hollow circular shaft

[khansaheb]

No.... Try
x = r * sin(I)

Incorrect....That should be: x = r * sin(u)..........................same as response #17

 

[logistic_guy]

New idea?!

You still haven't done a review, have you?

-Dan

i'm review everyday

is my steps correct?

Incorrect....That should be: x = r * sin(u)..........................same as response #17

i already do this

 

[topsquark]

i'm review everyday

No, you try to do practice problems every day. Go back and relearn what you don't remember how to do.

is my steps correct?


i already do this

No, you got the bounds of integration wrong and you didn't finish!

-Dan

 

[logistic_guy]

No, you try to do practice problems every day. Go back and relearn what you don't remember how to do.

i'll try to look up the thing i'm missing

No, you got the bounds of integration wrong and you didn't finish!

if the bounds wrong, give me the correct ones

$\displaystyle \int_{\sin^{-1}-1}^{\sin^{-1}1}\sqrt{r^2 - (r\sin u)^2} r\cos u du = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} r^2\cos^2 u du$

$\displaystyle = \frac{r^2}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 + \cos 2u du = \frac{r^2}{2}(u + \frac{1}{2}\sin 2u)\bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$

$\displaystyle = \frac{r^2}{2}(\frac{\pi}{2} + \frac{1}{2}\sin \pi - -\frac{\pi}{2} - \sin -\pi) = \frac{r^2}{2}(\frac{\pi}{2} + \frac{\pi}{2})$

$\displaystyle = \frac{r^2}{2}(\pi) = \frac{\pi r^2}{2}$

area of half a circle

 

[topsquark]

i'll try to look up the thing i'm missing


if the bounds wrong, give me the correct ones

$\displaystyle \int_{\sin^{-1}-1}^{\sin^{-1}1}\sqrt{r^2 - (r\sin u)^2} r\cos u du = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} r^2\cos^2 u du$

$\displaystyle = \frac{r^2}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 + \cos 2u du = \frac{r^2}{2}(u + \frac{1}{2}\sin 2u)\bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$

$\displaystyle = \frac{r^2}{2}(\frac{\pi}{2} + \frac{1}{2}\sin \pi - -\frac{\pi}{2} - \sin -\pi) = \frac{r^2}{2}(\frac{\pi}{2} + \frac{\pi}{2})$

$\displaystyle = \frac{r^2}{2}(\pi) = \frac{\pi r^2}{2}$

area of half a circle

Good!

-Dan

 

[khansaheb]

i'll try to look up the thing i'm missing


if the bounds wrong, give me the correct ones

$\displaystyle \int_{\sin^{-1}-1}^{\sin^{-1}1}\sqrt{r^2 - (r\sin u)^2} r\cos u du = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} r^2\cos^2 u du$

$\displaystyle = \frac{r^2}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 + \cos 2u du = \frac{r^2}{2}(u + \frac{1}{2}\sin 2u)\bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$

$\displaystyle = \frac{r^2}{2}(\frac{\pi}{2} + \frac{1}{2}\sin \pi - -\frac{\pi}{2} - \sin -\pi) = \frac{r^2}{2}(\frac{\pi}{2} + \frac{\pi}{2})$

$\displaystyle = \frac{r^2}{2}(\pi) = \frac{\pi r^2}{2}$

area of half a circle

OKay so now back to the original problem of calculating "area polar moment of inertia".

What is the "area polar moment of inertia" of a solid circular shaft?

 

[logistic_guy]

Good!

-Dan

thank

OKay so now back to the original problem of calculating "area polar moment of inertia".

What is the "area polar moment of inertia" of a solid circular shaft?

$\displaystyle J = \int r^2 dA = 2\int_{-r}^{r}x^2\sqrt{r^2 - x^2} + \frac{1}{3}(r^2 - x^2)^{\frac{3}{2}} dx$