Hi I'm kooltrainer - here is a simple engineering q i wan u engineers to solve (1 Viewer)

[Uncle]

An annular washer of constant surface density ρ occupies the region between the circles

x² + y² = a²

and

x² + y² = b²

where b > a

Find the moment of inertia of the washer about the x-axis.

 

[Uncle]

Oh don't worry guys, here is the picture and LaTeX solution:

[maths]I_{x}=\int \int_{\Omega}^{}y^{2} \rho dA[/maths]

[maths]I_{x}=\rho \int_{0}^{2\pi } \int_{a}^{b}(r sin\theta )^{2}rdrd\theta [/maths]

[maths]I_{x}=\rho \int_{0}^{2\pi } \int_{a}^{b}(r)^{3}(sin^{2}\theta) drd\theta[/maths]

[maths]I_{x}=\rho \int_{0}^{2\pi } [\frac{r^{4}}{4}]_{a}^{b}(sin^{2}\theta) d\theta [/maths]

[maths]I_{x}=\frac{\rho}{4}(b^{4}-a^{4}) \int_{0}^{2\pi } (\frac{1}{2}-\frac{1}{2}cos2\theta ) d\theta [/maths]

[maths]I_{x}=\frac{\rho}{8}(b^{4}-a^{4}) [\theta - \frac{1}{2}sin2\theta ]_{0}^{2\pi }[/maths]

[maths]I_{x}=\frac{\rho\pi }{4}(b^{4}-a^{4})[/maths]

I realised that rather than be a narutard and go ask the internet I realised I was better off solving the question myself.
No shit about that.
Thank you for your time and I really appreciate this.

 

[Omium]

An annular washer of constant surface density ρ occupies the region between the circles

x² + y² = a²

and

x² + y² = b²

where b > a

Find the moment of inertia of the washer about the x-axis.

Not to sure how to go about this but.

Consider a circle or arbitrary radius 'r'

Moment of inertial by defintion
I = r^2 dm

Constant surface density p

p = dm/dA.

pdA = dm.

so subbing this in

I = p dA r ^2

call this (1)


p is constant so it can be taken out of the integral.

Since we have a circle

A = Pi r^2

So differentiating dA/dr = 2 Pi r

so dA = 2 Pi r dr.

subbing this into (1)

I = p 2 Pi r^3 dr.

Integrate this with respect to r.

and we are given the radius of the circles is "b" and "a".

so we integrate from "b" to "a"

 

[Omium]

Oh don't worry guys, here is the picture and LaTeX solution:

[maths]I_{x}=\int \int_{\Omega}^{}y^{2} \rho dA[/maths]

[maths]I_{x}=\rho \int_{0}^{2\pi } \int_{a}^{b}(r sin\theta )^{2}rdrd\theta [/maths]

[maths]I_{x}=\rho \int_{0}^{2\pi } \int_{a}^{b}(r)^{3}(sin^{2}\theta) drd\theta[/maths]

[maths]I_{x}=\rho \int_{0}^{2\pi } [\frac{r^{4}}{4}]_{a}^{b}(sin^{2}\theta) d\theta [/maths]

[maths]I_{x}=\frac{\rho}{4}(b^{4}-a^{4}) \int_{0}^{2\pi } (\frac{1}{2}-\frac{1}{2}cos2\theta ) d\theta [/maths]

[maths]I_{x}=\frac{\rho}{8}(b^{4}-a^{4}) [\theta - \frac{1}{2}sin2\theta ]_{0}^{2\pi }[/maths]

[maths]I_{x}=\frac{\rho\pi }{4}(b^{4}-a^{4})[/maths]

I realised that rather than be a narutard and go ask the internet I realised I was better off solving the question myself.
No shit about that.
Thank you for your time and I really appreciate this.

Ah i got it right.

That way seems a little complex.

 

[Uncle]

Not to sure how to go about this but.

Consider a circle or arbitrary radius 'r'

Moment of inertial by defintion
I = r^2 dm

Constant surface density p

p = dm/dA.

pdA = dm.

so subbing this in

I = p dA r ^2

call this (1)


p is constant so it can be taken out of the integral.

Since we have a circle

A = Pi r^2

So differentiating dA/dr = 2 Pi r

so dA = 2 Pi r dr.

subbing this into (1)

I = p 2 Pi r^3 dr.

Integrate this with respect to r.

and we are given the radius of the circles is "b" and "a".

so we integrate from "b" to "a"

That is called mass moment of inertia.
It would be better to derive formulas for moments of inertia of any shape using the similar method I used.

Ah i got it right

That way seems a little complex.

It's just using double integration then shifting from Cartesian coordinates to polar coordinates to make life easier when encountering circular shapes.
You will hopefully learn this in Several Variable Calculus.

 

[Omium]

That is called mass moment of inertia.
It would be better to derive formulas for moments of inertia of any shape using the similar method I used.



It's just using double integration then shifting from Cartesian coordinates to polar coordinates to make life easier when encountering circular shapes.
You will hopefully learn this in Several Variable Calculus.

Ahhh.

We just started double integrals.

Kewl.

 

[Sir-BigBoyJames]

that same question was in my final statics exam

 

[lolokay]

finding I about centre;


the I about just the x-axis will just be half of this.


how come it's perfectly fine to ask for help and how to answer a question in the hsc forums, but not in the uni ones..? (yeah I know you should try and be more independent, but surely some help is fine?)

 

[Sir-BigBoyJames]

finding I about centre;


the I about just the x-axis will just be half of this.


how come it's perfectly fine to ask for help and how to answer a question in the hsc forums, but not in the uni ones..? (yeah I know you should try and be more independent, but surely some help is fine?)

ummm because,

 

[lolokay]

k

 

[Omium]

tbh i couldn't care less if you havent attempted questions, ill help you regardless.

However you'll usually get flamed if you've made no attempt at a question and simply want a solution.

If you at least try, everybody will help.

 

[tommykins]

finding I about centre;


the I about just the x-axis will just be half of this.


how come it's perfectly fine to ask for help and how to answer a question in the hsc forums, but not in the uni ones..? (yeah I know you should try and be more independent, but surely some help is fine?)

help is always there, but only if you've shown at least a decent attempt at the question

 

[Joel8945]

Oh don't worry guys, here is the picture and LaTeX solution:

[maths]I_{x}=\int \int_{\Omega}^{}y^{2} \rho dA[/maths]

[maths]I_{x}=\rho \int_{0}^{2\pi } \int_{a}^{b}(r sin\theta )^{2}rdrd\theta [/maths]

[maths]I_{x}=\rho \int_{0}^{2\pi } \int_{a}^{b}(r)^{3}(sin^{2}\theta) drd\theta[/maths]

[maths]I_{x}=\rho \int_{0}^{2\pi } [\frac{r^{4}}{4}]_{a}^{b}(sin^{2}\theta) d\theta [/maths]

[maths]I_{x}=\frac{\rho}{4}(b^{4}-a^{4}) \int_{0}^{2\pi } (\frac{1}{2}-\frac{1}{2}cos2\theta ) d\theta [/maths]

[maths]I_{x}=\frac{\rho}{8}(b^{4}-a^{4}) [\theta - \frac{1}{2}sin2\theta ]_{0}^{2\pi }[/maths]

[maths]I_{x}=\frac{\rho\pi }{4}(b^{4}-a^{4})[/maths]

I realised that rather than be a narutard and go ask the internet I realised I was better off solving the question myself.
No shit about that.
Thank you for your time and I really appreciate this.

The only double integrals I can solve are ones where the domain is rectangular i.e. R = [x_1,x_2] x [y_1,y_2]