Dynamics: Angular Momentum Calculations

[MarkL]

Could someone could show me where I am going wrong? Here is my work so far.
Given: F\cdot\nabla t = (5.4 N.s) j, Density = 1.2
So, M_{ring} = 1.696 kg, M_{rod} = 0.36 kg, M_{total} = 3.753 kg
Part a) V = 1.44 m/s j (which is correct)
Part b)

Moment of impulse:

H_{G} = r\times mv = (-0.225 m)\cdot (5.4 kg.m/s) i + (.225 m)\cdot (5.4 kg.m/s) k

Mass moments and product of Inertia:

I_{x} = 2\cdot (\frac{1}{2}M_{ring}\cdot r_c^2 + M_{ring}\cdot r_c^2 + M_{ring}\cdot (\frac{1}{2}\cdot L)^2) + \frac{1}{12}M_{rod}\cdot L^2

I_{y} = 2\cdot (M_{ring}\cdot r_c^2 + M_{ring}\cdot r_c^2) = 4\cdot (M_{ring}\cdot r_c^2)
I_{z} = 2\cdot (\frac{1}{2}M_{ring}\cdot r_c^2 + M_{ring}\cdot (\frac{1}{2} L)^2) + \frac{1}{12}M_{rod}\cdot L^2
I_{yz} = M_{ring}\cdot (0.15)\cdot (-0.225) + M_{ring}\cdot (-0.15)\cdot (0.225)
I_{xz}= I_{xy} = 0
H_x = (-)(.225 m)(5.4 kg.m/s) = I_{x}*/omega_{x} = 0.336*w_x
H_y = 0 = I_y*w_y - I_yz*w_z = 0.343*w_y + 0.114*w_z
H_z = (.225 m)*(5.4 kg.m/s) = - I_yz*w_y + I_x*w_x = 0.114*w_y + 0.164*w_z
Solving for w_x: w_x = (- 3.62 r/s) , which is wrong!
I won't bother with the rest.
The correct answer is (-3.55 r/s)i + (-3.2 r/s)j + (9.87 r/s)k
Note: I would use Latex, but it would not preview. Is there a way I can practice this offsite?
Thanks,
Mark

 

[Astronuc]

This problem seems to be a bit of beast.

As for LATEX resources, these have been recommended -

MiKTeX
http://www.ctan.org/tex-archive/systems/win32/miktex/?action=/installationadvice/

WinEdt (must pay)
http://www.winedt.com/

Tex -> Html
http://www.cse.ohio-state.edu/~gurari/TeX4ht/mn.html

Tex -> Pdf
http://www.tug.org/applications/pdftex/

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As for the problem, what \vec{r} are you using? It should have three components.

This would relate to the moment arm - what pivot point to you expect?

I take it that i, j, k are in x, y, z directions, respectively.

 

[MarkL]

r = (0.225 m) i + (0.15 m) j + (.225 m) k
through the center, G, to the point of impulse.

Yes, i, j and k are in x, y, z directions.

The pivot point is definitely at A.
But calculations are made on the principle axis through G. (is this right?)

Maybe my principle axis is incorrect since there are products of inertia.
But H_{x} should be independent of this. So,

I_{x} = 2\cdot (\frac{1}{2}M_{ring}\cdot r_c^2 + M_{ring}\cdot r_c^2 + M_{ring}\cdot (\frac{1}{2}L)^2) + \frac{1}{12}M_{rod}\cdot L^2

H_{x} = (-.225 m)(5.4 kg.m/s) = I_{x}*w_{x} = 0.336*w_{x}

I have a hunch the moment of inertia, I_{x}, is incorrect.

No hurry. I'll keep working on it and post the solution, if I get it.

 

[Astronuc]

Mark, I am curious, from what textbook is the problem taken?

The moment of inertia I of an object with respect to an axis other than the one through its center of mass if given by

I = I_CM + M d²

where,

I_CM in the moment of inertia of the object with respect to the axis passing through the center of mass,

M is the mass of the object

and d is the distance between the CM and the point of rotation on the other axis.

and the angular momentum, \vec{L} = \vec{r}\,\times\,m\vec{v}.

I wonder if this would help - http://scienceworld.wolfram.com/physics/AngularMomentum.html
http://scienceworld.wolfram.com/physics/MomentofInertia.html

 

[MarkL]

Textbook: Beer and Johnston, Statics and Dynamics, Chap 18

Use this link:

http://gaia.csus.edu/~grandajj/me115/me115index.htm"

The lecture is straight from the book, plus some example problems.

In three dimensions, \vec{L} is now \vec{H}

As they say in the book, the computation of the angular momentum H_{G} is now considerably more involved.

They aren't kidding.