Moment of inertia of rod about an axis inclined

[Raghav Gupta]

Homework Statement

What is the Moment of inertia of rod about an axis inclined?
It is given as $\frac{ML^2sin^2θ}{12}$ but how?

Homework Equations

I know MOI for axis passing through center along y-axis as
$\frac{ML^2}{12}$

The Attempt at a Solution

Shouldn't it be $\frac{ML^2}{12} = Icosθ$
by taking components?

 

[BvU]

Hello Raghav,

What is the axis of rotation in your picture ?

 

[Raghav Gupta]

Hello Raghav,

What is the axis of rotation in your picture ?

Sorry for the diagram,
The horizontal one is the rod.
The axis of rotation is the one which I have denoted by I.

 

[BvU]

The customary choice for $\theta$ is the angle between rod and axis of rotation. Perpendicular $\Rightarrow I = {ML^2/12}$ and $\theta = 0 \ \ \Rightarrow \ \ I = 0\ \$.

You can check by working out $I = \int r^2 dm$

 

[Raghav Gupta]

The customary choice for $\theta$ is the angle between rod and axis of rotation. Perpendicular $\Rightarrow I = {ML^2/12}$ and $\theta = 0 \ \ \Rightarrow \ \ I = 0\ \$.

You can check by working out $I = \int r^2 dm$

Yeah, the angle should be between rod and axis of rotation.
Because if we know the formula ML²sin²θ/12 . Then when θ= 90° the MOI is ML²/12.

Now in the first place how to derive that formula $I = ML^2sin^2θ/12$ ?
I know derivation of axis passing through center and perpendicular to rod.
The derivation goes like this.
Let $dm = \frac{Mdx}{L}$

$$ I = \int_{\frac{-L}{2}}^{\frac{L}{2}} x^2dm $$
Therefore on simplifying and integrating we get I = ML²/12
But how to go when inclined at angle θ?

 

[BvU]

You work out

$I = \int r^2 dm$

 

[Raghav Gupta]

You work out

View attachment 83690 $I = \int r^2 dm$

So, r = xtanθ and dm = Mdx/L ?

 

[BvU]

Yes and no. You should really do some work and check that yourself, instead of asking what to do !

One of the checks would be that $M = \int dm$ should work out OK !

 

[Raghav Gupta]

Yes and no. You should really do some work and check that yourself, instead of asking what to do !

One of the checks would be that $M = \int dm$ should work out OK !

Okay, getting dm = 2Mdx/L, as the length from centre to one end is half.
So $$ I = \int_0^{\frac{L}{2}} \frac{2x^2tan^2θMdx}{L} $$
But by this the answer coming is wrong.

And I have checked$$ M = \int_0^{\frac{L}{2}} \frac{2Mdx}{L} $$

 

[BvU]

Last check fails miserably! x does not run from 0 to L/2

 

[Raghav Gupta]

Last check fails miserably! x does not run from 0 to L/2

It can run from 0 to L/2.
By all this we would get half the MOI and then we can multiply it by 2.

 

[BvU]

No.

 

[Raghav Gupta]

No.

So are you trying to say from -L/2 to L/2. ?
Now you would say check it, but I am still arriving at wrong answer.

 

[BvU]

Take a ruler and measure half the length of the rod on your screen.
Then measure from 0 to the upper bound of the x integral.

 

[Raghav Gupta]

Take a ruler and measure half the length of the rod on your screen.
Then measure from 0 to the upper bound of the x integral.

So, you are trying to say one is base and other the hypotenuse?
Are you telling me to do trigonometry in a twisted manner?
Your every reply is looking a riddle, Don't know why it is not helping me.

Would take time to go to market and buy a ruler as at these times it is closed. Have not used a ruler from a long time.

 

[BvU]

It can also be made clear with two pencil marks on a piece of paper that x does not run from 0 to L/2. If $\theta=0$ it doesn't run at all !

So, you are trying to say one is base and other the hypotenuse

Yes! $\theta$ plays a role and you need it to get the integrals for M and also for I right !

 

[Raghav Gupta]

It can also be made clear with two pencil marks on a piece of paper that x does not run from 0 to L/2. If $\theta=0$ it doesn't run at all !

Yes! $\theta$ plays a role and you need it to get the integrals for M and also for I right !

Judging from your post 6th diagram cosθ = dm/dx
But I know it is wrong dimensionally.
I am not getting the approach for dm in terms of θ.
Can you tell finally?

 

[Raghav Gupta]

Okay, got it myself. Your diagram was confusing me of taking the triangle forming 90°

dm = Mdx/L
sinθ = r/x
r = xsinθ

$$ I = \int_{\frac{-L}{2}}^{\frac{L}{2}} \frac{x^2sin^2θMdx}{L} $$ Why were you telling not to take -L/2 to L/2 earlier ?
Solving we get the desired result,
I = $\frac{ML^2sin^2θ}{12}$

 

[BvU]

Judging from your post 6th diagram cosθ = dm/dx

By now I have enough faith in you that I start to think it must be me who is so completely unclear.
How can you possibly say this ? Anyone can see that the piece called dm is longer than the piece called dx ! So, if anything at all, then $dx = dm \cos\theta$ and not vice versa !
Of course the dimension is wrong, but that you fix by changing it to mass: multiply by $\rho$ where $\rho$ has the dimension of mass/length. In short, $\rho$ is M/L and then $\displaystyle dm = \rho \; dl = {M\over L} dl = {M\over L} {dx\over \cos\theta}$.
And if $l$ runs from 0 to L/2 in the integration, then $x$ runs from 0 to ... ?

