Moment of Inertia Calculation for a Horizontal Element

kieranl
Messages
22
Reaction score
0
can somebody please tell what i am doing wrong with this moment of area question?

question and working in the attached doc

thanks alot
 

Attachments

Physics news on Phys.org
To calculate the second moment about the x-axis, you have to choose a horizontal element dA, because only a horizontal element has a constant y. I don't know why you chose vertical strips and used their centers as the y value, but your method is not right.

To calculate Ix, express dA in terms of y and you'll get dA=(sqrt(ay)-y)dy. Integrate y^2 dA from 0 to a and you'll get the correct answer, a^4/28. I have no idea how your lecturer got to his equation for Ix, but it's probably unnecessarily complicated because the method I just described leads to the right answer very quickly.

To calculate Iy, choose vertical elements and express dA in terms of x. Then integrate x^2 dA from 0 to a.
 
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...

Similar threads

Replies
5
Views
4K
Replies
5
Views
958
Replies
4
Views
2K
Replies
2
Views
695
Replies
2
Views
2K
Back
Top