> I am not sure why this is half for the third cube, but all the way across for the first cube
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14:48:33 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 add the previous masses and then divide by the total mass
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14:59:10 ** 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 yes we have to multiply each LO fraction by the corresponding masses 1,8,27,then divide by the total
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15:27:07 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 --> The entire length of the canoe is 8m. We first need to calculate the center of mass when the woman is on the side of the canoe. so we have 1/8(45)+ 1/4 (60)=20.63 Now for the total 45+60=105, 20.625/105=.2 or about 1.6 m Now to find the center of mass when she is in the middle 1/4 45 + 1/4 60=26.26, divided by the total we get .25 which is about 2m so the canoe moved .4 m.
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15:30:15 ** 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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RESPONSE --> ok I an confused on how long the canoe is and the relative positions. How do we get the center of the canoe at the 2.5 mark as used in the first equatin. So we calculate from the left side, so I am assuming that the canoe is therefore 5 m long.
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