Finding COM, and inertia please help my answers don't make sense??by amyparker30 Tags: center of mass, inertia, phyics 

#1
Apr2012, 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 xaxis. 3. From this value, deduce the rotational inertia of the object about an axis parallel to the xaxis, 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 



#2
Apr2012, 08:02 AM

HW Helper
P: 4,716

Hi amyparker30! Where are your answers? Where is your working?




#3
Apr2012, 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 :( 



#4
Apr2012, 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. 



#5
Apr2012, 07:43 PM

HW Helper
P: 4,716

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 midway between them. Likewise, the CoM of the equal mass pair m3 and m4 is midway 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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