#$&* course Phy 121 7/1 at 7 029. Radian measure of angle; angular position, angular velocity*********************************************
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Given Solution: 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. STUDENT COMMENT: ok, I answered '40 m for one radian and 120m for three' I thought there was some formula and it couldn't be this easy to figure out. INSTRUCTOR RESPONSE It really is this easy, and this is the most important thing you need to remember about radian measure. There are formulas relating arc distance to radian measure (and arc velocity to angular velocity, and acceleration along the arc to angular acceleration), and they are simple formulas, but they are difficult to keep straight. It's difficult to remember what symbol goes with what, and when you multiply by r and when you divide, and when there's a 2 pi involved and whether you divide by that or multiply. Of course you haven't seen the formulas yet and don't yet know what all that means, but that last paragraph should alert you to one simple fact: There's one simple idea at work here, you just used it, you understand it, and if you never lose track of it there are a whole lot of confusing formulas that will just come down to common sense. (You'll also want to keep in mind that the circumference of a circle is 2 pi r, but it's assumed that you know this.) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: r = 40 m 1 radian = 2*pi*40m = 251 meters traveled around the circle. ½ radian = 251 meters / 2 = 125.5 meters ¼ radian = 251 meters / 4 = 62.8 meters confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT QUESTION A quarter of the arc would be pi/2 radians so, it would not be 40 pi/2 radians..I dont understand? INSTRUCTOR RESPONSE A quarter of the arc would be pi/2 radians, so the distance around the arc on this circle would be 40 pi/2 meters. Be careful not to confuse the distance along the arc, which is measured in meters and which you could actually walk, with the angle, which is measured in radians (if you were standing at the center you could turn through the angle, but you would do that without moving anywhere). You would have to go 80 pi meters to go around the whole circle, and this corresponds to an anglular displacement of 2 pi radians. You would only have to go 40 pi meters to get around half the circle, and you would have only half the angular displacement. The previous angular displacement was 2 pi radians, so halfway around would correspond to pi radians. To go a quarter of the way around you would travel 20 pi meters, and your angular displacement would be pi/2 radians. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3 radians = 1r(6m) * 3 r = 18 meters D = 'pi / 6 r * 6 m = 'pi / 36 D = 'pi / 4 r * 6 m = 'pi / 24 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT COMMENT I'm still not really clear on how to simplify the given equations. INSTRUCTOR RESPONSE You got the correct expressions, you just didn't simplify them. It might be that you didn't recognize these as fractional expressions. You should write these fractions out. For example a / c * b means 'multiply by a then divide by c, then multiply the result by b'. So a / c * b means (a / c) * b. (a/c) can be written as a fraction, with a in the numerator and b in the denominator. b can also be written as a fraction, with b in the numerator and 1 in the denominator. So (a / c) * b means (a / c) * (b / 1). When two fractions are multiplied, their numerators are multiplied, and their denominators are multiplied. So (a/c) * (b/1) = (a * b) / (c * 1), which simplifies to a * b / c. The expression 4 pi / 3 * 6 means (4 pi / 3 ) * 6, which means (4 pi * 6) / 3 = (24 pi / 3). Since 24 / 3 = 8, the expression (24 pi / 3) reduces to 8 * pi. pi / 6 * 6 means (pi * 6) / 6, or 6 pi / 6. Since 6 / 6 = 1, the expression (6 pi / 6) reduces to 1 * pi, or just pi. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok, fell victim to the same issue here, and didn't write the expression out correctly to multiply the fraction, clear on this now. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: r = 50 meters 4 radians * (1 r) 50 meters = 200 m 200 m / 8 s = 25 m/s v = 25 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. COMMON STUDENT SOLUTION 50 meters * 4 = 200 meters/8 seconds= 25 m/s INSTRUCTOR COMMENT I can see how you've thought this through and your thinking is absolutely correct. However the first expression 50 meters * 4 is not equal to you last expression 25 m/s. However these expressions are connected by a chain of equal signs and should therefore be equal (this is called the 'transitive property of multiplication'). What you clearly mean is 50 m * 4 = 200 meters 200 m / (8 sec) = 50 m/s. Your train of thought it clear and correct. However equal signs should be used only to indicate equality. Confusion can easily result when equal signs are used to indicate train of thought. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q005. Traveling at 3 radians / second around a circle of radius 20 meters, how fast would you have to be moving? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1 radian = 20 m 3r * 20 m = 60 m v = 60 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 60 meters /sec. NEARLY CORRECT STUDENT SOLUTION 20 meters * 3 = 60 meters/sec INSTRUCTOR COMMENT Your intent is clear and correct. However the units of your calculation don't work out: 20 meters * 3 = 60 meters, not 60 meters / second. What you know is that every second the object moves through an angular displacement of 3 radians, which on a circle of 20 m radius implies an arc distance of 60 meters. We conclude that the speed along the arc is 60 m/s. The problem with the notation can be fixed up as follows: 1 radian of angle corresponds to a distance equal to the radius. The units of this calculation could be expressed as 1 meter of radius * 1 radian of angle = 1 meter of arc distance. We can abbreviate this as 1 meter * 1 radian = 1 meter, understanding that the meter on the left corresponds to a meter of radius, while the meter on the right corresponds to 1 meter of arc distance. Then our calculation becomes 20 meters * 3 radians / second = (20 meters * 3 radians) / second = 60 meters / second, where we understand that the 20 meters on the left stands for radius and the 60 meters on the right stands for arc distance. