Assignment 9

course Mth 174

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???g??x??m??????Physics II

02-22-2007

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16:04:20

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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16:20:23

what is the moment of the rod?

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RESPONSE -->

Moment = Integral (from a to b) of x*delta(x)dx

= Integral (0 to 2) x*(2+6x) dx

= Integral (0 t0 2) 2x+6x^2 dx

= x^2+2x^3 (evaluated from 0 to 2)

= 4+16-0

= 20 gm.-m.

Supplementary note:

** To get the center of mass, which was not requested here but does appear in other problems, 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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16:20:35

What integral did you evaluate to get a moment?

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RESPONSE -->

See previous response.

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16:21:32

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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16:32:01

what is the total mass of the region?

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RESPONSE -->

Draw a vertical slice from g(x) to f(x) with width delta x.

Mass of slice = Area * mass density =

y * delta x * rho(x)

Riemann sum (from a to b):

Sum (i=1 to n) rho(x)*[f(x)-g(x)]*delta x

Mass of region = Integral (from a to b) rho(x)*[f(x)-g(x)] dx grams

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16:37:27

What integral did you evaluate to obtain this mass?

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RESPONSE -->

Integral in previous response--left in integral form.

Evaluated:

Mass = rho(x)*[F(x) - G(x)], evaluated from a to b

= rho (x)*[(F(b)-F(a))-(G(b)-G(a))] grams

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16:40:32

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

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RESPONSE -->

From first response,

mass of increment = area * mass density

= rho(x)* y * dx

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

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16:40:59

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 -->

See previous response.

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16:41:53

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

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RESPONSE -->

See first response for this problem--create Riemann sum of masses of increments, approximate with integral.

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16:49:35

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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17:00:04

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

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RESPONSE -->

Draw a slice parallel to circular bottom with height of delta x, at distance x from top of tank.

Volume of slice = pi*r^2*h = pi*(6^2)*delta x = 36pi*delta x feet^3

Force to lift slice = Density * Volume = 62.4 lb/ft^3 * 36pi*delta x ft^3 = 2246.4pi*delta x lb.

Distance slice to be lifted = x + 10 feet

Work to lift slice = Force * Distance

= 2246.4pi*delta x * (x+10)

Total Work to lift all water:

W = Integral (from 0 to 20) 2246.4pi*(x+10) dx

= 2246.4pi*Integral (from 0 to 20) (x+10) dx

= 2246.4pi*(x^2/2 + 10x), evaluated from 0 to 20

= 2246.4pi*(400/2+200)

= 898560pi

= 2,822,909.5 ft.-lb. (approx.)

excellent

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17:00:13

What integral did you evaluate to determine this work?

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RESPONSE -->

See previous response.

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17:01:21

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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See previous response. y=0 (I used x) is at the top of the tank.

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17:07:38

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

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RESPONSE -->

Taking the Riemann sum of the work expended on all of the segments gives the total work required to lift all the water. As the height of each segment approaches 0, the sum (total work) is approximated by taking the integral of the Riemann sum, evaluated from the top of the tank to the bottom (since the tank is full).

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17:19:46

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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20:40:09

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

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RESPONSE -->

I couldn't find this problem in the book. Based on the information given, I drew a cone with the apex at the bottom, a height of 10 cm, a diameter at the top of 10 cm, and the liquid to be lifted 5cm above top of cone. Also, I assumed the liquid to be water (or have the density of water).

r = radius of slice at height h from top (circular end) of cone.

Volume of slice = pi*r^2h = pi *[(10-h)/2]^2*delta h = (pi/2)*(100-20h-h^2)*delta h

Mass density of water = 1 g/cm^3

F(slice) = mass density * g * volume

F = 1 * g * (pi/2)*(100-20h-h^2)*delta h

Distance Raised = 5 + h

Work(slice) = F * D

= (pi*g/2)*(100-20h-h^2)*delta h * (5+h)

instead of 5 + h you should use 15 - h. The water is being raised from height h to height 15 feet.

= (pi*g/2)*(500-15h^2+h^3)*delta h

Total Work = Integral (from 0 to 10) (pi*g/2)*(500-15h^2+h^3) dh

=(pi*g/2)*(500h-5h^3+h^4/4), evaluated from 0 to 10

g = 9.8 m/sec^2 = 980 cm/sec^2

Work = 490pi*(5000-5000+2500)

= 490 pi* 2500

= 1,225,000pi

Since joules - kgm^2/sec^2 and I used g and cm, we divide answer by 10^3*10^2*10^2 = 10^7

Work = 1,225,000pi/10^7 = .1225pi joules

= 0.385 joules (approx.)

= 1

Very good, except possibly for the height through which each increment of water is being raised. Using the expression I give we get about .33 Joules.

For a ballpark comparison a 10 cm cube would contain 1 liter or 1 kg of water, weight 9.8 N, and when raised an average of 10 cm (midpt height from 5 cm to 15 cm) would require .98 Joules of energy. The cone contains less than 1/4 as much water, but the water is concentrated nearer the bottom, so the work is more than 1/4 as great.

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20:40:21

What integral did you evaluate to determine this work?

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RESPONSE -->

See previous response.

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20:41:26

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 -->

See previous response. I used h rather than y. h=0 is top of tank.

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20:41:42

How much drink is contained in the slice described above?

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See previous response.

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20:42:31

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

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See previous response. Cross-sectional area is the pi*r^2 part of volume.

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20:45:55

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

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RESPONSE -->

Cross-sectional area of slice times height (delta h) gave volume of slice. Voluume times density of water times g gives force required to move slice. Force times distance moved [h + (5 cm above top)] gives work required to move the slice to height 15 cm above bottom of cone. Riemann sum of work for slices gives total work. Integral of Reimann sum (as delta h approaches zero) approximates total work required.

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20:49:19

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

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RESPONSE -->

I'm amazed how integrals are essential to solving so many problems in physics that involve anything other than linear or constant functions.

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When you're in nonlinear situations integrals are the only way to go; of course not everything can be integrated in closed form, which makes life even more interesting.

Excellent work. See my one suggested correction on the cone problem, as well as a plausibility comparison.