# Kinetic Moment about the center of gravity

1. Apr 10, 2014

### Chaso

Hi I need some help explaining kinetic moment and moment of the center of mass.
This is an example:
http://session.masteringengineering.com/problemAsset/1529176/3/Hibbler.ch17.p107.jpg
Okay I know that when you do moment equations:
The Moment forces about some point is = to the kinetic moment about that point so for same.
M_g = KM_g (kinetic moment)
So if we do this for point C (where the wheel touches the ground)
Its M_C = Ig (moment of inertia about g, center of mass)*(angular acceleration) + kinetic moment about C which would be due to the acceleration of G.

But when you do that for for the Point G the Moment about G is always equal to the the Ig*a.
So my question is way is the moment about the center of mass always equal to the Ig*a when theres an acceleration at point c. wouldn't that affect the moment about G?

2. Apr 11, 2014

### Simon Bridge

I've heard of "kinetic moment" in fluid dynamics - but that does not seem to be what you mean.

A "moment" is usually how engineers refer to "torque" which would be "moment of force" ... it is the applied force multiplied by the moment-arm.
There are other moments - like a moment of area and so on.
http://en.wikipedia.org/wiki/Moment_(physics [Broken])

$M_g = KM_g$ looks odd - but you seem to be saying that "kinetic moment" and "moment of force" are defined to be the same thing. So why the different names?

From your examples, it looks like you are trying to calculate the moment of inertia about different points.
I suspect your confusion comes from being inexact about what you mean by "kinetic moment".
Start from a formal definition and a few simple examples.

Last edited by a moderator: May 6, 2017
3. Apr 11, 2014

### Chaso

No I meant kinetic moment. The moment of a point likes say c is equal to the kinetic moment of that point like angular acceleration*mass of inertia and acceleration*mass*distance

4. Apr 11, 2014

### Simon Bridge

I have not heard of "mass of inertia".
I am guessing you mean "moment of inertia".

I can only find "kinetic moment of inertia" is nuclear physics papers, where it relates to nuclear spin and shape.
There was an old wikipedia article mentioning it but only in comparison to the second moment of area as having units of length times area. The article is no longer current.

Taking "moment of mass" to mean "moment of inertia", lets see if I have understood you:
... lets see,

moment of inertia times angular acceleration is normally called "torque" or "moment of force". $\tau=I\alpha$
acceleration times mass times distance is normally called "work" ... $W=amd$ but it could be $mra$ which is the torque on a small mass accelerating around a circle radius r.
... it's looking an awful lot like you do mean "torque" here.

But you have insisted that kinetic moment is different to moment of force ... so please show me the difference.
Please provide a formal definition and a simple example.

points do not normally have moments anyway - masses may have moments and objects may have moments about points.

I think you need to clear up these ideas before you can go further.
I'm going to see if I can find someone who recognizes what you are trying to talk about.

Last edited: Apr 11, 2014
5. Apr 11, 2014

### Chaso

Yes, I meant moment of inertia sorry. So basically I'm learning this stuff in a dynamics class. For this subject it was under the chapter of general planar motion. So basically, in class my professor stated this equation:
M_P (Moment about point P of an rigid body) = I_G (Moment of inertia about center of mass) + m(mass) * [ (R_P/G (position vector of point G to P) X (cross product) (acceleration of point G)]

But at point G R_G/G makes the cross product 0 so:

M_G = I_G

and my question is why? I know in the equation it make sense but when I look at the example I posted about there's an acceleration at Point C doesn't this have an effect on the Kinetic Moment. So that:
M_G = I_G + m*a_C (acceleration of c)*(distance)

But this isn't the case:
M_G = I_G and my question is why.

6. Apr 11, 2014

### Simon Bridge

$M_P = I_{G}+m[\vec r_{PG}\times \vec a_G]$

That looks a bit like the regular torque... however: the dimensions do not work out.
The RHS is adding $ML^2$ to $ML^2T^{-2}$ which is not allowed.
Have you done dimensional analysis?

For an object mass m rotating about a point P other than it's center of mass G, the moment of inertia is adjusted according to the parallel axis theorem.

$I_P=I_G+mr_{GP}$

If a net force is then applied at the center of mass so the object undergoes angular acceleration $\alpha$ about point P, which would require something holding P in place, then:

$\tau = (I_G+mr_{GP})a/r_{GP}=|\vec r_{GP}\times \vec F|$ where $a=r\alpha$ is the instantaneous linear acceleration of the center of mass and $\vec r_{GP}$ is the vector pointing from G to P.

This starting to look familiar?

7. Apr 11, 2014

### Chaso

Um no to dimensional analysis. And the bottom is not looking familiar at all. I understand why the moment about C is equal to I_G + m(R x a). I don't understand why in for point g in the example I linked in my first post, the moment about point g is only equal to I_g*alpha
http://session.masteringengineering.com/problemAsset/1529176/3/Hibbler.ch17.p107.jpg

Last edited: Apr 11, 2014
8. Apr 11, 2014

### Simon Bridge

OK - last bit first:

The torque about any arbitrary point P is $\tau_P=I_P\alpha_P$
(using the subscript to indicate the point the rotationas are taken about

So the torque about point G is going to be the same equation but with a G everywhere you see a P.
They just left the G off the angular acceleration.

(I am explicitly using "torque" instead of "moment" here to avoid the earlier confusion.)

"Dimensional analysis"
http://www.usciences.edu/~lvas/Math422/Dimensions.pdf
... it's basically a fancy way of saying that the units have to make sense.

IG has units of kg.m2
m(Rxa) has units of kg. m2/s2

Saying that M=IG+m(Rxa) is like saying M = 5 carrots + 3 oranges.
It's nonsense: you can only add like to like.

9. Apr 13, 2014

### vela

Staff Emeritus
It would help if you didn't keep writing an incorrect equation. As Simon repeatedly pointed out, the units don't work out. If you checked your textbook, you'll see that Hibbeler writes that $\sum M_G = I_G \alpha$, not $\sum M_G = I_G$. You probably just incorrectly copied what your professor wrote down, or your professor made a mistake.

Anyway, the answer to your question is that it's because Newton's second law describes how the forces that act on a body cause the center of mass to move, not any arbitrary point on the body. You have
\begin{align*}
\sum \vec{F}_i &= m \vec{a}_G \\
\sum M_G &= \sum (\vec{r}_i \times \vec{F}_i) = I_G \alpha
\end{align*} where $\vec{r}_i$ is the moment arm from the center of mass $G$ to where the force $\vec{F}_i$ acts. Now suppose instead you want to calculate about point $O$ instead. You still have
$$\sum \vec{F}_i = m \vec{a}_G$$ but the rotational equation of motion becomes
$$\sum M_O = \sum [(\vec{r}_i+\vec{r}_{GO}) \times \vec{F}_i] = I_O \alpha$$ where $\vec{r}_{GO}$ is the displacement of the center of mass from point $O$. Distributing the cross product gives you
$$\sum M_O = \sum \vec{r}_i\times\vec{F}_i + \sum \vec{r}_{GO} \times \vec{F}_i = I_G\alpha + \vec{r}_{GO}\times \sum \vec{F}_i = I_G\alpha + \vec{r}_{GO}\times (m\vec{a}_G).$$ The term $\vec{r}_{GO}\times (m\vec{a}_G)$ comes from the fact that the net force describes how the center of mass accelerates, not any arbitrary point $C$ on the body.