assignment 9

course Mth 174

c?r???~????????assignment #009?????S??B+???Physics II

03-12-2009

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19:42:14

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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RESPONSE -->

The idea of a moment:

If a horizontal rod is balanced on a point, then a mass placed on the rod to one side of the pivot or the other will tend to make the entire system rotate. The effect is greater if the mass is placed further from the balance point. And of course the effect is greater if a greater mass is used.

It turns out that the rotational effect of a mass is the product of the mass and its distance from the pivot point. This quantity is called the moment of the mass.

In general the moment of a mass about a point is equal to the product of the mass and its distance from the point.

To find the moment of inertia we partition the interval from one end of the rod to the other, and find the contribution to the moment of inertia of the ith interval:

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).

Summary:

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.

Units:

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

Additional information:

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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19:42:16

what is the moment of the rod?

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19:42:18

What integral did you evaluate to get a moment?

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19:52:37

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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RESPONSE -->

The integral from g(x) to f(x) of the changes in y dx

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19:53:16

what is the total mass of the region?

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RESPONSE -->

Total Mass =

The integral from g(x) to f(x) of the changes in y dx

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19:57:25

What integral did you evaluate to obtain this mass?

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RESPONSE -->

The integral from g(x) to f(x) of the changes in y dx

I really wasn't sure how to approach this problem but I used the integral from g(x) to f(x) because the problem stated that the density varies only with x, so since y is constant I took the integral of the changes in y.

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20:05:39

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

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RESPONSE -->

mass of the ith piece =

mi is approximatly theta(xi) change in x

center of mass =

the sum of ximi divided by the sum of mi

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20:18:26

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

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RESPONSE -->

Area of an increment= Ax(x)change in x is approximatly (1- x) change in x

Center of mass is given by:

integral of (x theta Ax(x) dx) / Mass

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20:20:16

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

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RESPONSE -->

I'm not sure how to find total mass from the mass of the increment.

** You have parts of the solution. See below for the details of assembling these ideas into a Riemann sum then an integral:

You want to think of the mass of a typical interval 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 ‘height’ of the region on this interval is approximately

‘height’ of ith region: 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 of ith interval = 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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assignment #010

?????S??B+???Physics II

04-01-2009

????l????????????assignment #009

?????S??B+???Physics II

04-01-2009

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19:42:13

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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20:11:05

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

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RESPONSE -->

You first slice the oil horizontaly because each part moves in the same vertical distance.

radius = 6

height = 20 + 10 ( feet past the rim)

water is at h=20

Density of water = 1g/ cu cm.

Volume of slice= 6 * change in h

Force of gravity on slice = Density * g * Volume=

1 * 6 * change in h nt (I'm not sure what ""g"" is)

Work done on slice = Force * Distance=

6g * w^2 * ( 30 - h) change in h joules

w = w/h = 6/30 = 5h

Work on strip = 6g (5h) ^2 (30 - h) change in h joules

The integral with the upper limit = 20 (depth of water)

Total work = limit as change in x approaches 0 , the sum of 60gh^2 (30 - h) change in h = integral from 20 to 0 is 30gh^2 (30-h) oules

g = 9.8 m/sec^2

Your approach is sound. You have a couple of errors in detail, but you did a good job here.

We partition the vertical interval from 0 to 20 ft. All the water is contained within this interval. Each interval of the partition corresponds to a horizontal ‘slice’ of the water. The water in the entire horizontal ‘slice’ is lifted the same distance.

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 (A = pi r^2 = 36 `pi ft^2), 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 water at height y_i getting pumped to height 30 ft, thereby rising distance 30 – y_i.

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.

The result should be around 2.8 million ft*lb.

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20:14:04

What integral did you evaluate to determine this work?

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RESPONSE -->

Total work = limit as change in x approaches 0 , the sum of 60gh^2 (30 - h) change in h = integral from 20 to 0 is 30gh^2 (30-h) joules

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20:15:30

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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I do not understand the relation of the y coordinate y in this problem.

