Moment of inertia of T bar about 3 axes

In summary, the conversation discusses the use of an equation to calculate the centre of mass for a 3D object. The equation in question is I = Icm + md^2, where d is the distance from the centre of mass to the axis. However, there is confusion about which axis to use and the correct result for the centre of mass. The conversation also touches on the use of Latex formatting for equations and the presence of multiple axes of symmetry for the object in question.
  • #1
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Homework Statement
A bar of length 10cm and diameter 2cm is positioned vertically, it has another bar of length 5cm and diameter 2cm attached at a right angle to its middle. The density is 800 kg m^-3. Find the position of the centre of mass. Then find the moments of inertia about 3 perpendicular axis through its centre of mass.
Relevant Equations
Xcm = m1x1 + m2x2 / m1 + m2

I = mr^2/2 (for a bar)
Using the equation above I get Xcm = 0.022 m. I set the origin be at the left of the vertical rod parallel to its centre of mass as in the diagram. But I’m not sure if the equation is correct for 3d.

FE34595A-4E91-4D8E-B601-ED7ADCC50F24.jpeg

for the moments of inertia I am using

I = Icm + md^2
= (mr^2)/2 + md^2

where d is the distance from the centre of mass to the axis. Does this equation seem correct?
 
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  • #2
so_gr_lo said:
I = Icm + md^2
= (mr^2)/2 + md^2

where d is the distance from the centre of mass to the axis. Does this equation seem correct?
I don't believe so. Please elaborate on your calculations. Also, figure out how to format your mathematics using Latex so it is easily read.

##\bar{x}## looks good.

See LaTeX Guide
 
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  • #3
so_gr_lo said:
I = mr^2/2 (for a bar)
About what axis?
 
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  • #4
haruspex said:
About what axis?
Aha! good point.
 
  • #5
haruspex said:
About what axis?
thats the result for a cylinder about its symmetry axis, not sure if that’s the axis I’m supposed to use
 
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  • #6
so_gr_lo said:
thats the result for a cylinder about its symmetry axis, not sure if that’s the axis I’m supposed to use
It's not a "which one" its multiple.
 
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  • #7
so_gr_lo said:
thats the result for a cylinder about its symmetry axis, not sure if that’s the axis I’m supposed to use
I assume you mean its axis of rotational symmetry. There are also two axes of reflective symmetry normal to that one and to each other.
If the rotation is about the axis of rotational symmetry for one, which of its axes is the other rotating about?
 
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  • #8
The diagram shows one axis not crossing the center of mass.
Locating that center should be the first step.
 
  • #9
Lnewqban said:
The diagram shows one axis not crossing the center of mass.
Locating that center should be the first step.
The question specifically asks for the centre of mass and the OP found it correctly in post #1.
The axes the solver chooses for the purpose of analysis need not be those about which the MoI is requested.
 
  • #10
haruspex said:
The question specifically asks for the centre of mass and the OP found it correctly in post #1.
The axes the solver chooses for the purpose of analysis need not be those about which the MoI is requested.
It seems to me that the shown result Xcm = 0.022 m in post #1 is not numerically correct.
Also, according to the shown disposition of the axes in the diagram, the center of mass should not be spatially located on the x axis.

Could you please, explain your following statement a little more for me?:
“There are also two axes of reflective symmetry normal to that one and to each other.“
Sorry, I can only see one.
 
  • #11
Lnewqban said:
It seems to me that the shown result Xcm = 0.022 m in post #1 is not numerically correct.
Also, according to the shown disposition of the axes in the diagram, the center of mass should not be spatially located on the x axis.

Could you please, explain your following statement a little more for me?:
“There are also two axes of reflective symmetry normal to that one and to each other.“
Sorry, I can only see one.
Oh I see. ##x_{cm}## is not the proper label. Per the diagram they found ##y_{cm}##. I think we are assuming ##x_{cm} = 0 = z_{cm}##.
 
  • #12
erobz said:
Oh I see. ##x_{cm}## is not the proper label. Per the diagram they found ##y_{cm}##. I think we are assuming ##x_{cm} = 0 = z_{cm}##.
Yes.
I may be wrong, but I am getting ##y=1.83~cm## as the location for CM measured from the left surface of the 10 cm vertical rod (plane x-z).
 
