Moment of inertia

I wrote this article about “Moment of inertia” in a “moment of weakness” …

Physics defies language, simple example:

1. “I has” the dimension different from the dimensions of my mass.

2. “I is” moment of inertia.

3. “The I” of a rigid body is not difficult to calculate.

I will post something if you want to learn pretty much everything you have to learn “about I”,

Pretty much all the ideas of moment of inertia is packed in the following:

(from the result
$\inline \dpi{150} \bg_green I=mr^2$
you can integrate or sum this to get more complicated rigid bodies, define from C.O.M. etc)

Without knowing, I solved problem 1.5.5 of mathematical physics text Arfken, 6th edition, above, you can copy this and rearrange and you have an answer. I was just trying to help a kid with his Physics, he asked on facebook. For the general case in Arfken you have to be careful somewhere, figure, perhaps when you make the replacement

$\inline \dpi{150} \bg_green \vec{r} \times \frac{d}{dt}\vec\omega = \vec0$

using orthogonality of $\inline \dpi{150} \bg_green \vec{r}$ and $\inline \dpi{150} \bg_green \vec{\omega}$ .

You can think of it this way:

(you have to understand many concepts such as vector products, derivatives, vector operations such as tripple products and the ideas in these equation, if you do not understand something feel free to ask)

(.) is dot product of vectors.

$\inline \dpi{150} \bg_green \vec p_{line} = m \vec v\vspace{1} \\ \vec p_{angular} = I \vec{\omega}$

$\inline \dpi{150} \bg_green \frac{d}{dt}\vec L = \frac{d\vec r}{dt} \times \vec p + \vec r \times \frac{d\vec p}{dt} \implies \vec F_{ang} = \vec v \times \vec p + \vec r \times \vec F_{lin}, \vec F_{ang} = \frac{\vec p_{ang}}{dt}$

force is always time derivative of momentum

p = momentum. so linear momentum = $\inline \dpi{150} \bg_green \vec p_{line}$ , corresponds to mass & velocity,
angular momentum = $\inline \dpi{150} \bg_green \vec p_{angular}$ , in terms of moment of inertia I and angular velocity $\inline \dpi{150} \bg_green \vec \omega$ ,
now $\inline \dpi{150} \bg_green \vec p_{angular} = \vec L = \vec r \times \vec p$ , a cross product. if you differentiate wrt time this becomes

$\inline \dpi{150} \bg_green = \frac {d(I\omega)}{dt} = I \frac {d\omega}{dt}$ , as I is constant in time but w is not constant in time.
also F_ang = v X (p) + r X (ma) , using F_ang = v X p + r X F_lin and F_lin = m x a (as you wrote)
= v X [mv] + r X ma = zero + r X ma,
SO r X ma = I dw/dt
or I = [mr X a]/[(d/dt) w] = [m r X d/dt (d/dt) r] / [ (d/dt) w ]
(wrote earlier: v = L w, correct v=rw)

So above becomes
I = [m r X d/dt (v)] / [ dw/dt ] or I = [m r X (dr/dt x w + r x dw/dt)] / [ dw/dt ]
Again since r X v = zero, they are same direction, 1st term dr/dt x w vanishes after cross product from left
SO, I = m r X [ r x dw/dt ] / [dw/dt], if you expand by using the identity [ A (B X C) = B(A.C) – C(A.B) ] and use the fact that r and dw/dt are perpendicular since r and theta are perpendicular (rotation is perpendicular to radial direction) you get the fact that
I = m x r x r (mrr)

Moment of inertia

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Moment of inertia

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