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We know that one radian along the arc of a circle is equal to the radius of the circle; therefore, we take the radians and do a conversion by multiplying by the given radius. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We would simply divide the distance an object traveled along the arc of a circle by the radius, this would yield the number of corresponding radians. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: An arc distance which is equal to the radius corresponds to a radian. Therefore if we divide the arc distance by the radius we obtain the number of radians. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q008. If you know the time required for an object to travel a given number of radians along the arc of a circle of known radius, then how do you find the average speed of the object? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We find the total distance traveled from a given number of radians by multiplying the number of radians by the known radius. Then, apply vAve = 'ds / 'ds confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Angular speed of the object in radians/second is calculated by taking the velocity of an object along the arc of a circle, and simply dividing it by the known radius of the circle. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The speed of the object is the distance it travels along the arc per unit of time. The angular velocity is the number of radians through which the object travels per unit of time. The distance traveled and the number radians are related by the fact that the distance is equal to the number of radians multiplied by the radius. So if the distance traveled in a unit of time is divided by the radius, we get the number of radians in a unit of time. So the angular speed is found by dividing the speed along the arc by the radius. STUDENT ANSWER: change in radians (distance) /time INSTRUCTOR RESPONSE Right idea; one small correction: • the denominator should be change in clock time, or time interval, not just 'time' The angular speed on an interval is the angular distance during that interval (which you clearly indicated as 'change in radians (distance)', divided by change in clock time ortime interval. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q010. We usually let `d`theta stand for the angular displacement in radians between two points on the arc of the circle. We usually let `omega stand for the angular velocity in radians / second. We let `ds stand for the distance traveled along the arc of a circle, and we let r stand for the radius of the circle. If we know the radius r and the arc distance `ds, what is the angular displacement `d`theta, in radians? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 'd theta = 'ds / r confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since an angular displacement of 1 radian corresponds to an arc distance equal to the radius, the angular 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. STUDENT QUESTION: Omega (velocity in radians/second)/ radius = angular speed= angular displacement/time '(velocity in radians / second) / radius' implies that you divide the number of radians by the radius to get the angular displacement, when in fact you would divide the angular displacement by the radius to get the number of radians. `d`theta = `ds / r I'm getting confused by this now. INSTRUCTOR RESPONSE The angle (number of radians) corresponding to an arc distance is found by dividing arc distance by radius. As you say: `dTheta = `ds / r. That is, angular displacement is arc displacement divided by radius. So if you know the arc displacement during a certain time interval, the angular displacement during that same interval is found by dividing the arc displacement by theradius. Thus angular displacement per unit of time is equal to arc displacement per unit of time, divided by radius: omega = v / r. Rule of thumb: If it's along the arc you divide by the radius to get the angular quantity; if it involves angle you multiply by radius to get the arc quantity. This follows directly from the reasoning you used in the very first problem of this document. Applying the rule of thumb: v is along the arc, omega involves angle. We divide arc displacement by radius to get angle in radians. More specifically: v is velocity along the arc, so v / r involves (among other things) dividing an arc displacement by radius. The result is an angular displacement. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: velocity = radius * 'omega confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The angular velocity is the number of radians per second. The velocity is the distance traveled per second along the arc. Since an angular displacement of 1 radian corresponds to an arc distance equal to the radius, if we multiply the number of radians per second by the radius we get the distance traveled per second. Thus v = `omega * r. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q012. We can change an angle in degrees to radians, or vice versa, by recalling that a complete circle consists of 360 degrees or 2 `pi radians. A half-circle is 180 degrees or `pi radians, so 180 degrees = `pi radians. How many radians does it take to make 30 degrees, how many to make 45 degrees, and how many to make 60 degrees? