Finding centre of mass of a semicircular lamina

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SUMMARY

The discussion focuses on calculating the center of mass of a semicircular lamina with a density function proportional to the distance from the origin. The lamina is defined as the right half of a circular disk of radius r centered at the origin. The density function is expressed as δ(x,y) = &sqrt;(x2 + y2). To find the center of mass coordinates (xg, yg), the mass M must first be computed using the double integral M = ∫∫L δ(x,y) dx dy, followed by the coordinates calculations using xg = (1/M) ∫∫L x δ(x,y) dx dy and yg = (1/M) ∫∫L y δ(x,y) dx dy.

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Suppose the density of any point on a semicircular lamina {(x, y) : x >= 0,$$ x^2 + y^2$$ <= r^2 is proportional to its distance from the origin. Compute the centre of mass.

I am trying to figure out what is the density function for this case.
 
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Re: finding centre of mass of a semicircular lamina

The lamina here is the right half of a circular disk of radius $r$ centered at the origin.

What is the density function $\rho(x)$ and what kind of symmetry may we use?
 
Re: finding centre of mass of a semicircular lamina

MarkFL said:
The lamina here is the right half of a circular disk of radius $r$ centered at the origin.

What is the density function $\rho(x)$ and what kind of symmetry may we use?
Yaa.. that's what I am trying to figure out.. the density function.. and after that its pretty straight forward.
 
Re: finding centre of mass of a semicircular lamina

We are told the density at some point $P(x,y)$ is proportional to the point's distance from the origin. What is the distance of $P$ from the origin?
 
Re: finding centre of mass of a semicircular lamina

candy said:
Suppose the density of any point on a semicircular lamina {(x, y) : x >= 0,$$ x^2 + y^2$$ <= r^2 is proportional to its distance from the origin. Compute the centre of mass.

I am trying to figure out what is the density function for this case.

If the task is to find the coordinates x and y of the center of mass, the density function can be supposed to be... $\displaystyle \delta (x,y)= \sqrt{x^{2}+ y^{2}} $ (1)

Remember that You have to perform the following steps...

a) compute the mass of the lamina... $\displaystyle M= \int \int_{L} \delta (x,y)\ dx dy$ (2)b) compute the coordinates of the center of mass... $\displaystyle x_{g}= \frac{1}{M}\ \int \int_{L} x\ \delta(x,y)\ dx dy$

$\displaystyle y_{g}= \frac{1}{M}\ \int \int_{L} y\ \delta(x,y)\ dx dy$ (3)

Kind regards

$\chi$ $\sigma$
 
Last edited:
Re: finding centre of mass of a semicircular lamina

candy said:
Suppose the density of any point on a semicircular lamina {(x, y) : x >= 0,$$ x^2 + y^2$$ <= r^2 is proportional to its distance from the origin. Compute the centre of mass.

I am trying to figure out what is the density function for this case.
Are you saying you do not know what "proportional" means? Or that you do not know what the distance from the origin is?

I would suggest changing to polar coordinates.
 

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