Assignment 9

course Mth 174

I am resending this assignment.

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Physics II

06-14-2009

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18:26:25

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 -->

moment of mass = (2 + 6 xi) * xi

I assumed the axis of rotation is x=0

I used:

integral from 2 to 0 of (x(2 +6x) = integral of 2x + 6x^2= x^2 + 2x^3 = (2)^2 + 2(2)^3= 20

Right. More detail:

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. **

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18:26:41

what is the moment of the rod?

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RESPONSE -->

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18:26:55

What integral did you evaluate to get a moment?

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RESPONSE -->

integral from 2 to 0 of 2x + 6x^2

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18:28:02

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 -->

integral from g(x) to f(x) change in y dx

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).

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18:28:35

what is the total mass of the region?

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RESPONSE -->

Total mass is the integral from g(x) to f(x) change in y dx

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18:28:48

What integral did you evaluate to obtain this mass?

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RESPONSE -->

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18:29:36

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

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RESPONSE -->

mass of the ith peice = m is approximatly theta(xi) change in x

the center of mass = the sum of xi*mi divided by the sum of mi

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18:30:27

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= ax(x) change in x is approximatly (1-x) change in x

Center of mass is the integral of (x theta Ax(x) dx) / mass

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18:32:28

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

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RESPONSE -->

mass of ith intergal= area * density= f(ci) - g(ci) * 'dx * rho(ci)= rho(ci) f(ci) - g(ci) * 'dx

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18:37:44

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 -->

radius = 6

height = 20 + 10

density of wanter= 1

Volume of slice = 6 change in h

Force = density * g* volume

= 1 * 6 * g change in h

work done on slice= force * distance

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

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

integral from 20 to 0 is 30gh^2(30 - h)

STUDENT SOLUTION:

delta W = delta F(30-y)
delta W = (62.4)(volume)(30-y)
delta W = 62.436pidelta 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.

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18:38:00

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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18:38:38

What integral did you evaluate to determine this work?

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RESPONSE -->

the limit as change in x approaches 0 the um of 60 gh^2 (30 - h) change in h

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18:39:11

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 -->

I do not understand how to do this problem where y=0.

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18:39:39

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

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RESPONSE -->

The problems can be solved similarily

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18:43:23

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 -->

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

density of oil= 800 kg m^3

force= 800 * gw^2 change in h

work on slice= 800g * gw^2 (19 - h) change in h

w= w/h= 5/19

total work= limit as change in x approaches 0 is the sum of 55.3 gh^2 (19 - h) change in h

= integral of 55.3 gh^2(19 - h)

Im not sure what to do from here.

** 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.

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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. **

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18:45:43

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

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RESPONSE -->

work on slice = force * distance= 800g *gw^2 (8 - h) change in h

total work = limit as change in x approaches 0 is teh susm of 55.3 gh^2 (8-h) change in h

= integral of 55.3 gh^2 (8 - h)

Im not sure where to go from here.

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18:46:16

What integral did you evaluate to determine this work?

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RESPONSE -->

the limit as change in x approaches 0 is the sum of 55.3g h^2 (19 - h) change in h

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18:46:45

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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RESPONSE -->

I am still uncertain of how to work this part of the problem in relation to y

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18:47:16

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

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18:47:58

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

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RESPONSE -->

Volume = 6 * change in h

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18:48:30

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

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RESPONSE -->

You use the work found done on each slice to find the integral

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18:48:32

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

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Assignment 9

&#This looks good. See my notes. Let me know if you have any questions. &#