course MTH 174 11:08 a.m. 6/29/10 174assignment # 9
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********************************************* Question: problem 8.2.6 moment of 2 meter rod with density `rho(x) = 2 + 6x g/m YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The mass is found by evaluating the integral of the density over the length of the rod:∫ (2+6x dx)= 2x+3x^2 from 0 to 2 then gives: 16 g. The moment is found by multiplying density by x and then evaluating that integral over the length of the rod: ∫ (x(2+6x) dx) = ∫ (2x+6x^2) = x^2+ 2x^3 and then evaluate this over the length to get a moment of: 20 g*m. The center of mass is found by dividing the moment by the mass and this gives: 20/16 = 5/4 m. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Given Solution: The mass of an increment of length `dx, with sample point x_i, is (2 + 6 x_i ) `dx. The moment is mass * distance from axis of rotation. Assuming axis of rotation x = 0: The moment of the mass in the increment is (2 + 6 x_i) * x_i. Setting up a Riemann sum and allowing the increment to approach zero, we obtain the integral • moment = int(x(2+6x), x, 0, 2). Thus the integrand is 2x + 6 x^2. An antiderivative is F(x) = x^2 + 2 x^3, so the definite integral is • moment = int(x(2+6x), x, 0, 2) = F(2) - F(0) = 20. The moment of the typical increment has units of mass/unit length * length * distance from axis, or (g / m) * m * m = g * m The units of the integral are therefore g * m, and the moment of this object is 20 g * m (i.e., 20 gram * meters). ADDITIONAL INFORMATION (finding center of mass): To get the center of mass relative to x = 0 (this was not requested here but you should know how to do this), divide the moment about x = 0 by the mass of the object: The mass of the object is easily found to be int((2+6x), x, 0, 2) (see above for the mass increment, which leads to the Riemann sum then to the integral). The center of mass is therefore • center of mass = moment / mass = int(x(2+6x), x, 0, 2) / int((2+6x), x, 0, 2). The integrand for the denominator is 2 + 6 x, antiderivative G(x) = 2x + 3 x^2 and definite integral G(2) - G(0) = 16 - 0 = 16, meaning that the object has mass 16 grams (more specifcally the units of the denominator will be units of mass / unit length * length = mass, or in this case g / m * m = g). So the center of mass is at x = 20 g * m /(16 g) = 5/4 (g * m) / g = 5/4 g. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: problem 8.2.12 mass between graph of f(x) and g(x), f > g, density `rho(x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The mass of each function is found by multiplying the area between the curves by the density. Creating the ∆x increments the area of each slice will be A_i = (f(x)-g(x)) ∆x then the mass of each slice: rho(x)[f(x)-g(x)] ∆x. The reman sum ist then : ∑ rho(x)[f(x)-g(x)] ∆x and taking the limit of that we arrive at the definite integral we need: ∫ (rho(x)[f(x)-g(x)] dx, a,b). Confidence Assessment: 3
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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 #$&* ********************************************* Question: What is the mass of an increment at x coordinate x with width `dx? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The mass is found by area x density, so here we have: (f(x)-g(x))∆x for the area and rho(x) for the density and multiplication gives: m_i= rho(x)[f(x)-g(x)] ∆x. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* ********************************************* Question: problem 8.3.6 cylinder 20 ft high rad 6 ft full of water
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the work we need to find the force and the distance equations. For the distance we have: d_i =30-h_i. We need to know volume and density to find the force and the density of water is 62.4 lb/ft^3 and the volume is: V= π r^2 ∆h. The force equation is thus: F_i = 62.4(π r^2 ∆h). Since this is a cylinder the radius is constant and doesn't need to be expressed in terms of the height like a lot of other problems. The work for any given strip is then found by W_1= d_i * F_i = (30-h_1)(62.4π r^2) ∆h. For the sum: ∑ (30-h_i)(62.4π r^2) ∆h and taking the limit gives us the integral we need to evaluate: ∫ (30-h)(62.4π r^2) dh = (30h- 1/2 h^2)(62.4π 36). We evaluate this over the interval of 0 to 20 for the depth of the water and get: 2.82x10^6 ft*lb. Confidence Assessment:
