Question

A wire shaped like the first-quadrant portion of the circle \(x^2+y^2=a^2\) has density \(\delta=kxy\) at the point \((x,y)\). I need to find its moment of inertia around the x axis!

Answer 1

So I have \[\vec{r}(t)=\left\langle a\cos t, a\sin t, 0 \right\rangle\quad 0\leq t\leq \frac{\pi}{2}\]and therefore \[ds=\sqrt{a^2\cos^2t+a^2\sin^2t}\,dt=a\,dt\]so the mass is\[m=\int^{\pi/2}_0\delta\,ds=ka^3\int^{\pi/2}_0\cos t\sin t\,dt=\frac{ka^3}{2}\]

Answer 2

now how do I find the moment of inertia. I have the formula \[I=\int_C p^2\,dm\] where p is the perpendicular distance from the axis in question. So I believe that would be \[p=a^2\sin t\]\[p^2=a^4\sin^2 t\]plugging this in I get \[I_x=\int^{\pi/2}_{0}a^4\sin^2 t\,dm=ka^7\int^{\pi/2}_{0}\sin^3t\cos t\,dt\]

Answer 3

I feel that step is where I am going wrong but I can't see why

Answer 4

oh damn this should be in math!

Answer 5

Mostly everything seems good, but \(p=a\sin (t)\), not \(p=a^2\sin (t)\). Otherwise I think you're good.

Answer 6

ok so then I would have\[I_x=a^4\int^{\pi/2}_0\sin^3t\cos t\,dt\] which is still incorrect.

Answer 7

sorry that should be \(a^5\) because \[p^2=a^2\sin^2t\] and \[dm=\delta\,dt=a^3\sin t\cos t\]

Answer 8

i think you need a k still....

Answer 9

yeah sorry. still doesn't get me the answer! the answer is... \[I_x=\int^{\pi/2}_0ka^5\sin^2t\cos t\,dt\] where am I getting this extra \(\sin t\) from?

Answer 10

dm = kxy = ka^2 sin t cost
p = y = a sin t
ds = a dt

so I = int p^2 dm = ka^5 sin^3t cos t dt

I think I agree with you......

Answer 11

yup turns out we were correct. there was a typo in the textbook!