Inertia from four masses connected by massless rigid rods

[aivilo775]

Homework Statement

The four masses shown in (Figure 1) are connected by massless, rigid rods. Assume that m = 230 g. Find the moment of inertia about an axis that passes through mass A and is perpendicular to the plane of the image.

Relevant Equations

Ia = kg * m^2

I found xcm = 6.2cm, ycm = 4.9cm, but am unsure of how to tackle the inertia part of the problem.

 

[PhDeezNutz]

Assuming your $x_{cm}$ and $y_{cm}$ are right (I haven't checked). Then you can merely apply the "parallel axis theorem" which in plain english states

"The moment of inertia about a given axis (presumably not the center of mass) is equal to the moment of inertia about the center of mass plus $Mh^2$ where $M$ is the total mass and $h$ is the distance from the center of mass to the axis of interest.

$I = I_{cm} + Mh^2$

You've already done the hard part. The answer to your second question is seconds away.

So I suggest you find the Moment of Inertia about the center of mass that you calculated and just add $Mh^2$

 

[PhDeezNutz]

I'm going to do it both manually and by using the parallel axis theorem just to make sure I'm not going crazy.

You should do the same.

Remember $I = \sum \limits_{i=1}^{n} m_i r_i^2$ for each of the $n$ particles and the axis of interest.

 

[aivilo775]

Remember I=∑i=1nmiri2 for each of the n particles and the axis of interest.

So I did I = Ia +Ib+Ic+Id. Since it's rotating through mass A, I set Ia = 0. I then did I = (.5)(230/1000)(8/100)^2 + (.5)(300/1000)(sqrt[(10/100)^2 + (8/100)^2)])^2 + (.5)(230/1000)(10/100)^2

I got that I = 0.004346 kg * m^2

Does this seem right?

Or I suppose I shouldn't use the .5 so
I = (230/1000)(8/100)^2 + (300/1000)(sqrt[(10/100)^2 + (8/100)^2)])^2 + (230/1000)(10/100)^2 = 0.008692

 

[kuruman]

So I did I = Ia +Ib+Ic+Id. Since it's rotating through mass A, I set Ia = 0. I then did I = (.5)(230/1000)(8/100)^2 + (.5)(300/1000)(sqrt[(10/100)^2 + (8/100)^2)])^2 + (.5)(230/1000)(10/100)^2

I got that I = 0.004346 kg * m^2

Does this seem right?

It doesn't seem right. Why are you multiplying each term by 0.5? Please post your equations in symbolic form then substitute the numbers at the very end. If you don't know how to use LaTeX for posting equations, click on the link "LaTeX Guide", lower left.

 

[aivilo775]

View attachment 316186

Removing the 1/2 I used the equation above and got I = 0.008692

 

[kuruman]

Still incorrect. There are three masses to consider, so there are three terms to add. I see only two. Also, only one mass is in your expression, the one in the second term.

 

[haruspex]

Homework Statement:: The four masses shown in (Figure 1) are connected by massless, rigid rods. Assume that m = 230 g. Find the moment of inertia about an axis that passes through mass A and is perpendicular to the plane of the image.
Relevant Equations:: Ia = kg * m^2

I found xcm = 6.2cm, ycm = 4.9cm

It is rarely helpful to find the common mass centre of an assemblage. Moments of inertia are additive, as shown in post #3. This gives you a much easier way.