Proof Questions.

**dadon** · April 20th, 2006, 02:59 AM

Need help on two proof questions I hope you guys can help!

There is alot on the internet but it's hard the way they explain it.

1) Prove that the centre of mass of a uniform wire in the form of a semi-circle of radius r is at a distance 2r/pie (sorry don't know how to insert symbol) from the centre.

2) Prove that the centre of mass of a uniform semi-circular laminar of radius r is at a distance 4r/3pie from the centre.

Thanks

**dadon** · April 21st, 2006, 06:16 AM

I guess everyone hates doing proof questions.

**CaptainBlack** · April 21st, 2006, 07:51 AM

Originally Posted by dadon

1) Prove that the centre of mass of a uniform wire in the form of a semi-circle of radius r is at a distance 2r/pie (sorry don't know how to insert symbol) from the centre.

The centre of mass is the average position of the mass of a body.

In the case of the semi-circle of $m\ \mbox{kg/m}$ , the total
mass is $m.r. \pi\ \mbox{kg}$ .

Introduce $\theta$ as in the diagram, then an element of the wire
has length $r.d\theta$ and the centre of mass is:

$<br /> R=\frac{1}{m.r. \pi}\int_{wire} \rm{r}$ $\ \ dm=\frac{1}{m.r. \pi}\int_0^{\pi} r {\sin(\theta) \choose \cos(\theta)} m.r d\theta=\frac{r}{\pi} \int_0^{\pi} {\sin(\theta) \choose \cos(\theta)} d\theta$ $=\frac{r}{\pi} {2 \choose 1} d\theta<br />$ .

I'm sure you can complete the problem from here.

RonL

**earboth** · April 21st, 2006, 07:54 AM

Originally Posted by dadon

...
1) Prove that the centre of mass of a uniform wire in the form of a semi-circle of radius r is at a distance 2r/pie (sorry don't know how to insert symbol) from the centre.
2) Prove that the centre of mass of a uniform semi-circular laminar of radius r is at a distance 4r/3pie from the centre.
Thanks

Hello,

to 1) Put the semicircle into a coordinate system, so that the diameter lies on the x-Axis and the centre of the semicircle is the origin. Then the semicircle is described by:
$y=\sqrt{r^2-x^2}$

The centroid of a homogenous plane area, which has a graph of a function and the x-axis as boundaries, can be calculated in this problem here by:

$x_{c}=\frac{\int^r_{-r}{xydx}}{\int^r_{-r}{y dx}}$ and

$y_{c}=\frac{\int^r_{-r}{y^2dx}}{2 \cdot \int^r_{-r}{y dx}}$

$x_{c}=\frac{\int^r_{-r}{x\cdot \sqrt{r^2-x^2}dx}}{\int^r_{-r}{\sqrt{r^2-x^2} dx}}$

Using the substitution method on the integral of the numerator, you'll get the value 0 (zero). So the x-coordinate of the centroid is 0, which could be expected.

$y_{c}=\frac{\int^r_{-r}{(r^2-x^2)dx}}{2 \cdot \int^r_{-r}{\sqrt{r^2-x^2} dx}}$

you'll get:

$y_{c}=\frac{\left[r^2 \cdot x -\frac{1}{3} x^3 \right]^r_{-r}} {\left[ x \cdot \sqrt{r^2-x^2} +r^2 \cdot \arcsin \left( \frac{x}{r} \right) \right]^r_{-r} }$

$y_{c}=\frac{2r^3}{\frac{\pi}{2} r^2 - \left(- \frac{\pi}{2} r^2 \right)}=\frac{2r}{\pi}$

I'm awfully sorry, but I'm a little bit in hurry. I hope this was of some help.

Greetings

EB

**dadon** · April 21st, 2006, 08:31 AM

Thank you! You were a great help!

I will try both problems this way and ask for help if I still need it. I’ll post my solutions as well to get your feedback. Thanks again

Take care,

dadon

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