Find the centre of mass of the system?

[emily081715]

Homework Statement

All three disks are made of sheet metal of the same material, and the diameters are 1.0 m , 2.0 m , and3.0 m . Assume that the x-axis has its origin at the left-most point of the left-most object and it points to the right.Determine the location of the center of mass of the system shown below .

Homework Equations

x_cm=mx₁+mx₂+mx₃/m1+m2+m3

The Attempt at a Solution

so we were never shown how to do a question like this, so i am very lost on where to begin, and need to find an answer for my mastering physics assignment.
i used the area of a circle (πr²) to find the mass of each object. I found the radius by dividing the given diameters by two.
m1=0.78593 kg
m2=3.14159 kg
m3=7.06858 kg
using theses masses i plugged values into the equation x_cm=mx₁+mx₂+mx₃/m1+m2+m3
x_cm=0.78593(0.5)+3.14159(1)+ 7.06858(1.5)/0.78593+3.14159+7.06858
x_cm=1.3m

 

[PhanthomJay]

In determining the center of mass from the formula, the values of x1, x2, and x3 are not the radius of each disc. What distances do each represent?

 

[emily081715]

is it the diameter?

 

[emily081715]

is it the diameter?

 

[emily081715]

In determining the center of mass from the formula, the values of x1, x2, and x3 are not the radius of each disc. What distances do each represent?

the x in the equation is the centre of mass, would this not just be half the diameter of the circle

 

[gneill]

It's probably best to work symbolically rather than trying to plug in (and lug around) decimal numbers.

Define the radius of the first circle to be r and its mass to be M. You know that the radius is directly proportional to the diameter (D = 2r after all), and that mass varies with the area which varies as the square of the radius. So if the first disk has mass M, what's the mass of the second disk in terms of M? And the third? Next pencil in the locations of their centers of mass in terms of r.

 

[emily081715]

It's probably best to work symbolically rather than trying to plug in (and lug around) decimal numbers.

Define the radius of the first circle to be r and its mass to be M. You know that the radius is directly proportional to the diameter (D = 2r after all), and that mass varies with the area which varies as the square of the radius. So if the first disk has mass M, what's the mass of the second disk in terms of M? And the third? Next pencil in the locations of their centers of mass in terms of r.

Your reply confused me. I'm unsure how to do what you asked

 

[PhanthomJay]

the x in the equation is the centre of mass, would this not just be half the diameter of the circle

no , you are trying to find the center of mass of the entire three disk system. The center of mass of each disk is at its center. You need to determine the distance of each center from a chosen point when calculating x1 , x2, and x3

 

[gneill]

Your reply confused me. I'm unsure how to do what you asked

Okay, if the first disk has radius r and mass M, and if mass is proportional to r², then if the next disk in line has a radius 2r it's mass must be 4M. Let's do that in detail:

Suppose the density per unit area is $\rho$. Then for radius $r$ the mass is $M = \rho \pi r^2$. If we double the radius we find:

$M_2 = \rho \pi (2r)^2 = \rho \pi 4 r^2 = 4(\rho \pi r^2) = 4M$

The mass goes as the square of the radius, so double the radius yields four times the mass. Triple the radius and what do you get for the mass?

 

[emily081715]

no , you are trying to find the center of mass of the entire three disk system. The center of mass of each disk is at its center. You need to determine the distance of each center from a chosen point when calculating x1 , x2, and x3

I'm sorry can you explain that again.

 

[I like Serena]

What are the x coordinates of the centers of those disks?

 

[emily081715]

What are the x coordinates of the centers of those disks?

wouldn't the first one behalf way through the first circle which would be 0.5. would that make the second one 1.5 and the third be 2.5?

 

[emily081715]

Okay, if the first disk has radius r and mass M, and if mass is proportional to r², then if the next disk in line has a radius 2r it's mass must be 4M. Let's do that in detail:

Suppose the density per unit area is $\rho$. Then for radius $r$ the mass is $M = \rho \pi r^2$. If we double the radius we find:

$M_2 = \rho \pi (2r)^2 = \rho \pi 4 r^2 = 4(\rho \pi r^2) = 4M$

The mass goes as the square of the radius, so double the radius yields four times the mass. Triple the radius and what do you get for the mass?

9M?

 

[I like Serena]

wouldn't the first one behalf way through the first circle which would be 0.5. would that make the second one 1.5 and the third be 2.5?

Let's make a drawing:
https://dl.dropboxusercontent.com/u/14301878/Math/mass_center_3_disks.png
See what x1, x2, and x3 must be?

 

[gneill]

9M?

Yup!

Next list the locations of the disk's centers of mass in terms of multiples of r:

 

[emily081715]

Let's make a drawing:
https://dl.dropboxusercontent.com/u/14301878/Math/mass_center_3_disks.svg
See what x1, x2, and x3 must be?

x1=0.5
x2=2
x3=4.5

 

[emily081715]

Yup!

Next list the locations of the disk's centers of mass in terms of multiples of r:

View attachment 107188

3r
9r?

 

[gneill]

3r
9r?

Nope. Use the diagram I provided. You should be able to count off the r's. The first one is located at x = r.

 

[I like Serena]

x1=0.5
x2=2
x3=4.5

Correct!
Substitute in the formula with the masses you already had and presto.
(Those masses effectively assume a density of 1, but that is okay'ish, since we're dividing again by the mass anyway.)

 

[emily081715]

Nope. Use the diagram I provided. You should be able to count off the r's. The first one is located at x = r.

15r, 25r?

 

[gneill]

15r, 25r?

I don't see how you're getting those numbers. What's the width (diameter) of the first disk in terms of r? How about the second disk? The center of mass of the second disk is located at the center of the second disk. How many r's is that from the origin? Should should be able to add up the r lengths from the origin to the center of the second disk. Do the same for the third disk.

 

[emily081715]

I don't see how you're getting those numbers. What's the width (diameter) of the first disk in terms of r? How about the second disk? The center of mass of the second disk is located at the center of the second disk. How many r's is that from the origin? Should should be able to add up the r lengths from the origin to the center of the second disk. Do the same for the third disk.

thank you for the help, i reached the answer. its 3.5 m

 

[David Lewis]

Hope I'm not giving bad advice but, due to laziness, I'd denominate the x-axis in metres (as per post #14), not r.

 

[gneill]

Hope I'm not giving bad advice but, due to laziness, I'd denominate the x-axis in metres (as per post #14), not r.

I just find it easier to manipulate values symbolically rather than push around a lot of decimal places. It's also much easier to find errors in the algebra that way; numbers tend to lose their identity in a long formula and are subject to copying errors over the course of a derivation.