ASSN9

course Mth 174

Physics II03-13-2008

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00:07:44

query problem 8.4.3 (8.3.4 3d edition) was 8.2.6) moment of 2 meter rod with density `rho(x) = 2 + 6x g/m

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00:21:04

what is the moment of the rod?

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The rod weighs about 16.75 grams

The mass of an increment of length `dx, with sample point x_i, is (2 + 6 x_i ) `dx.

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.

This will approach the integral

int(x(2+6x), x, 0, 2).

Using rho(x) for the density function:

To get the moment you integrate x * rho(x).

The integrand for the numerator is 2x + 6 x^2, antiderivative F(x) = x^2 + 2 x^3 and definite integral 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

**

** To get the center of mass, which was not requested here, you would integrate x * p(x) and divide by mass, which is the integral of rho(x).

You get 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.

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 m. **

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00:21:55

What integral did you evaluate to get a moment?

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I used mid point riemann sums and the given formula with a n= .5 m to get the integral.

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00:22:12

query problem 8.4.12 (was 8.2.12) mass between graph of f(x) and g(x), f > g, density `rho(x)

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00:28:56

what is the total mass of the region?

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The mass would be f(x) - g(x) 'dx

** You want to think of this as a simple product, just area * density.

Assuming a typical partition with into n subintervals with the ith interval containing sample point c_i, the width of the region on this interval is approximately f(c_i) - g(c_i) The area of the region is height * width, or approximately (f(c_i) - g(c_i) ) * `dx. The density at the sample point c_i is `rho(c_i) so you get the approximation

mass = area * density = (f(c_i) - g(c_i) ) * 'dx * `rho(c_i) = `rho(c_i) (f(c_i) - g(c_i) ) * 'dx.

The sum of the `rho(c_i) ( f(c_i) - g(c_i) ) `dx increments is

sum(`rho(c_i) ( f(c_i) - g(c_i) ) `dx, i from 1 to n) , which approaches the integral

integral( rho(x) (f(x) - g(x) dx, x from a to b)

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00:31:12

What integral did you evaluate to obtain this mass?

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the area between the graphs shows the changing of density over the change in x axis so by multiplying this by 'dx we can find the mass of the cardboard.

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00:33:38

What is the mass of an increment at x coordinate x with width `dx?

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('df(x)) - ('dg(x)) 'dx

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00:34:41

What is the area of the increment, and how do we obtain the expression for the mass from this area?

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the area of this area is like what is in the given example by lines a and b, by only using intervals between this increment we can find the mass inside this parameter.

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00:45:31

How to we use the mass of the increment to obtain the integral for the total mass?

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because the mass on the increment is just like the mass for the whole figure except we only use it on a small interval, if we widen our interval out to include the whole figure then we can find the whole mass.

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00:45:39

query problem 8.5.13 (8.4.10 3d edition; was 8.3.6) cylinder 20 ft high rad 6 ft full of water

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00:56:45

how much work is required to pump all the water to a height of 10 ft?

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2,945,052.6 ftlbs

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01:08:26

What integral did you evaluate to determine this work?

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V= 'pi r^2 h = 2261.9 ft^3

for every 113 ft^3 pumped out the water drops 1 ft and thus 1ft is added the pump height.

Work = MAD (no pun intended)

water weighs about 62lb/ft^3 so assuming an accelreation of 1ft/s

The equation for a 113ft^3 pumped out is:

W= 7012.03D

Since we have a constantly changing distance depending on the water pumped out we work out this equation from 30ft the maximum pumped and 10ft the minimum pumped.

sumason 30->10 = 420

7012.03 (420)= Work = 2945052.6 ftlb of work

Good attempt, but we need to put everything into the notation of integrals.

** 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 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.

I believe the result differs a little from your estimate. One succinct solution from a student, which is consistent with the above and which gives the result I obtain, is

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

**

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01:20:08

Approximately how much work is required to pump the water in a slice of thickness `dy near y coordinate y? Describe where y = 0 in relation to the tank.

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y= 0 when the tank is full,

W= 7012.03 (D)

when D= sum of all numbers in the interval

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01:22:12

Explain how your answer to the previous question leads to your integral.

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It is because the depth and therefore the work changes as you pump more water out, so you have to account for that, it is the same constant of work / ft drop weather you're pumping 1ft of depth of 20, you must add all the footage together because that would be the total amount of footage that the pump must pump.

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01:34:33

query problem 8.3.18 CHANGE TO 8.5.15 work to empty glass (ht 15 cm from apex of cone 10 cm high, top width 10 cm)

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I had a question very similar to the last question except oil instead of water and different tank demensions for this problem. I had a total work of 518,370.4 ftlb

I don't have a copy of the text so I don't know the dimensions or density involved in your problem. It should however have been solved my means similar tot he following:

** 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 done in my head. Check them. However I believe that the process is correct. **

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01:34:46

how much work is required to raise all the drink to a height of 15 cm?

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01:34:55

What integral did you evaluate to determine this work?

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01:35:03

Approximately how much work is required to raise the drink in a slice of thickness `dy near y coordinate y? Describe where y = 0 in relation to the tank.

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01:35:10

How much drink is contained in the slice described above?

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01:35:18

What are the cross-sectional area and volume of the slice?

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01:35:25

Explain how your answer to the previous questions lead to your integral.

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question->

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01:39:51

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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I think that this was hard in some ways and someways not. It was easy to undertand what needed to be done but the process of doing it was very indepth and if one thing was wrong then the whole process was ruined. It reminded me of a very long deminsional analysis problem from chemistry converting from Molality to percent mass then to number of atoms in that percent, a several step process with multiple conversion factors.

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See my notes and be sure you understand the process of partitioning an interval, setting up a Riemann sum, and taking it to the limit to get the integral.