Assignment 9

course Mth 174

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assignment #009
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Physics II
07-03-2008

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12:28:52
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 -->
Part a) Write a Riemann sum approximating the total mass of the rod: Since the density at a distance 'dx from the left of the rod is given by 2 + 6x. My R-sum is 2 + 6x 'dx
Part b) Find the exact mass by converting the sum into an integral: I as 'dx approaches 0, from [0,2] int( 2 + 6x 'dx) = 2x + 6x^2 I then used the Fundamental Theorem to find the mass, F(b) - F(a). 2(2) + 6(2)^2 = 4 + 24 = 28. F(b) - F(a) = 28 - 0.

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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12:29:18

what is the moment of the rod?

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RESPONSE -->

Not real sure what the 'moment' of the rod really means.

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.

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12:30:55

What integral did you evaluate to get a moment?

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I just realized that my original integral was wrong. I would like to use this answer to correct it.

My integral should have been int(2 + 6x) = 2x + 3x^2

Which would have made the mass 16

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12:37:13

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 -->

With left and right bounds of x = a and x = b, upward bound of f(x), and lower bound of g(x) the mass can be found:

In terms of f(x) and g(x): int(f(x)) - int(g(x)). Since f(x) is larger than g(x), the difference in area underneath the two would yield the area in between the two functions, which is the mass.

In terms of Density: int[ R-sum('rho(x) 'dx)

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12:37:50

what is the total mass of the region?

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RESPONSE -->

Numerically I do not know, but int(f(x)) - int(g(x)) = mass of the region

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12:38:39

What integral did you evaluate to obtain this mass?

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RESPONSE -->

I didn't really find an integral to evaluate.

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12:47:33

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

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RESPONSE -->

Mass = the integral of the R-sum of 'rho(x) within the interval (x, 'dx)

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12:52:35

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 -->

I am not completely sure, but I would say the area of the increment would be smaller interval [x, 'dx] over which we can find a more solid density on which to base our R-sum. With this R-sum we can find the mass.

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12:54:51

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

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RESPONSE -->

mass of increment =

[int from a -> b (x * 'rho(x) 'dx)] / [int from a -> b ('rho(x) 'dx)]

The numerator of your expression is the total moment about x = 0.

The denominator is the total mass.

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13:59:51

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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RESPONSE -->

Problem 12 asked to find how much work is required to pump all of the water over the rim, which is from 10ft to 20ft. Problem 13 asks how much work is required to pump all of the water from the same, full cylinder to a point 10 ft over the cylinder.

I first cut out a slice of the cylinder with a width of 12 and a height of 'dh. I then found the Volume of this slice = 'pi * r^2 * 'dh = 'pi * w^2 * 'dh

The Density of water is 1000 kg/m^2. Force of gravity on the slice = Density * g * Volume = 1000 * g * 'pi * w^2 * 'dh. Since we are trying to move every piece of water to 10 ft above the cylinder, that includes moving that water at the bottom of the cylinder also, we must move all of the water 30 ft (20ft of the cylinder + 10 ft above).

Work = Force * Distance

Force = 1000 * g * 'pi * w^2 * 'dh

Distance = (30 - h)

I then found w in terms of h: (w / h = 12 / 20) w = .6 *h

Work = 1000 * g * 'pi * .36h^2 * (30 - h) * 'dh

I then summed up the slices and took the limit as 'dh -> 0.

R-sum [360 * g * 'pi * (30h^2 - h^3)]

When g = 9.8

R-sum [360 * g * 'pi * (30h^2 - h^3)]

From b = 30 and a = 0; int [360 * g * 'pi * (30h^2 - h^3)]

= 'pi * 360 * p * ((30/3)h^3 - (1/4)h^4)

= 'pi * 360 * 9.8 * ((30/3)h^3 - (1/4)h^4)

= 'pi * 3528 * ((30/3)h^3 - (1/4)h^4)

F(b) - F(a) = 748188874.5 = Work

I looked it up, and this is wrong, but I am not sure where I went wrong.

You've made a good attempt. However the width of the cylinder does not change with its height; your solution was probably based on an example using a cone, in which case the width would depend on the height.

You have

Force = 1000 * g * 'pi * w^2 * 'dh
Distance = (30 - h),

both of which are OK. However w corresponds to the radius of the horizontal cross-section and is in this case constant, equal to half the diameter of the cylinder.

In the solution given below I use y_i for the y coordinate at the sample point of the ith slice. I leave the density rho, the acceleration g of gravity, and the cross-sectional area A as constants.

** 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.
**

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13:59:56

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

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RESPONSE -->

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13:59:58

What integral did you evaluate to determine this work?

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RESPONSE -->

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14:02:19

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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RESPONSE -->

Work = lim('dh -> 0) R-sum [ Force * Distance] = lim('dh -> 0) R-sum [ (Density * g * Volume) * Distance]

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14:02:33

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

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14:26:19

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 -->

I do not see this question in the book. I am not sure what to do.

The question here is how much work it takes to empty a conical glass. The cone opens upward, is 10 cm high and at its top is 10 cm wide. Water is pumped to a pont 15 cm above the bottom of the glass, which is the apex of the cone.

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

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