  • #13
Lnewqban said:
Yes.
I may be wrong, but I am getting ##y=1.83~cm## as the location for CM measured from the left surface of the 10 cm vertical rod (plane x-z).
Do you mean ##18.3 \rm{cm}##? I got ## y_{cm} = 21.7 \rm{cm}## How are you assuming they are connected ( I'm assuming they are basically set next to each other and welded)?
 
  • #14
Lnewqban said:
Could you please, explain your following statement a little more for me?:
“There are also two axes of reflective symmetry normal to that one and to each other.“
I should have written "a plane of rotational symmetry and two planes of reflective symmetry ".
Lnewqban said:
It seems to me that the shown result Xcm = 0.022 m in post #1
I had not noticed @so_gr_lo's confusion between the x and y axes.
 
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  • #15
erobz said:
Do you mean ##18.3 \rm{cm}##? I got ## y_{cm} = 21.7 \rm{cm}## How are you assuming they are connected ( I'm assuming they are basically set next to each other and welded)?
It is not entirely clear whether the cross section is circular or square. It says "diameter", but I would have called them rods if circular; bar suggests square. And welding them together would be tricky if rods.
 
  • #16
erobz said:
Do you mean ##18.3 \rm{cm}##? I got ## y_{cm} = 21.7 \rm{cm}## How are you assuming they are connected ( I'm assuming they are basically set next to each other and welded)?
As I understand it (again, perhaps wrongly), the biggest dimensions of the assembled “T” are 7 cm and 10 cm.

D0921E8C-90C3-40BA-BF2D-2BFC1427E3D8.jpeg
 
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  • #17
haruspex said:
It is not entirely clear whether the cross section is circular or square. It says "diameter", but I would have called them rods if circular; bar suggests square. And welding them together would be tricky if rods.
I assume circular rods (as in cylinders).

And welding them together would be tricky if rods

I worked in a steel mill... believe me when I say...It can be done quite easily!I think another interpretation is that this is a solid formed piece where the end of the horizontal rod is inserted into the vertical rod up to its midpoint.
 
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  • #18
Lnewqban said:
As I understand it (again, perhaps wrongly), the biggest dimensions of the assembled “T” are 7 cm and 10 cm.
No, that's how I assumed it too. Maybe I made a computational mistake.
 
  • #19
erobz said:
No, that's how I assumed it too. Maybe I made a computational mistake.
The mass of the vertical rod is two times the mass of the horizontal one.
 
  • #20
So, ignoring the weld. Here is what I did:

$$ y_{cm} = \frac{\rho \frac{\pi D^2}{4} \cdot 10 \cdot 1 + \rho \frac{\pi D^2}{4} \cdot 5 \cdot ( 2 + 2.5) }{ \rho \frac{\pi D^2}{4} ( 10 + 5 ) } = \frac{10+5\cdot(2+2.5)}{15} \approx 2.17 \rm{cm}$$

I did make a mistake. It was ##2.17 \rm{cm}## not ##21.7 \rm{cm}##. We still get different results as you are getting ##1.83 \rm{cm}## ?
 
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  • #21
haruspex said:
I should have written "a plane of rotational symmetry and two planes of reflective symmetry ".
Thank you much, @haruspex
 
  • #22
erobz said:
So, ignoring the weld. Here is what I did:

$$ y_{cm} = \frac{\rho \frac{\pi D^2}{4} \cdot 10 \cdot 1 + \rho \frac{\pi D^2}{4} \cdot 5 \cdot ( 2 + 2.5) }{ \rho \frac{\pi D^2}{4} ( 10 + 5 ) } = \frac{10+5\cdot(2+2.5)}{15} \approx 2.17 \rm{cm}$$

I did make a mistake. It was ##2.17 \rm{cm}## not ##21.7 \rm{cm}##. We still get different results as you are getting ##1.83 \rm{cm}## ?
Your calculation is correct; mine was not.
Please, see attached scaled diagram (PDF format below).
Thank you very much, @erobz
Bars COM.jpg
 

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