course Phy 121 Ȟ˞z㰹cRStudent Name:
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14:48:42 `q001. Note that this assignment contains 15 questions. If an object moves a distance along the arc of a circle equal to the radius of the circle, it is said to move through one radian of angle. If a circle has a radius of 40 meters, then how far would you have to walk along the arc of the circle to move through one radian of angle? How far would you have to walk to move through 3 radians?
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RESPONSE --> r = 40 meters rad = 3 In order to find the distance traveled over an angle of 3 radians on a circle with a radius of 40 meters, we must multiply the number of radians by the radius of the circle. `ds = 40m * 3 = 120 meters
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14:49:11 Since 1 radian of angle corresponds to the distance along the arc which is equal to the radius, if the radius of the circle is 40 meters then a 1 radian angle would correspond to a distance of 40 meters along the arc. An angle of 3 radians would correspond to a distance of 3 * 40 meters = 120 meters along the arc. Each radian corresponds to a distance of 40 meters along the arc.
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RESPONSE --> I understand this problem. NOTE: One radian of angle corresponds to the distance along the arc which is equal to the radius.
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14:53:03 `q002. On a circle of radius 40 meters, how far would you have to walk to go all the way around the circle, and through how many radians of angle would you therefore travel? Through how many radians would you travel if you walked halfway around the circle? Through how many radians would you travel if you walked a quarter of the way around the circle?
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RESPONSE --> Around whole: d = 2r = 2 * 40m = 80 meters C = pi * d = pi * 80m = 251.33 m Since 40 meters is the radius, we can tell from the equation, 2*pi*40m that it travels 2*pi radians. Halfway: This distance would be half of the circumference. Therefore (pi*80m)/2 = pi*40m. 40 meters is the radius, so we know that we travel pi radians. Quarter: This distance would be a quarter of the circumference: (pi*80m)/4 = pi*20m. 20meters is half of the radius. Therefore, this distance is 1/2(pi*40m). We walked 1/2*pi radians.
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14:53:29 The circumference of a circle is the product of `pi and its diameter, or in terms of the radius r, which is half the diameter, C = 2 `pi r. The circumference of this circle is therefore 2 `pi * 40 meters = 80 `pi meters. This distance can be left in this form, which is exact, or if appropriate this distance can be approximated as 80 * 3.14 meters = 251 meters (approx). The exact distance 2 `pi * 40 meters is 2 `pi times the radius of the circle, so it corresponds to 2 `pi radians of arc. Half the arc of the circle would correspond to a distance of half the circumference, or to 1/2 ( 80 `pi meters) = 40 `pi meters. This is `pi times the radius so corresponds to `pi radians of angle. A quarter of an arc would correspond to half the preceding angle, or `pi/2 radians.
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RESPONSE --> I understand.
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14:55:23 `q003. On a circle of radius 6 meters, what distance along the arc would correspond to 3 radians? What distance would correspond to `pi / 6 radians? What distance would correspond to 4 `pi / 3 radians?
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RESPONSE --> r = 6m 3 radians: `ds = 3*r = 3*6m = 18 meters pi/6 radians: `ds = (pi/6)(6m) = pi meters = 3.14 meters 4*pi/3 radians: `ds = (4pi/3)(6m) = 8*pi meters
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14:55:59 3 radians along the arc would correspond to an arc distance of 3 times the radius, or 3 * 6 meters, or 18 meters. `pi / 6 radians would correspond to `pi / 6 times the radius, or `pi / 6 * 6 meters = `pi meters. 4 `pi / 3 radians would correspond to 4 `pi / 3 * 6 meters = 8 `pi meters.
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RESPONSE --> I understand this problem.
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14:58:48 `q004. If you were traveling around a circle of radius 50 meters, and if you traveled through 4 radians in 8 seconds, then how fast would you have to be moving?
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RESPONSE --> r = 50m `ds = 4 radians * 50m = 200 meters `dt = 8 seconds v = `ds/`dt = 200m / 8s = 25 m/s
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14:58:57 If you travel 4 radians along the arc you half traveled an arc distance of 4 times the radius, or 4 * 50 meters = 200 meters. If you traveled this distance in 8 seconds your average speed would be 200 meters / (8 seconds) = 25 m/s.
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RESPONSE --> I understand this problem.
