Moment of Inertia of Thin Plate: x^2+4y^2=4

Nyasha
Messages
127
Reaction score
0
Consider a thin plate of constant density which occupies the region in the first quadrant inside the curve:

x^2+4y^2=4


Find moment of inertia about line x=-3


Attempt to solution:


y=\frac{\sqrt{4-x^2}}{2}

I(x=-3)=\frac{1\rho}{2}\int_0^2(x+3)^2\sqrt{4-x^2}


\text{Is there any easier way of integrating this thing without having to expand} (x+3)^2 \text{and then multiply it with}\sqrt{4-x^2}
 

Attachments

  • untitled.JPG
    untitled.JPG
    3.4 KB · Views: 404
Last edited:
Physics news on Phys.org
In your solution for y(x), where did the 1/2 come from?

I don't see any options except to do it piece by piece.
 
Dr.D said:
In your solution for y(x), where did the 1/2 come from?

I don't see any options except to do it piece by piece.


y=\sqrt{\frac{4-x^2}{4}}
 
There is a conflict in your posted information. On the figure, it agrees with what you have used
y(x) = (1/2) * sqrt(4-x^2)
I see that now.
In your original post, you also wrote
LaTeX Code: x^2+y^2=4
which leads to
y(x) = sqrt(4-x^2)
That is why I asked the question.
 
Dr.D said:
There is a conflict in your posted information. On the figure, it agrees with what you have used
y(x) = (1/2) * sqrt(4-x^2)
In your original post, you also wrote
LaTeX Code: x^2+y^2=4
which leads to
y(x) = sqrt(4-x^2)
That is why I asked the question.

Thanks for pointing out that mistake :smile:
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top