Finding COM, and inertia please help my answers don't make sense??


by amyparker30
Tags: center of mass, inertia, phyics
amyparker30
amyparker30 is offline
#1
Apr20-12, 04:51 AM
P: 13
1. The problem statement, all variables and given/known data

Find the center of mass of the object shown in the figure below?

LINK TO FIGURE:
http://s1168.photobucket.com/albums/...t=Untitled.jpg

1. Find the center of mass of the object shown in the figure below.

2. Calculate the rotational inertia of the object about the x-axis.

3. From this value, deduce the rotational inertia of the object about an axis parallel to the x-axis, and going through the center of mass.

m1=2.5kg , m2=5kg, m3=2.5kg , m4=5kg

2. Relevant equations

COM

Inertia?

3. The attempt at a solution

my answers don't make sense

for COM I got 2
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NascentOxygen
NascentOxygen is offline
#2
Apr20-12, 08:02 AM
HW Helper
P: 4,715
Hi amyparker30! Where are your answers? Where is your working?
amyparker30
amyparker30 is offline
#3
Apr20-12, 10:01 AM
P: 13
ok, my answers are totaly wrong but for COM 1 got (2(2.5+5+2.5+5))/15 = 2

ICOM=Rotational Inertia = 2^2(2.5+5+5+2.5) = 60

i bet these are 100% wrong what should I do, my teacher wouldn't help me and the people from class just skipped this question :(

joseamck
joseamck is offline
#4
Apr20-12, 11:14 AM
P: 12

Finding COM, and inertia please help my answers don't make sense??


Ok first you need to understand the COM formula and see how it works. Basically you need to specify a coordinate to tell where the COM is. That means you need to find the center of mass in the x direction, y direction, and z direction of the system or object.
So we start by separating everything in components( x, y and z). Lets do the x direction first.

We let [itex]d_{nx} [/itex] represent the distance in the x direction for each particle (n=1,2,3,4)
Then
[itex]X_{com} = \frac{m_2 d_{2x}+m_3 d_{3x} + m_1 d_{1x}+m_4 d_{4x}}{M}[/itex]
where M is the total mass. Note that
[itex]d_{2x}=d_{1x}=0[/itex]
and
[itex]d_{3x}=d_{4x}=2 m[/itex]

Similarly for y:
[itex] Y_{com} = \frac{m_2 d_{2y}+m_3 d_{3y} + m_1 d_{1y}+m_4 d_{4y}}{M}[/itex]
Note that
[itex]d_{2y}=d_{3y}=0[/itex]
[itex]d_{1y}=d_{4y} = 2 m[/itex]

Similarly for z:
[itex] Z_{com} = \frac{m_2 d_{2z}+m_3 d_{3z} + m_1 d_{1z}+m_4 d_{4z}}{M}[/itex]
But Note that
[itex]d_{1z} = d_{2z} = d_{3z} = d_{4z} = 0[/itex]
then
[itex]Z_{com} = 0[/itex]

Then at the end you get the COM to be at [itex](X_{com},Y_{com},Z_{com})[/itex]

Now try to plug in the numbers and see what you get. (Easy!!! =))

Do the same thing for the next part (moment of inertia), first understand the formula/equation and just follow it slowly.
NascentOxygen
NascentOxygen is offline
#5
Apr20-12, 07:43 PM
HW Helper
P: 4,715
I think the first part of this question would be a good one for a multiple choice test.

By symmetry, the CoM of the two equal masses m1 and m2 is mid-way between them. Likewise, the CoM of the equal mass pair m3 and m4 is mid-way between m3 and m4. And you wouldn't believe our luck but those two CoM locations coincide!

So that surely must be the CoM for the 4 bodies.


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