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14:59:48 `q005. Traveling at 3 radians / second around a circle of radius 20 meters, how fast would you have to be moving?
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RESPONSE --> v = 3 rad/s r = 20m If I am moving 3 radians per second, I am moving a distance of 3 times the radius per second. Therefore the velocity is (3*20m)/second = 60m/s.
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15:00:12 3 radians along the arc is a distance of 3 times the radius, or 3 * 20 meters = 60 meters. Moving at 3 radians/second, then, the speed along the arc must be 3 * 20 meters / sec = 60 meters /sec.
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RESPONSE --> I understand this problem.
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15:02:14 `q006. If you know how many radians an object travels along the arc of a circle, and if you know the radius of the circle, how do you find the distance traveled along the arc? Explain the entire reasoning process.
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RESPONSE --> If we know the radius of a circle and how many radians are traveled. We multiply the two together to get the distance traveled along the arc of the circle.
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15:02:26 The distance traveled along the arc of circle is 1 radius for every radian. Therefore we multiply the number of radians by the radius of the circle to get the arc distance.
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RESPONSE --> I understand this problem.
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15:03:30 `q007. If you know the distance an object travels along the arc of a circle, and if you know the radius of the circle, how do you find the corresponding number of radians?
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RESPONSE --> If we know the distance traveled and the radius of the circle we can find the number of radians traveled if we divide the distance by the radius.
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15:03:40 An arc distance which is equal to the radius corresponds to a radian. Therefore if we divide the arc distance by the radius we obtain the number of radians.
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RESPONSE --> I understand.
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15:07:39 `q008. If you know the time required for an object to travel a given number of radians along the arc of a circle of known radius, then how do you find the average speed of the object?
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RESPONSE --> If we know the time interval during the travel of a certain number of radians over a circle of a known radius, we can find the average speed by first multiplying the number of radians by the radius. This gives us the distance traveled. Then we divide the distance by the time interval to attain the average speed.
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15:07:49 If you know the number of radians you can multiply the number of radians by the radius to get the distance traveled along the arc. Dividing this distance traveled along the arc by the time required gives the average speed of the object traveling along the arc.
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RESPONSE --> I understand this concept.
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15:10:33 `q009. If you know the speed of an object along the arc of a circle and you know the radius of the circle, how do you find the angular speed of the object in radians/second?
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RESPONSE --> If we know the speed of an object and the radius of the circular path it follows, we can find the radians traveled per second. First we divide the speed by the radius to get the number of radians traveled per unit of time. This is the angular speed in radians/second.
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15:10:41 The speed of the object is the distance it travels along the arc per unit of time. The angular velocity is the number of radians through which the object travels per unit of time. The distance traveled and the number radians are related by the fact that the distance is equal to the number of radians multiplied by the radius. So if the distance traveled in a unit of time is divided by the radius, we get the number of radians in a unit of time. So the angular speed is found by dividing the speed along the arc by the radius.
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RESPONSE --> Ok. I did not mention this, but I do understand that the speed of the object is the distance it travels along the arc per unit of time and the angular velocity is the number of radians through which the object travels per unit of time.
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15:14:09 `q010. We usually let `d`theta stand for the anglular displacement in radians between two points on the arc of the circle. We usually let `omega stand for the angular velocity in radians / second. We let `ds stand for the distance traveled along the arc of a circle, and we let r stand for the radius of the circle. If we know the radius r and the arc distance `ds, what is the anglular displacement `d`theta, in radians?
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RESPONSE --> `d`theta = angular displacement, radians `omega = angular velocity, rad/sec `ds = distance r = radius of the circle r * `d`theta = `ds We can rearrange the equation to solve for the angular displacement. `d`theta = `ds/r
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15:14:20 Since an angular displacement of 1 radian corresponds to an arc distance equal to the radius, the anglular displacement `theta in radians is equal to the number of radii in the arc distance `ds. This quantity is easily found by dividing the arc distance by the radius. Thus `d`theta = `ds / r.
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RESPONSE --> I understand this concept.
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15:16:03 `q011. If we know the radius r of a circle and the angular velocity `omega, how do we find the velocity v of the object as it moves around the arc of the circle?
