# Centre of Mass problem

• Dec 6th 2010, 10:57 AM
craig
Centre of Mass problem
Got a problem here dealing with finding the centre of mass which I'm a bit stuck on.

A uniform lamina of mass m is bounded by the curve with equation

$\displaystyle y = \frac{x^2}{a^2}(a-x)$

and that part of the x-axis between $\displaystyle x = 0$ and $\displaystyle x = a$, where a is a positive constant. You are given that the area of the lamina is $\displaystyle \frac{1}{12}a^2$. By considering thin strips of thickness $\displaystyle \delta x$ parallel to $\displaystyle Oy$ find the position of the centre of mass $\displaystyle G$of the lamina.

For the $\displaystyle x$ coordinate:

$\displaystyle \bar{x} = \frac{\int^a_0 x(\frac{x^2}{a} - \frac{x^3}{a^2}) dx}{\frac{1}{12}a^2}$

$\displaystyle \frac{1}{12}a^2 \bar{x} = \int^a_0 x(\frac{x^2}{a} - \frac{x^3}{a^2}) dx$

This is because the mass of the strip would be $\displaystyle \delta x$ multiplied by the height, $\displaystyle y$. The distance from the origin would then be $\displaystyle x$, giving us the above equation.

So after working out the integral we're left with:

$\displaystyle \bar{x} = \frac{12}{a^2} \left( \frac{a^3}{20}\right) = \frac{3a}{5}$ which I believe is right?

However when I come to work out the $\displaystyle y$ coordinate it get's a bit confusing.

As far as I'm aware, the right hand side of the equation should be the integral of the mass multiplied by the distance from your origin, over the integral of the range of masses? This is how I proceeded with the first equation. This is what I get:

$\displaystyle \frac{1}{12}a^2 \bar{y} = \int^a_0 (\frac{x^2}{a} - \frac{x^3}{a^2}).(\frac{x^2}{a} - \frac{x^3}{a^2}) dx$

Which simplifies to:

$\displaystyle \frac{1}{12}a^2 \bar{y} = \int^a_0 (\frac{x^2}{a} - \frac{x^3}{a^2})^2 dx$

This is because the mass of the strip would again be $\displaystyle (\delta x)y$, and the distance from the origin in the $\displaystyle y$ direction would be $\displaystyle y$, giving us the above equation.

However in the answers provided, they have the integral as:

$\displaystyle \frac{1}{12}a^2 \bar{y} = \int^a_0 \frac{1}{2} \left(\frac{x^2}{a} - \frac{x^3}{a^2}\right)^2 dx$

I'm just not sure where that $\displaystyle \frac{1}{2}$ has come from?

• Dec 8th 2010, 12:38 AM
craig
Anyone got any ideas on this?
• Dec 8th 2010, 05:26 AM
Ackbeet
Consider a thin strip like you have there, with a dx width, and y = f(x) height. Where is its center of mass?
• Dec 8th 2010, 10:46 AM
craig
The COM would be halfway up the strip, hence the half y...

Thank you very much for pointing out the obvious for me there, sometimes you spend so much time looking for a complicated reason you just overlook the obvious!
• Dec 8th 2010, 10:51 AM
Ackbeet
Yeah. Been there, done that. You're welcome, and have a good one!