Math in Maxwell needle method- the formula the moment of inertia

[Outrageous]

1. Homework Statement [
How do I get I1( the moment of inertia of the picture b) ?is just a math problem but really no idea.

Homework Equations

I=mr^2

The Attempt at a Solution

maybe centre of mass .

Guide will do . please help , thank

 

[haruspex]

I cannot work out what your picture is showing, and there's no definition of the variables in the second attachment.
Edit: Just noticed there's a tiny 3rd attachment explaining the variables...

 

[haruspex]

What part of the equation do you not understand? Is it the 3L/8?

 

[Outrageous]

yes , exactly that part.

 

[haruspex]

It's just the parallel axis theorem. The centre of mass of the cylinder is 3L/8 from the axis of rotation.

 

[Outrageous]

Thank you. One more to ask,
I know that moment of inertia I=mr^2
For a system, I= m(r1)^2 + m(r2)^2 + m(r3)^2 +...
or we can write I=Ʃmr^2
where the m= mass of a particle , r1 and r2 are not the same value.
This is the only thing that I understand but I=∫r^2 dm=mr^2 from this link
http://easycalculation.com/theorems/parallel-axis-moment-of-inertia.php

How can I=∫r^2 dm=mr^2 ?
here mean the r is constant ? m is the total mass?

Thank you.

 

[haruspex]

That link is saying that if the MoI about the CoM is I_c = ∫x².dm, where x varies, and you want the MoI I_r about a an axis displaced by an amount r then Ir = ∫(x+r)².dm = ∫(x²+2xr+r²).dm = ∫x².dm+2∫xr.dm+∫r².dm = I_c + 2∫xr.dm + ∫r².dm
Since r is a constant, I_r = I_c + 2r∫x.dm + r²∫.dm
Since x is displacement from centre of mass, by definition ∫x.dm = 0, and ∫.dm is just the total mass, M:
I_r = I_c + r²M

 

[Outrageous]

That link is saying that if the MoI about the CoM is I_c = ∫x².dm, where x varies, and you want the MoI I_r about a an axis displaced by an amount r then Ir = ∫(x+r)².dm = ∫(x²+2xr+r²).dm = ∫x².dm+2∫xr.dm+∫r².dm = I_c + 2∫xr.dm + ∫r².dm
Since r is a constant, I_r = I_c + 2r∫x.dm + r²∫.dm
Since x is displacement from centre of mass, by definition ∫x.dm = 0, and ∫.dm is just the total mass, M:
I_r = I_c + r²M

I_c = ∫x².dm is same as
For a system, I= m(r1)^2 + m(r2)^2 + m(r3)^2 +...
or we can write I=Ʃmr^2
where the m= mass of a particle ?
Why I=integral x^2 dm , x varies. Then for I =∫x dm ,it become zero.

 

[haruspex]

I_c = ∫x².dm is same as
For a system, I= m(r1)^2 + m(r2)^2 + m(r3)^2 +...
or we can write I=Ʃmr^2
where the m= mass of a particle ?

Yes. Strictly, I= m₁(r₁)^2 + m₂(r₂)^2 + ...

Why I=integral x^2 dm , x varies. Then for I =∫x dm ,it become zero.

In the analysis, x is defined to be the displacement (in the x coordinate) of the element dm from the centre of mass. By definition of centre of mass, ∫x dm =0.

 

[Outrageous]

By definition of centre of mass, ∫x dm =0.

Really thank you^^