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Given Solution: STUDENT SOLUTION: delta W = delta F*(30-y) delta W = (62.4)(volume)*(30-y) delta W = 62.4*36*pi*delta y*(30-y) delta W = (211718.211 - 7057.2737y)delta y integral [0,20] (211718.211 - 7057.2737y)dy = 211718.211y - 3528.63685y^2 from 0 to 20 = 2,822,909,.48 ft-lb INSTRUCTOR COMMENTARY ** For the ith interval, using F_i and dist_i for the force required to lift the water and the distance is must be lifted, `rho for the density and A for the cross-sectional area 36 `pi, the work is • Fi * dist_i = `rho g A 'dyi * (30 - y_i) = `rho g A (30 - yi) 'dy_i. We thus have a Riemann sum of terms `rho g A (30 - y_i) 'dy_i. This sum approaches the integral • int(`rho g A (30 - y) dy between y = 0 and y = 20). The limits are y = 0 and y = 20 because that's where the fluid represented by the terms `rho g A (30 - y_i) 'dyi is (note that the tank for Problem 6 is full of water, not half full). The 30-y_i is because it's getting pumped to height 30 ft. Your antiderivative is `rho g A ( 30 y - y^2 / 2). At this point you can plug in your values for `rho, g and A, then evaluate 30 y - y^2 / 2 at the limits to get your answer. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* ********************************************* Question: **** query problem 8.4.24 What is the kinetic energy of a record mass of mass 50 g rad 10 cm rotating at 33 1/3 revolutions per minute? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The density of the record is 50/(2π 10)≈ .159 g/cm^2. Dividing the record up into rings whose areas are: A_i = 2π r_i ∆r and whose mass is then: m = .159(2π r_i) ∆r ≈ .318π r_i ∆r. The velocity of each slice is : v_i = 2π r_i (33 1/3)/ 60 ≈ 1.11π r_i. The kinetic energy of each slice is then KE_i= 1/2(.318π r_i ∆r)(1.11π r_i)^2. Forming the sum: ∑ (.196π^3) r_i^3 ∆r and taking the limit gives the integral: ∫ (.196π^3) r^3 dr= .(196π^3)/4 r^4. Evaluating this over the interval 0 to 10 gives: 15193 g*cm^2 / s^2. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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; integrating from r = 0 to r = 10 we get approximately 15,000 gram cm^2 / s^2. The process shown here is correct, but the calculations represented here are not numerically accurate to any degree of precision; you should compare with your results with the results obtained here and if necessary rework these calculations using more accurate values. The value of the integral, using .16 * 2 pi r in place of the approximation 1.0 r, is 14 526.72443 g cm^2, to 10 significant figures. This is a ridiculously precise value, considering that the radius of a pressed disk is consistent to only within perhaps +-.001 cm, with even more significant uncertainty in the mass. A more reasonable figure would be 14 500 g cm^2 +- 100 g cm^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* 3 ********************************************* Question: problem 8.3.18 work to empty glass (ht 15 cm from apex of cone 10 cm high, top width 10 cm) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The density of water is rho= 1 g/cm^3. The volume of a slice of the cone is: V_i = π r^2 ∆h. Expressing r in terms of h using proportionality gives h/2 = r. Substituting this for r and multiplying by the density and the acceleration due to gravity gives (with simplification): F_i = 245π h^2 ∆h dynes. The work is therefore: W_i= 245π(15-h)h^2 ∆h= 245π (15h^2-h^3) ∆h. Forming the sum and the taking the limit yields the integral: ∫ (245π(15h-h^2) dh = 245π (5h^3- 1/4 h^4). Evaluating this over the interval 0 to 10 gives: 1.92x10^7 ergs. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given 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): The units messed me up here because at first I didn't convert the 9.8 m/s^2 to 980 m/s^2 and I had to look up what that was even called and what the final work units would be.