Centre of mass problem

**drew.walker** · May 31st 2009, 02:46 AM

I've been given the following problem to solve:

The region above the interval [1, 2] of the x-axis and below the curve y = (3x - x^2 - 2)^1/2 is rotated around the x-axis giving a solid of revolution. This solid of revolution has density, depending only on the x-coordinate, this is given by p(x) = 3x. Where is the centre off mass of this solid?

So far I've worked out the following (which may or may not be correct):

$y = \sqrt{3x - x^2 - 2}$

$A(x) = \pi(\sqrt{3x - x^2 - 2})^2 = \pi(3x - x^2 - 2)$

$Volume = \int^2_1 A(x) dx = \pi \int^2_1 3x - x^2 - 2 dx = \pi \left [ {\frac {3x^2} {2} - \frac {x^3} {3} - 2x} \right ]^2_1$

I think I'm on the right track, but I don't know how I can take this information about the volume and apply it to the density function to produce the mass, then calculate the centre of mass. Any help would be greatly appreciated.

**Laurent** · May 31st 2009, 05:50 AM

Originally Posted by drew.walker

I've been given the following problem to solve:

The region above the interval [1, 2] of the x-axis and below the curve y = (3x - x^2 - 2)^1/2 is rotated around the x-axis giving a solid of revolution. This solid of revolution has density, depending only on the x-coordinate, this is given by p(x) = 3x. Where is the centre off mass of this solid?

So far I've worked out the following (which may or may not be correct):

$y = \sqrt{3x - x^2 - 2}$

$A(x) = \pi(\sqrt{3x - x^2 - 2})^2 = \pi(3x - x^2 - 2)$

$Volume = \int^2_1 A(x) dx = \pi \int^2_1 3x - x^2 - 2 dx = \pi \left [ {\frac {3x^2} {2} - \frac {x^3} {3} - 2x} \right ]^2_1$

I think I'm on the right track, but I don't know how I can take this information about the volume and apply it to the density function to produce the mass, then calculate the centre of mass. Any help would be greatly appreciated.

This looks correct to me. For the mass, you have to integrate $A(x)dx$ (the volume of an infinitesimal slice) times $p(x)$ (the density of that slice):
${\rm Mass} = \int_1^2 p(x)A(x)dx$ .

Then the position of the center of mass is obtained the same way:
$x_{\rm center of mass}=\frac{1}{\rm Mass}\int_1^2 x p(x)A(x)dx$

**skeeter** · May 31st 2009, 06:03 AM

Originally Posted by drew.walker

I've been given the following problem to solve:

The region above the interval [1, 2] of the x-axis and below the curve y = (3x - x^2 - 2)^1/2 is rotated around the x-axis giving a solid of revolution. This solid of revolution has density, depending only on the x-coordinate, this is given by p(x) = 3x. Where is the centre off mass of this solid?

So far I've worked out the following (which may or may not be correct):

$y = \sqrt{3x - x^2 - 2}$

$A(x) = \pi(\sqrt{3x - x^2 - 2})^2 = \pi(3x - x^2 - 2)$

$Volume = \int^2_1 A(x) dx = \pi \int^2_1 3x - x^2 - 2 dx = \pi \left [ {\frac {3x^2} {2} - \frac {x^3} {3} - 2x} \right ]^2_1$

I think I'm on the right track, but I don't know how I can take this information about the volume and apply it to the density function to produce the mass, then calculate the centre of mass. Any help would be greatly appreciated.

the center of mass lies on the x-axis because of symmetry.

$\bar{x} = \frac{\int x \, dm}{\int dm}<br />$

$\rho = \frac{M}{V} = \frac{dm}{dV}$

$dm = \rho \, dV = 3x \cdot \pi [f(x)]^2 \, dx$

so ...

$\bar{x} = \frac{\int_1^2 3x^2 \cdot \pi (3x-x^2-2) \, dx}{\int_1^2 3x \cdot \pi (3x-x^2-2) \, dx}$

**drew.walker** · June 1st 2009, 02:12 AM

Okay, it took me the better part of a day to work out the relationship between both replies and make progress, but I think my answer (or working) is still incorrect. Here is my new worked solution:

$y = \sqrt{3x - x^2 - 2}$

${\rm Density} = p(x) = 3x$

${\rm Area} = A(x) = \pi(\sqrt{3x - x^2 - 2})^2 = \pi(3x - x^2 - 2)$

${\rm Mass} = M = \int^2_1 {p(x)A(x)} dx = \int^2_1 {3x \cdot \pi(3x - x^2 - 2)} dx$

$\bar{x} = \frac {1} {M} \int^2_1 {x \cdot p(x) \cdot A(x)} {dx} = \frac {1}{M} \int^2_1 {x \cdot 3x \cdot \pi(3x - x^2 - 2)} dx$

$= \frac {\int^2_1 {3x^2 \cdot \pi(3x - x^2 - 2)} dx} {\int^2_1 {3x \cdot \pi(3x - x^2 - 2)} dx}$

I think solving these integrals is where I start having problems. The next step in my working is:

$= \frac {3\pi \int^2_1 {x^2(3x - x^2 - 2)} dx}{3\pi \int^2_1 {x(3x - x^2 -2)} dx} = \frac {\int^2_1 {x^2(3x - x^2 - 2)} dx}{\int^2_1 {x(3x - x^2 -2)} dx} = \frac {\int^2_1 {3x^3 - x^4 - 2x^2} dx} {\int^2_1 {3x^2 - x^3 - 2x} dx}$

$= \frac {\left [ \frac{3x^4}{4} - \frac{x^5}{5} - \frac{2x^3}{3}\right ]^2_1}{\left [ \frac{3x^3}{3} - \frac{x^4}{4} - \frac{2x^2}{2}\right ]^2_1}$

I've left out the numeric calculations for brevity, but I end up with:

$= \frac{\frac{23}{60}}{\frac{1}{4}} = \frac{23}{15} = 1.5\dot3$

This seems like a plausible answer, but I have almost no confidence, because 90% of the time (or more) I make mistakes. Hopefully someone can confirm that I've done the working the correct way and whether the answer is correct.

**skeeter** · June 1st 2009, 04:23 AM

your plausible solution is correct.

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