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RESPONSE --> velocity = `ds/second `omega = angular velocity = rad/second `ds = r * rad `ds/second = rad/second * radius v = `omega * r
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15:16:11 The angular velocity is the number of radians per second. The velocity is the distance traveled per second along the arc. Since an angular displacement of 1 radian corresponds to an arc distance equal to the radius, if we multiply the number of radians per second by the radius we get the distance traveled per second. Thus v = `omega * r.
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RESPONSE --> I understand.
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15:17:37 `q012. We can change an angle in degrees to radians, or vice versa, by recalling that a complete circle consists of 360 degrees or 2 `pi radians. A half-circle is 180 degrees or `pi radians, so 180 degrees = `pi radians. How many radians does it take to make 30 degrees, how many to make 45 degrees, and how many to make 60 degrees?
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RESPONSE --> 360 degrees : 2`pi radians 180 degrees : `pi radians 30 deg / 180 deg = 1/6 1/6*`pi radians 45 deg/180 deg = 1/4 1/4*`pi radians 60deg/180deg = 1/3 1/3*`pi radians
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15:17:46 30 degrees is 1/6 of 180 degrees and therefore corresponds to 1/6 * `pi radians, usually written as `pi/6 radians. 45 degrees is 1/4 of 180 degrees and therefore corresponds to 1/4 * `pi radians, or `pi/4 radians. 60 degrees is 1/3 of 180 degrees and therefore corresponds to 1/3 * `pi radians, or `pi/3 radians.
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RESPONSE --> I understand this problem.
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15:21:30 `q013. Since 180 deg = `pi rad, we can convert an angle from degrees to radians or vice versa if we multiply the angle by either `pi rad / (180 deg) or by 180 deg / (`pi rad). Use this idea to formally convert 30 deg, 45 deg and 60 deg to radians.
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RESPONSE --> 30 deg * (pi rad/180 deg) = 1/6 *`pi rad 45 deg * (pi rad/180 deg) = 1/4 *`pi rad 60 deg * (pi rad/180 deg) = 1/3 *`pi rad
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15:21:45 To convert 30 degrees to radians, we multiply by the rad / deg conversion factor, obtaining 30 deg * ( `pi rad / 180 deg) = (30 deg / (180 deg) ) * `pi rad = 1/6 * `pi rad = pi/6 rad. To convert 45 degrees to radians we use the same strategy: {}45 deg * (`pi rad / 180 deg) = ( 45 deg / ( 180 deg) ) * `pi rad = 1/4 * `pi rad = `pi/4 rad. To convert 60 degrees: 60 deg * (`pi rad / 180 deg) = ( 60 deg / ( 180 deg) ) * `pi rad = 1/3 * `pi rad = `pi/3 rad.
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RESPONSE --> I understand this concept.
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15:23:11 `q014. Convert 50 deg and 78 deg to radians.
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RESPONSE --> 50 deg * (pi rad/180 deg) = 5/18*`pi rad 78 deg * (pi rad/180 deg) = 13/30*`pi rad
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15:23:22 50 deg * (`pi rad / 180 deg) = ( 50 deg / ( 180 deg) ) * `pi rad = 5/18 * `pi rad = (5 `pi/ 18) rad. 78 deg * (`pi rad / 180 deg) = ( 78 deg / ( 180 deg) ) * `pi rad = 78/180 * `pi rad = (13 `pi/ 30) rad.
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RESPONSE --> I understand this problem.
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15:24:15 `q015. Convert (14 `pi / 9) rad to degrees.
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RESPONSE --> (14 pi / 9) rad * (180 deg/ pi rad) = 280 degrees
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15:24:24 Since the angle is given radians, we need to multiply by deg / rad to get the angle in degrees. (14 `pi / 9) rad * ( 180 deg / (`pi rad)) = ( 14 `pi / 9 ) * (180 / `pi ) deg = ( 14 * 180 / 9) * (`pi / `pi) deg = 14 * 20 deg = 280 deg.
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RESPONSE --> I understand this problem.
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偙U{ۢ°cm assignment #029 029. `query 29 Physics I 07-17-2006
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22:51:05 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 --> vo = 0m/s angular velocity = change in angular position / time interval angular acceleration = change in angular velocity / time interval Since the initial velocity is 0m/s, we can say that the average velocity is half of the final velocity. We know the given angle through which the object travels in a certian time interval. If we divide that angular distance by the time interval, we get the angular velocity. Then we divide the angular velocity by the time interval to get the angular acceleration.