Why were you telling not to take -L/2 to L/2 earlier

because it is not correct $x$ does not run from 0 to L/2 or from -L/2 to L/2.

sinθ = r/x

Shame on you ! You had the correct " $r = x \tan\theta$ " in post #7 !

Your question was

So, r = xtanθ and dm = Mdx/L ?

and the answer in post 8 was and still is

Yes and no.

 

[Raghav Gupta]

That was my big mistake.
Yes you are correct and clear. I also have faith in you and I admit cosθ = dm/dx was my typo.
Actually I typed it fast as it was looking dimensionally incorrect that how we are dividing length and mass.
But you told me the reason in above post by converting dm to length part.

Okay x should run from 0 to Lcosθ.

So $$ I = \int_0^{Lcosθ} \frac{x^2tan^2θMdx}{Lcosθ} $$ -----1)

And I have checked$$ M = \int_0^{Lcosθ} \frac{Mdx}{Lcosθ} $$

Integrating ---1) I am getting,

$I = \frac{ML^2sin^2θ}{3}$
But I am supposed to get 12 in denominator.

 

[BvU]

Okay x should run from 0 to Lcosθ

No. From -Lcosθ/2 to Lcosθ/2 .

 

[Raghav Gupta]

A twist in a tale!
You were saying in post 19

because it is not correct $x$ does not run from 0 to L/2 or from -L/2 to L/2.

 

[BvU]

Yeah, just had dinner and have to run now. Small oversight, corrected by editing .

And: if you miss a factor of 1/4 you should be able to locate the causes all by yourself ! It's not rocket science (but rod science ).

 

[Raghav Gupta]

And if $l$ runs from 0 to L/2 in the integration, then $x$ runs from 0 to ... ?

You were also saying that x from 0 to ..
Is this also a twist
And then I also have checked, as you were saying earlier$$ M = \int_0^{Lcosθ} \frac{Mdx}{Lcosθ} $$

Though your limits also give same value.
How was I suppose to know that?

 

[Raghav Gupta]

How was I suppose to know the factor of 1/4 if in the first place I don't know the formula?

 

[BvU]

You were also saying that x from 0 to ..
Is this also a twist
And then I also have checked, as you were saying earlier$$ M = \int_0^{Lcosθ} \frac{Mdx}{Lcosθ} $$

Though your limits also give same value.
How was I suppose to know that?

I expected you had already understood that for both integrals you have$$
\int_{-{L\cos\theta\over 2}}^{L\cos\theta\over 2} = 2 \int_0^{L\cos\theta\over 2}
$$

The integral bounds $$ M = \int_0^{Lcosθ} \frac{Mdx}{Lcosθ} $$work without problem for $M$, but not for $I$. For $M$ it's as if the rod is shifted lengthwise over half its length. For $I$ you would also have to change the expression for r.

How was I suppose to know the factor of 1/4 if in the first place I don't know the formula

Which formula did you not know ?
$$M = \int dm \\ I = \int r^2 dm $$or yet another one ?
A factor of 4 is something like $2^2$, or in this case $2 / 2^3$. and then you go hunting for the factors 2.

 

[haruspex]

I hate to interfere with all this lovely integration, but it really isn't necessary.
The contribution of a mass element to the MoI depends only on its distance from the axis. So you can squash the object parallel to the axis, as long as you preserve masses. In this case, you will have a shorter, fatter (or higher density) rod. The mass will not change. What will the new length be?

 

[Raghav Gupta]

Thanks BvU , got it.

I hate to interfere with all this lovely integration, but it really isn't necessary.
The contribution of a mass element to the MoI depends only on its distance from the axis. So you can squash the object parallel to the axis, as long as you preserve masses. In this case, you will have a shorter, fatter (or higher density) rod. The mass will not change. What will the new length be?

I'm not able to imagine the situation that how by squashing rod we will get it parallel to the axis?

 

[Raghav Gupta]

Thanks BvU , got it.

I hate to interfere with all this lovely integration, but it really isn't necessary.
The contribution of a mass element to the MoI depends only on its distance from the axis. So you can squash the object parallel to the axis, as long as you preserve masses. In this case, you will have a shorter, fatter (or higher density) rod. The mass will not change. What will the new length be?

I'm not able to imagine the situation that how by squashing rod we will get it parallel to the axis?

Edit - Sorry, don't know why the post repeated.

 

[haruspex]

I'm not able to imagine the situation that how by squashing rod we will get it parallel to the axis?

No, not squash it to make it parallel to the axis, squash it in a direction parallel to the axis (making it perpendicular to the axis). Each bit of the rod should stay the same distance from the axis in the squashing process.