How to get a # out of moment of the parallel axis theorm

[pinkfishegg]

The question is : four masses shown in figure 12.14 (Knight 12.14) are connected by massless, rigid rods. (see attached fig)
A) find the coordinates of the center of mass
B) find the moment of inertia about a diagonal mass that passed through B and D
so I got part a right by setting the x-y axis at A and using
X-bar=(200g(10cm)+200(10))/700g=(40/7)cm=5.7cm
Y-Bar=5.7cm

I'm unsure how to do part B
I can use the parallel axis theorum

I=[/cm]+M[d[/2]
but I'm unsure how to use the MD^2 part. M i just the sum of all the masses right? D would (√x^2+y^2)^2 which is just x^2 plus y^2. But they're askng for a number for moment of inertia and i don't know which point they are looking at on the diagonal rod. What am i missing here?

 

[mfb]

I can use the parallel axis theorum

You could, but this is much easier if you just apply the usual formula for the moment of inertia. You know the axis and you can calculate the distances to this axis.

and i don't know which point they are looking at on the diagonal rod.

I think the diagonal rod is the axis? The phrasing of the problem statement is a bit odd.

 

[pinkfishegg]

You could, but this is much easier if you just apply the usual formula for the moment of inertia. You know the axis and you can calculate the distances to this axis.
I think the diagonal rod is the axis? The phrasing of the problem statement is a bit odd.

Yeah the diagonal rod is the axis and the center of mass is at (5.7 cm , 5.7 cm)
What confused me is which part of the axis i start using the moment of inertia formula from.
It's Icm=∑Mr^2. So M is just the sum of the masses. Is r the distance from the axis to the center of mass then?this changes depending on your position. so i guess you make an integral but then what can you plug into the integral?

 

[mfb]

M are the individual masses, not their sum (because the sum sign is running over all masses).

Is r the distance from the axis to the center of mass then?

It is the distance from the axis to the individual masses. It will be different for each mass.
There is nothing to integrate. You are overthinking this, the solution can be written down in a single line.

 

[pinkfishegg]

M are the individual masses, not their sum (because the sum sign is running over all masses).It is the distance from the axis to the individual masses. It will be different for each mass.
There is nothing to integrate. You are overthinking this, the solution can be written down in a single line.

Wait can i just pick any point on the axis and plug in

I=sum(MD^2)

 

[mfb]

No you cannot pick a point. You have to calculate the distance to the axis. This is defined as the distance to the closest point (which you can find).

 

[pinkfishegg]

No you cannot pick a point. You have to calculate the distance to the axis. This is defined as the distance to the closest point (which you can find).

Oh i see so the closest distance to points B and D is just 0 because the line is on both B and D. The closest point to points A and C would be in the center of the diagonal (x=5 cm y=5cm+ and the distance to the other points would be (sqrt (5-0)cm+(5-0)cm)^2 and (sqrt (10-5)cm+(10-5)cm)^2 = sqrt(50)cm=2sqrt(5cm) in both cases?

 

[mfb]

I think the square root got misplaced. Apart from that, right.

 

[pinkfishegg]

I think the square root got misplaced. Apart from that, right.

oh i see yeah i completely over-complicated this thanks.