Center of Mass Problem: Finding Coordinates of a Half-Disk with Uniform Density

[Hexagram1000]

The Problem is the following:
One half of a uniform circular disk of radisu 1 meter lies in the xy-plane with its diameter along the y-axis, its center at the origin, and x>0. The mass of the hallf-disk is 3 kg. Find(xcoord of center of mass, y-coord of center of mass)
The equation of the center of mass of the coord is the following:
x coord of center of mass = $$\int$$x * density(x)AreaofSlice(x)dx$$\div$$$$\int$$density(x)dx

or
x coord = $$\int$$massofslice * x $$\div$$$$\div$$ massofslice

How would you do this problem

 

[mjsd]

new to PF?

How would you do this problem

it is more than just how we would do this problem but we wish to know how you may tackle it too.


hint: perhaps first sort out what those entries in the equation mean and then go fromt there

 

[tacman]

?

To solve this problem, we first need to understand the concept of center of mass. The center of mass of an object is the point at which the mass of the object is evenly distributed, meaning the object would balance perfectly at that point if it were placed on a pivot. In this problem, we have a half-disk with uniform density, which means the mass is evenly distributed throughout the half-disk.

To find the coordinates of the center of mass, we can use the equation given above. The first step is to find the density of the half-disk. Since the mass of the half-disk is 3 kg and the area of the half-disk is half of the area of a full disk (πr^2), we can say that the density is 3/ (1/2πr^2) = 6/π kg/m^2.

Next, we need to find the area of each slice of the half-disk, which can be done by taking the integral of the equation for a circle, πr^2, and dividing it by 2 since we are dealing with a half-disk. This gives us the area of each slice as (πx^2)/2.

Now, we can plug in the values into the equation for the x coordinate of the center of mass:
x coord of center of mass = \intx * density(x)AreaofSlice(x)dx\div\intdensity(x)dx

= \int_0^1 x * (6/π) * [(πx^2)/2] dx \div \int_0^1 (6/π) * [(πx^2)/2] dx

= (3/π) * \int_0^1 x^3 dx \div (3/π) * \int_0^1 x^2 dx

= (3/π) * [(x^4)/4]_0^1 \div (3/π) * [(x^3)/3]_0^1

= (3/π) * (1/4) \div (3/π) * (1/3)

= 1/4

Therefore, the x coordinate of the center of mass is 1/4 meter. Similarly, we can find the y coordinate by using the same equation and integrating with respect to y instead of x. The y coordinate will be the same