Center of mass

• Jul 19th 2009, 01:40 PM
arbolis
Center of mass
A lattice is on the portion of the disk $x^2+y^2 \leq 1$ in the first quadrant. Find its center of mass if the density at any point is proportional to its distance with the $x$ axis.
My attempt : I first calculated the total mass of the lattice which gave me $\frac{k}{3}$, where $k$ is the constant of proportionality.
Then I found out that the center of mass is located at $\left( \frac{3}{8},\frac{3}{16} \right)$.

I just want to know if this is the correct result because I find it quite strange.
• Jul 19th 2009, 03:25 PM
Jester
I agree with your total mass and the x coordinate of the c of m. How did you arrive at the y coord.?
• Jul 19th 2009, 03:40 PM
arbolis
Quote:

Originally Posted by Danny
I agree with your total mass and the x coordinate of the c of m. How did you arrive at the y coord.?

I've made many errors and now I'm not able to read my notes. Is $y=\frac{3\pi}{16}$?
$y=\frac{k \int_0^{\pi/2} \int_0^{1} \rho ^3 \cos ^2 (\theta) d\rho d\theta}{\frac{k}{3}}=\frac{3\pi}{16}$.
• Jul 19th 2009, 03:43 PM
Jester
Quote:

Originally Posted by arbolis
I've made many errors and now I'm not able to read my notes. Is $y=\frac{3\pi}{16}$?
$y=\frac{k \int_0^{\pi/2} \int_0^{1} \rho ^3 \cos ^2 (\theta) d\rho d\theta}{\frac{k}{3}}=\frac{3\pi}{16}$.

Yes, that's what I obtained. Side note - most people use $r$ and not $\rho$ for the radius in 2D problems.
• Jul 19th 2009, 03:53 PM
arbolis
Quote:

Originally Posted by Danny
Yes, that's what I obtained. Side note - most people use $r$ and not $\rho$ for the radius in 2D problems.

You're right. I'm the odd one. (Punch)(Giggle)
• Jul 20th 2009, 02:43 AM
HallsofIvy
Two things I notice immediately are that the figure and density function are both symmetric in x and y. The x and y coordinates of the center of mass must be the same.
• Jul 20th 2009, 04:37 AM
Jester
Quote:

Originally Posted by arbolis
A lattice is on the portion of the disk $x^2+y^2 \leq 1$ in the first quadrant. Find its center of mass if the density at any point is proportional to its distance with the $\color{red}{x}$ axis.
My attempt : I first calculated the total mass of the lattice which gave me $\frac{k}{3}$, where $k$ is the constant of proportionality.
Then I found out that the center of mass is located at $\left( \frac{3}{8},\frac{3}{16} \right)$.

I just want to know if this is the correct result because I find it quite strange.

Quote:

Originally Posted by HallsofIvy
Two things I notice immediately are that the figure and density function are both symmetric in x and y. The x and y coordinates of the center of mass must be the same.

I beg to differ.