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2 pi radians* 1/12 = 30 degrees 2 pi radians * 1/8 = 45 degrees 2 pi radians * 1/6 = 60 degrees confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT QUESTION I mathematically solved for a numerical value of how many radians there were. For example, 30 deg is 1/6 of 180 deg, so it's pi / 6 radians. I got .78 radians, which I rounded to .8 radians. Based on the answer in the given solution, I should not have done this. Answers were left as ‘pi/6 radians, ‘pi/4 radians, and ‘pi/3 radians. Is the answer technically wrong solved the way that I did it??? Or does it need to be written the way it was in the given solution. INSTRUCTOR RESPONSE Only the multiple-of-pi results are exact. Anything else is a rounded approximation. For the special angles (angles which are multiples of pi/4 and pi/6), we can find their exact sines and cosines. So we use the expressions for the exact angles. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 30 deg. * pi / 180 = 0.5 radians 45 deg. * pi / 180 = 0.8 radians 60 deg. * pi / 180 = 1.0 radians confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To convert 30 degrees to radians, we multiply by the rad / deg conversion factor, obtaining 30 deg * ( `pi rad / 180 deg) = (30 deg / (180 deg) ) * `pi rad = 1/6 * `pi rad = pi/6 rad. To convert 45 degrees to radians we use the same strategy: 45 deg * (`pi rad / 180 deg) = ( 45 deg / ( 180 deg) ) * `pi rad = 1/4 * `pi rad = `pi/4 rad. To convert 60 degrees: 60 deg * (`pi rad / 180 deg) = ( 60 deg / ( 180 deg) ) * `pi rad = 1/3 * `pi rad = `pi/3 rad. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q014. Convert 50 deg and 78 deg to radians. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 50 deg. * pi / 180 = 0.9 radians 78 deg. * pi / 180 = 1.4 radians confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q015. Convert (14 `pi / 9) rad to degrees. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 14 'pi / 9 radians = degrees? 4.89 radians * 180 / pi = 280 degrees confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since the angle is given radians, we need to multiply by deg / rad to get the angle in degrees. (14 `pi / 9) rad * ( 180 deg / (`pi rad)) = ( 14 `pi / 9 ) * (180 / `pi ) deg = ( 14 * 180 / 9) * (`pi / `pi) deg = 14 * 20 deg = 280 deg. STUDENT QUESTION The only problem I ran into when solving is that I was unable to mathematically remove radians from the equation that I set up. I know that they are not supposed to be part of the answer to the problem, but I’m not sure how to get rid of it from the answer. INSTRUCTOR RESPONSE As in your calculation 14 pi/9 rad * 180 degrees / (pi rad), the units come down to rad * deg / rad. The radians divide out, being present in both numerator and denominator, leaving degrees. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Phy 121 7/1 ~8 029. `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): I forgot to include the necessary doubling of the average angular velocity before dividing by 'ds to find 'alpha. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qQuery problem 7.50 3 cubes sides L0, 2L0 and 3L0; center of mass. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This problem is not in my textbook, and I could not find anything similar in the notes. 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): Indeed after reading this answer I could find nothing at all like this in the book. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat is the mass of the second cube as a multiple of the mass of the first? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: N/A confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the mass of the third cube as a multiple of the mass of the first? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: N/A confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The mass of the third cube is 3^3 = 27 times the mass of the first. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qHow far from the outside edge of the first cube is its center of mass? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: N/A confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qHow far from the outside edge of the first cube is the center of mass of the second cube? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: N/A confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** In the x direction the distance is L0 + L0 (the L0 across the first cube, another L0 to the center of the second), or 2 L0. In the y direction the distance is L0 (the center of the second cube). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qHow far from the outside edge of the first cube is the center of mass of the third cube? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: n/a confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ok ********************************************* 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: n/a confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** In the x direction you have moment m1 * L0/2 + 8 m1 * (2 L0) + 27 M1 * (9/2 L0) = 276 m1 L0 / 2 = 138 m1 L0. The total mass is m1 + 8 m1 + 27 m1 = 36 m1 so the center of mass is at 138 m1 L0 / (36 m1) = 3.83 L0. In the y direction the moment is m1 * L0/2 + 8 m1 * (L0) + 27 m1 * ( 3/2 L0) = 45 m1 L0 so the center of mass is at 45 m1 L0 / (36 m1) = 1.25 L0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `qUniv. 8.94 (8.82 10th edition). 45 kg woman 60 kg canoe walk starting 1 m from left end to 1 m from right end, moving 3 meters closer to the right end. How far does the canoe move? Water resistance negligible. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Not required confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!