Assignment 9

course MTH 174

RVܩKÃassignment #009

|hrW

Physics II

10-13-2008

......!!!!!!!!...................................

19:34:29

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

......!!!!!!!!...................................

RESPONSE -->

a) Riemann Sum: `sigma 2 + 6x g/m * dx

b) integral from 0 to 2 (2 + 6x) dx

2x + 6x^2 / 2 from 0 to 2

2(2) + 6(2)^2 / 2 - 2(0) + 6(0)^2 / 2

4 + 12 = 16 g

.................................................

......!!!!!!!!...................................

19:46:29

what is the moment of the rod?

......!!!!!!!!...................................

RESPONSE -->

The moment of the rod would be:

integral from 0 to 2 x(2 + 6x) dx

integral from 0 to 2 (2x + 6x^2) dx

2x^2 / 2 + 6x^3 / 3 from 0 to 2

x^2 + 2x^3 from 0 to 2

2(2)^2 + 2(2)^3 - 2(0)^2 + 2(0)^3

8 + 16 = 24 g * m is the moment of the rod

I believe there's a small error in your next-to-last step. x^2 doesn't have a coefficient of 2.

Complete solution, for reference and comparison:

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.

.................................................

......!!!!!!!!...................................

19:46:45

What integral did you evaluate to get a moment?

......!!!!!!!!...................................

RESPONSE -->

integral from 0 to 2 x(2 + 6x) dx

.................................................

......!!!!!!!!...................................

20:05:00

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

......!!!!!!!!...................................

RESPONSE -->

Area is integral from a to b of [f(x) - g(x)]dx

Mass is integral from a to b of `rho(x) * [f(x) - g(x)] dx

.................................................

......!!!!!!!!...................................

20:05:08

what is the total mass of the region?

......!!!!!!!!...................................

RESPONSE -->

Mass is integral from a to b of `rho(x) * [f(x) - g(x)] dx

.................................................

......!!!!!!!!...................................

20:05:17

What integral did you evaluate to obtain this mass?

......!!!!!!!!...................................

RESPONSE -->

Mass is integral from a to b of `rho(x) * [f(x) - g(x)] dx

.................................................

......!!!!!!!!...................................

20:06:31

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

......!!!!!!!!...................................

RESPONSE -->

Mass of an increment of x with width `dx is

`rho(x) * [f(x) - g(x)] `dx

.................................................

......!!!!!!!!...................................

20:07:35

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

......!!!!!!!!...................................

RESPONSE -->

Area of the increment is [f(x) - g(x)] * `dx

The area of this increment times the density function will determine the mass of the area.

.................................................

......!!!!!!!!...................................

20:08:47

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

......!!!!!!!!...................................

RESPONSE -->

The mass of the increment is used as the basis for the Riemann sum of all masses and a limit is taken giving the integral to be evaluated.

You should start your analysis of problems of this nature with partitions and Riemann sums. I expect that you are doing so; in these examples the process is pretty obvious and it's not difficult to move straight to the integral. However it's a good habit to routinely write out these steps, and I will want to see the steps on tests.

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)

.................................................

......!!!!!!!!...................................

20:19:59

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

......!!!!!!!!...................................

RESPONSE -->

Slicing the water in strips parallel to x axis:

Volume of a strip is `pi * (6ft)^2 * `dy

Force on a strip 62.4 lb/ft^3 * `pi * 36ft^2 * `dy

Work done in pumping the water strip:

62.4 lb/ft^3 * `pi * 36ft^2 * (30 - y) * `dy

The integral will be from 0 to 20

62.4 lb/ft^3 * `pi * 36ft^2 * (30 - y) dy

62.4 lb/ft^3 * `pi * 36ft^2 * [30(20) - (20)^2 / 2]

62.4 lb/ft^3 * `pi * 36ft^2 * [600 - 200] = 898560 * `pi ft-lb.

It's probably preferable to do the work symbolically before plugging in the numbers. The symbolic solution below is identical to yours, except that the substitutions and final evaluation aren't included.

** 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.

**

.................................................

......!!!!!!!!...................................

