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

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.

I=mr^2

## The Attempt at a Solution

maybe centre of mass .

#### Attachments

• 17.7 KB Views: 1,699
• 21.8 KB Views: 967
• 32 KB Views: 675

Related Introductory Physics Homework Help News on Phys.org
haruspex
Homework Helper
Gold Member
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
Homework Helper
Gold Member
What part of the equation do you not understand? Is it the 3L/8?

yes , exactly that part.

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

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
Homework Helper
Gold Member
That link is saying that if the MoI about the CoM is Ic = ∫x2.dm, where x varies, and you want the MoI Ir 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 = Ic + 2∫xr.dm + ∫r2.dm
Since r is a constant, Ir = Ic + 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:
Ir = Ic + r2M

That link is saying that if the MoI about the CoM is Ic = ∫x2.dm, where x varies, and you want the MoI Ir 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 = Ic + 2∫xr.dm + ∫r2.dm
Since r is a constant, Ir = Ic + 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:
Ir = Ic + r2M
Ic = ∫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
Homework Helper
Gold Member
Ic = ∫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.

By definition of centre of mass, ∫x dm =0.
Really thank you^^