# Dynamics: Angular Momentum

1. Oct 23, 2005

### 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

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Last edited: Oct 24, 2005
2. Oct 25, 2005

### Astronuc

Staff Emeritus
This problem seems to be a bit of beast.

As for LATEX resources, these have been recommended -

MiKTeX

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.

Last edited by a moderator: May 2, 2017
3. Oct 25, 2005

### 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.

Last edited: Oct 25, 2005
4. Oct 26, 2005

### Astronuc

Staff Emeritus
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 = ICM + M d2

where,

ICM 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

5. Oct 26, 2005

### MarkL

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

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.