- #1

Haths

- 33

- 0

I remember that for any point particle in the disk its MOI is going to be:

mr

^{2}

I can show this because if we consider a point mass at some distance r from a point of rotation, it will still subscribe to F=ma in the tangential direction. Knowing that:

**T**= I [alpha]

Where

**T**is the torque, I the MOI, and alpha the angular acceleration.

**T**=Fr

Therefore:

**T**=mar

As

[alpha] = a / r

mar=I a/r

I=mr

^{2}

Hence I'm certain of that. Thus I can appreciate that for a solid disk I am looking at a summation of the MOI's between the axis and the edge of the disk. Hence I need an integration factor such that;

I

_{T}= Integ

^{r}

_{0}{ dm/dr r

^{2}} dr

I think. Yes? Hence I need an expression for the rate of change of mass with respect to radial displacement. Hence I need an area integral, and this is where I'm left blank, because I have no idea how to express the circle's mass with respect to increasing radius. Unless...

I make my integral:

I

_{T}= Integ

^{2PI}

_{0}{ dm/dr r

^{2}} d[theta]

Where dm/dr = r

^{2}PI

Therefore dm = 1/3 r

^{3}PI |

^{r}

_{0}hence:

I

_{T}= Integ

^{2PI}

_{0}{ 1/3 r

^{3}PI r

^{2}} d[theta]

= 1/3 PI r

^{5}[theta] |

^{2PI}

_{0}

= 2/3 PI

^{2}r

^{5}

Which I'm certain is wrong. I'm not asking why it's wrong. But how should I construct my dm/dr and integrate that through with reasoning why that's a good method. Because I have a feeling I'm forgetting something fundamental in this part of the calculation.

Cheers,

Haths