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20:16:44

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

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RESPONSE -->

You find total mass in a similar way.

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20:28:06

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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RESPONSE -->

Oil to surface = h=25 ft

radius = 5 ft

Density of oil= 800 kg m^3

Oil is 6ft deep

Volume of slice = 6 * change in h m^3

Force of gravity on slice = Density * g * Volume =

800g * 6 w^2 change in h nt

Work on slice = Force * Distance =

800g * g w^2 (19 - h) change in h joules

w= w/h = 5/19 = .263h

Work on slice = 800g (.263)^2 (19-h) change in h joules

Total work = limit as change in x approaches 0 is the sum of 55.335g h^2 (19 - h) change in h =

integral of 55.335 gh^2(19 - h) joules

The oil in the tank is 6 feet deep, so we partition this 6-foot vertical interval.

A typical interval of length `dy and sample point y_i* corresponds to a horizontal slice of the oil, which is a cylinder with radius 5 ft and altitude `dy. The volume of the slice is A * `dy, where A = pi ( 5 ft)^2 = 25 ft^2 * pi is the cross-sectional area (i.e., the area of the base). At 50 lb / ft^3 the oil in the slice has weight 25 ft^2 * 50 lb / ft^3 * pi * `dy = 1250 lb / ft * pi * `dy.

The bottom of the barrel is 25 ft below the surface; the oil in this slice is at height y_i* above the bottom of the barrel. To get the oil in the slice to the surface it must be raised a distance of 25 ft - y_i*. This requires

work to raise oil in ith interval = weight * distance raised = 1250 lb/ft * pi * `dy * (25 ft - y_i*).

Summing over all intervals and taking the limit as the interval width approaches zero we obtain the integral

int( 1250 pi * (25 - y), y from 0 to 6) = 1250 pi int(25 - y), y from 0 to 6).

where the units have not been indicated in the integral. However it is easy to see that the units will be ft * lb.

Using antiderivative 25 y - y^2 / 2 we obtain final result

1250 pi ( (25 * 6 - 6^2 / 2) - (25 * 0 - 0^2 / 2) ) = 1250 pi * 132,

or approximately 500,000 ft * lb.

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20:29:41

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

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RESPONSE -->

Oil to surface = h=25 ft

radius = 5 ft

Density of oil= 800 kg m^3

Oil is 6ft deep

15 ft - 6ft = 8 ft rise

Volume of slice = 6 * change in h m^3

Force of gravity on slice = Density * g * Volume =

800g * 6 w^2 change in h nt

Work on slice = Force * Distance =

800g * g w^2 (8 - h) change in h joules

w= w/h = 5/19 = .263h

Work on slice = 800g (.263)^2 (8-h) change in h joules

Total work = limit as change in x approaches 0 is the sum of 55.335g h^2 (8 - h) change in h =

integral of 55.335 gh^2(8 - h) joules

(I don't know how to get to the very last step)

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20:30:01

What integral did you evaluate to determine this work?

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RESPONSE -->

Total work = limit as change in x approaches 0 is the sum of 55.335g h^2 (19 - h) change in h =

integral of 55.335 gh^2(19 - h) joules

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20:30:28

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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Again, I'm not sure how to relate this integral with y coordinate y

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20:30:57

How much drink is contained in the slice described above?

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RESPONSE -->

Work on slice = 800g (.263)^2 (19-h) change in h joules

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20:31:22

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

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RESPONSE -->

Oil to surface = h=25 ft

radius = 5 ft

Density of oil= 800 kg m^3

Oil is 6ft deep

Volume of slice = 6 * change in h m^3

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20:31:50

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

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RESPONSE -->

The work found of each slice leads to the integral formed to find total work.

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20:32:28

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

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RESPONSE -->

The main problem I had was figuring out the very last step in finding total work.

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You followed correct procedures on the last couple of problems; you missed some details but your approach is sound.

&#See my notes and let me know if you have questions. &#