20:20:32

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

......!!!!!!!!...................................

RESPONSE -->

Work required is 898560 * `pi ft-lb.

.................................................

......!!!!!!!!...................................

20:20:48

What integral did you evaluate to determine this work?

......!!!!!!!!...................................

RESPONSE -->

The integral will be from 0 to 20

62.4 lb/ft^3 * `pi * 36ft^2 * (30 - y) dy

.................................................

......!!!!!!!!...................................

20:21:50

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.

......!!!!!!!!...................................

RESPONSE -->

Work done in pumping the water strip:

62.4 lb/ft^3 * `pi * 36ft^2 * (30 - y) * `dy

y=0 is the bottom of the tank.

.................................................

......!!!!!!!!...................................

20:23:17

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

......!!!!!!!!...................................

RESPONSE -->

The work is done over the distance 30 - y and is the distance in the force * displacement equation for work.

.................................................

......!!!!!!!!...................................

20:38:17

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)

......!!!!!!!!...................................

RESPONSE -->

g = 980cm/s^2

Slicing the water in slices parallel to the x-axis:

For a slice:

Volume = `pi [1/2 (8 - y)]^2 `dy

Force = 1.2 * `pi [1/2 (8 - y)]^2 `dy

Work is:

1.2 * `pi [1/2 (8 - y)]^2 * 980 * (15 - y) * `dy

Integral from 0 to 8

1.2 * `pi [1/2 (8 - y)]^2 * 980 * (15 - y) dy

Multiplying out the integrand and pulling out factors to be in front of the integral:

294 * `pi * integral 960 - 304y + 79y^2 - y^3

294 * `pi * [960y - 304y^2 / 2 + 79y^3 / 3 - y^4 / 4]

294 * `pi * [960 * 8 - 304 * 8^2 / 2 + 79 * 8^3 / 3 - 8^4 / 4]

294 * `pi * [7680 - 9728 + 13482 * 2/3 - 1024] = 9615585 dynes * 10^-5 N/1dyne = approx 96.15N

We might have the apex at different points. I don't see anything wrong with your solution, assuming the apex is at the top.

Compare with the following, where the apex is at the bottom:

** 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. **

.................................................

......!!!!!!!!...................................

20:38:39

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

......!!!!!!!!...................................

RESPONSE -->

9615585 dynes * 10^-5 N/1dyne = approx 96.15N

.................................................

......!!!!!!!!...................................

20:39:12

What integral did you evaluate to determine this work?

......!!!!!!!!...................................

RESPONSE -->

294 * `pi * integral 960 - 304y + 79y^2 - y^3 dy

.................................................

......!!!!!!!!...................................

20:40:13

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.

......!!!!!!!!...................................

RESPONSE -->

Work for one slice is:

1.2 * `pi [1/2 (8 - y)]^2 * 980 * (15 - y) * `dy

y = 0 is the bottom of the glass.

.................................................

......!!!!!!!!...................................

20:40:44

How much drink is contained in the slice described above?

......!!!!!!!!...................................

RESPONSE -->

Volume of slice = `pi [1/2 (8 - y)]^2 `dy

.................................................

......!!!!!!!!...................................

20:41:54

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

......!!!!!!!!...................................

RESPONSE -->

Volume of slice is `pi [1/2 (8 - y)]^2 `dy

Cross-sectional area of the slice `pi [1/2 (8 - y)]^2

.................................................

......!!!!!!!!...................................

20:46:41

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

......!!!!!!!!...................................

RESPONSE -->

Volume = `pi [1/2 (8 - y)]^2 `dy

Force = 1.2 * `pi [1/2 (8 - y)]^2 `dy

Work is:

1.2 * `pi [1/2 (8 - y)]^2 * 980 * (15 - y) * `dy

The work is the force applied through the interval.

The Force is the Volume * density.

Volume is Cross-sectional area * `dy

Cross-sectional area is found by using similar triangles to find the width of the strip. I then took half that for the radius.

.................................................

......!!!!!!!!...................................

20:46:44

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

......!!!!!!!!...................................

RESPONSE -->

.................................................

"

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