#$&* course Mth 174 7/27/13 around 12pm A point mass m at position x has torque or moment x * m about the origin.
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Given Solution: The mass of an increment of length `dx, with sample point x_i, is (2 + 6 x_i ) `dx. The moment is mass * distance from axis of rotation. Assuming axis of rotation x = 0: The moment of the mass in the increment is (2 + 6 x_i) * x_i. Setting up a Riemann sum and allowing the increment to approach zero, we obtain the integral • moment = int(x(2+6x), x, 0, 2). Thus the integrand is 2x + 6 x^2. An antiderivative is F(x) = x^2 + 2 x^3, so the definite integral is • moment = int(x(2+6x), x, 0, 2) = F(2) - F(0) = 20. The moment of the typical increment has units of mass/unit length * length * distance from axis, or (g / m) * m * m = g * m. The units of the integral are therefore g * m, and the moment of this object is 20 g * m (i.e., 20 gram * meters). ADDITIONAL INFORMATION (finding center of mass): To get the center of mass relative to x = 0 (this was not requested here but you should know how to do this), divide the moment about x = 0 by the mass of the object: The mass of the object is easily found to be int((2+6x), x, 0, 2) (see above for the mass increment, which leads to the Riemann sum then to the integral). The center of mass is therefore • center of mass = moment / mass = int(x(2+6x), x, 0, 2) / int((2+6x), x, 0, 2). The integrand for the denominator is 2 + 6 x, antiderivative G(x) = 2x + 3 x^2 and definite integral G(2) - G(0) = 16 - 0 = 16, meaning that the object has mass 16 grams (more specifcally the units of the denominator will be units of mass / unit length * length = mass, or in this case g / m * m = g). So the center of mass is at x = 20 g * m /(16 g) = 5/4 (g * m) / g = 5/4 g. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Units are g*m ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: problem 8.4.14 (4th edition 8.4.12 3d edition 8.3.12) mass between graph of f(x) and g(x), f > g, density `rho(x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Mass = Area*Density The area is equal to f(x) - g(x) So the mass = (f(x) - g(x)) `dx* density, or `rho(x) Riemann’s Sum would be: Int((f(x) - g(x))`dx*`rho(x) x, a, b) confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: First you find the mass of a typical increment of width `dx, with sample point x within the interval. The mass is just area * density. The area of the region is height * width, or approximately (f(x) - g(x) ) * `dx. The density is `rho(x) so you get the approximation • `dm = area * density • = (f(x) - g(x) ) * 'dx * `rho(x) • = `rho(x) (f(x) - g(x) ) * 'dx. The Riemann sum is the sum of all such mass increments for a partition of the interval, and at the interval width `dx approaches 0 this sum approaches the integral int( rho(x) * (f(x) - g(x)), x, a, b). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: What is the mass of an increment at x coordinate x with width `dx? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: = mas* density Density = width * height = f(x) - h(x) * `dx * rho(x) = `rho(x)*(f(x) - h(x)) `dx confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** You want to think of this as a simple product, just area * density. The area of the region is height * width, or approximately (f(x) - g(x) ) * `dx. The density is `rho(x) so you get the approximation mass = area * density = (f(x) - g(x) ) * 'dx * `rho(x) = `rho(x) (f(x) - g(x) ) * 'dx. Note that `dx stands for delta-x, a finite but small interval and that it's f - g, not f + g. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: problem 8.5.13 (4th edition 8.5. 12) (3d edition 8.4.12) 8.3.6 cylinder 20 ft high rad 6 ft full of water
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13:19:02 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I don’t know what I am supposed to be finding here or how I start this problem.
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Given Solution:
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Given Solution: ** We partition the interval 0 <= r <= 10 cm between the center of the record and its rim. • An small interval of a partition will correspond to an interval of r. • The part of the record for which the radius is within the partition consists of a thin ring of the disk. For example if 3.4 cm < r < 3.5 cm is an interval of the partition, then the corresponding region of the disk is the ring which is also described by 3.4 cm < r < 3.5 cm. This ring lies between the circle r = 3.4 cm and r = 3.5 cm; its 'width' is .1 cm and its 'average circumference' is somewhere between 2 pi * 3.4 cm (the circumference of the 'inner' circle) and 2 pi * 3.5 cm (the circumference of the 'outer' circle). The 'area density' of the record (mass / unit area) is • area density = total mass / total area = 50 grams / total area = 50 grams / (pi * (10 cm)^2) = .16 grams / cm^2, approx. For a partition with interval width `dr, considering a typical interval with sample point r_i*: • The corresponding 'ring' would have radius r_i* and width `dr. • Its area would be approximately circumference * width = 2 pi r_i* `dr. • The mass of the typical slice would be area * density = .16 * 2 pi r_i* `dr = 1.0 r_i* `dr, approx.. • The speed of a point on the slice would be dist / time = 2 pi r_i* (33 1/3 / (60 sec)) = 3.4 r_i*, with speed in cm/s when radius is in cm. • The KE of the slice is therefore .5 m v^2 = .5 ( 1.0 r_i* `dr) * (3.4 r_i*)^2, with KE in gram cm^2 / s^2. The Riemann sum of all KE contributions would, as `dr -> 0, approach the an integral which represents the total KE: • total KE = integral of .5 ( 1.0 r) (3.4 r)^2 with respect to r, from r = 0 to r = 10. The simplified form of this expression is approximately 6 r^3; integra
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Given Solution: The given solution solves the following problem: A conical glass is 10 cm high, and at the top its width is 10 cm. How much work is required to empty the glass by raising the liquid through a straw to a height of 15 cm from the base? Solution: ** The diameter of the top of the cone is equal to the vertical distance y from the apex to the top. At height y the diameter of the cone is easily seen to be equal to y, so the cross-section at height y has radius .5 y and therefore area A = `pi ( .5 y ) ^ 2 = `pi / 4 * y^2. A slice of thickness `dy at height y has approximate volume A * `dy = `pi/4 * y^2 * `dy. This area is in cm^3 so its mass is equal to the volume and the weight in dynes is 980 * mass = 245 `pi y^2 `dy. This weight is raised from height y to height 15, a distance of 15 - y. So the work to raise the slice is force * distance = 245 `pi y^2 `dy * ( 15 - y ) = 245 ( 15 - y ) `pi y^2 `dy = 245 `pi ( 15 y^2 - y^3 ) `pi `dy Slices go from y = 0 to y = 10 cm so the integral is 245 `pi ( 15 y^2 - y^3 ) `pi dy, evaluated from y = 0 to y = 10. We get 245 `pi * (5 y^3 - y^4 / 4) evaluated between 0 and 10. The result is 245 `pi * 2500 ergs, close to 2 million ergs. Most calculations were mentally so check the precise numbers. The process is correct. ** . STUDENT COMMENT: I am stuck at a point close to the end on this problem. The integral I have is from 0 to 10 'rho g A (15-y) dy INSTRUCTOR RESPONSE: ** Good, but A is a function of y because the glass is tapering. A = `pi r^2. What is the radius r at height y? Just draw a picture--two straight lines for the outline of the glass--and use proportionalities. Draw some similar triangles if necessary. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!