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22:52:07 **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 understand.
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22:53:46 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 --> First we find the total mass. 1.0kg + 1.5kg + 1.1kg = 3.6 kg Then we multiply each mass by the position at which it is found. If we assume that the furthest left is at 0 meters, then 1.0kg * 0m = 0 kg*m 1.5 kg * (0 + 0.5 m) = 0.75 kg*m 1.1 kg * (0.5m + 0.25m) = 0.825 kg*m Now we add the mass positions together. 0kg*m + 0.75kg*m + 0.825kg*m = 1.575kg*m We have all the information we need to find the center of mass: CM = 1.575kg*m / 3.6kg = 0.4375 meters
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22:54:09 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 --> I understand.
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22:55:47 Query problem 7.50 3 cubes sides L0, 2L0 and 3L0; center of mass.
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RESPONSE --> The mass of the cubes can be found by (Length of side)^3. Therefore, the first cube is 1L0^3 = L0, we will call this m1 The second is 2L0^3 = 8 L0 = 8m1 The third is 3L0^3 = 27 L0 = 27m1 Total mass = m1 + 8m1 + 27m1 = 36m1 For the first cube: Xcm = ycm = 1/2L0 For the second cube: ycm = 1/2(2L0) = L0 Xcm = 1/2(2L0) + L0 = 2L0 For the third cube: ycm = 1/2(3L0) = 3/2L0 xcm = 3/2(L0) + 2L0 + L0 = 4.5L0 In order to find the center of mass for the entire system, we must do so in both horizontal and vertical directions. (page 183). We use the formulas: Xcm = (m1x1 + m2x2 + m3x3)/ m total ycm = (m1y1 + m2y2 + m3y3)/ m total Xcm = (m1*1/2L0 + 8m1*2L0 + 27m1*4.5L0) / 36m1 = 3.8333 L0 ycm = (m1*1/2L0 + 8m1*L0 + 27 m1*3/2 L0) / 36m1 = 1.3611 L0
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22:57:09 ** 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 --> So in order to find the center of mass we must find it in the y direction and the x direction. Would you state this point as (3.83 L0, 1.25 L0)?
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22:58:34 What is the mass of the second cube as a multiple of the mass of the first?
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RESPONSE --> We know that the mass of the first is L0 and the mass of the second is 8L0. Therefore, the second is 8 times the mass of the first cube.
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22:58:47 ** 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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22:58:56 What is the mass of the third cube as a multiple of the mass of the first?
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RESPONSE --> The first cube has a mass of L0 while the third has a mass of 27L0. Therefore, the third has 27 times the mass of the first.
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22:59:08 ** 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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22:59:24 How far from the outside edge of the first cube is its center of mass?
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RESPONSE --> The center of mass would be at the midpoint of the length. Therefore, the CM is at 1/2L0 horizontally and vertically.
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22:59:34 ** 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 --> I understand.
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23:00:12 How far from the outside edge of the first cube is the center of mass of the second cube?
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RESPONSE --> The CM of the second cube is 1/2(2L0) = L0 in the vertical direction. The CM is 1/2(2L0) + L0 = 2L0 in the horizontal direction.
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23:00:23 ** 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 understand this problem.
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23:00:49 How far from the outside edge of the first cube is the center of mass of the third cube?
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RESPONSE --> The CM of the third cube is 1/2(3L0) = 3/2L0 in the vertical direction. The CM is 3/2(L0) + 2L0 + L0 = 4.5L0 in the horizontal direction.
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23:01:06 ** 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 understand this problem.
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23:02:43 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 --> First we find the total mass: m1 + 8m1 + 27m1 = 36 m1 Xcm = (m1x1 + m2x2 + m3x3)/ m total = (m1*1/2L0 + 8m1*2L0 + 27m1*4.5L0) / 36m1 = 3.8333 L0 ycm = (m1y1 + m2y2 + m3y3)/ m total = (m1*1/2L0 + 8m1*L0 + 27 m1*3/2 L0) / 36m1 = 1.3611 L0
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23:02:49 ** 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 --> Where the solution says 45m1, I got 49m1. Where did I make a mistake?
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23:03:03 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 --> I am not required to answer this problem.
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23:03:09 ** 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 -->
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