Where did ##\large \frac {4R}{3π}## come from?

  • Thread starter Benjamin_harsh
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In summary: In post #14 above you linked to a problem which, according to your post #1 in that thread, asked for the MoI about "the centroidal axis".In summary, the authors calculated the moment of inertia about the centroid of a semi-circle section parallel to the base, the centroid of a circle with radius R, and the moment of inertia about the centroid of a semi-circle section parallel to the base. They also calculated the moment of inertia about the center of mass and parallel to the axis of rotation (line AB).
  • #1
Benjamin_harsh
211
5
Homework Statement
Find the Moment of inertia of the shaded region w.r.t the line AB.
Relevant Equations
##\large \frac {4R}{3π} \normalsize = 1.7 cm##
245391

Sol: ##I_{AB} = I_{AB1} + I_{AB2} - I_{AB3}##
245392


##I_{AB1} = I_{G1x} + A_{1}.Y_{2}##

moment of inertia of a triangle,##I_{XX} = \large \frac{bh^3}{36}##

moment of inertia of a triangle,##I_{YY} = \large\frac{hb^3}{36}##

##I_{AB1} = \large \frac {8*8^{3}}{36} + (\frac {1}{2})\normalsize*8*8 (\large\frac {8}{3})^{2} \normalsize = 341.33 cm^{4}##
245395


##I_{AB2} = I_{G2x} + A_{1}.Y_{2}##

##\large \frac {4R}{3π} \normalsize = 1.7 cm##

Moment of inertia of semi circle section parallel to the base, ##I_{AB2} = 0.11*R^{4}##

##I_{AB2} = 0.11*44 + (\large \frac{π*4^4}{2})^{2}\normalsize (1.7)^{2}##

##I_{AB2} = 100.79 cm^{4}##

moment of inertia of a circle w.r.t centroidal X or centroidal Y: ##\frac {π}{64}d^{4}##

245396

##I_{AB3} = I_{G2x} + A_{1}.Y^{2}##

##I_{AB3} = \large\frac{π}{64}\normalsize*4^{4}+(π*2^{2}).0##

##I_{AB3} = 12.56 cm^{4}##

##I_{AB} = I_{AB1} + I_{AB2} - I_{AB3} ##

##I_{AB} = 341.33 + 100.79 – 12.56 = 429.55 cm^{4}##
 
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  • #2
They are starting with the moment of inertia about the centroid of the semicular section, and then applying the parallel axis theorem. The centroid is located ##\frac{4R}{3\pi}## below the line AB.
 
  • #3
So centroid of any semicircular section is always located at ##\frac{4R}{3\pi}##?
 
  • #5
In this equation, ##I_{AB1} = \large \frac {8*8^{3}}{36} + (\frac {1}{2})\normalsize*8*8 (\large\frac {8}{3})^{2} \normalsize = 341.33 cm^{4}## did they use both ##I_{XX}## and ##I_{YY}## values in it?
 
  • #6
Benjamin_harsh said:
did they use both ##I_{XX}## and ##I_{YY}## values in it?
No. Why would they do that?
 
  • #7
Doc Al said:
No. Why would they do that?
So what values have they substitute in this formulae: ##I_{G1x}##?
 
  • #8
In all cases, they are applying the parallel axis theorem. They take the moment of inertia about the center of mass and parallel to the axis of rotation (the line AB) then add ##mD^2##.
 
  • #9
In this equation for full circle, ##I_{AB3} = \large\frac{π}{64}\normalsize*4^{4}+(π*2^{2}).0##

Why they took zero as distance between centroid and parallel axis ? I am failing at measuring the distance between the parallel axis and centroid. .Please explain clearly. :cry:
 
  • #10
Benjamin_harsh said:
Why they took zero as distance between centroid and parallel axis ?
The center of the circle is on the axis.
 
  • #11
Doc Al said:
The center of the circle is on the axis.
how can you tell?
 
  • #12
Benjamin_harsh said:
how can you tell?
Just by looking at the diagram.
 
  • #13
Doc Al said:
Just by looking at the diagram.

In circles, axis always lies on the centroid?
 
  • #14
Why they didn't calculate ##I_{XX}## and ##I_{YY}## for each 3 sections; right angled triangle, semicircle & full circle? In this problem, ##I_{XX}## and ##I_{YY}## calculated for each given 3 sections.

When should we calculate ##I_{XX}## and ##I_{YY}## for each sections?
 
  • #15
Benjamin_harsh said:
Why they didn't calculate ##I_{XX}## and ##I_{YY}## for each 3 sections; right angled triangle, semicircle & full circle? In this problem, ##I_{XX}## and ##I_{YY}## calculated for each given 3 sections.

When should we calculate ##I_{XX}## and ##I_{YY}## for each sections?
You have a lamina in the XY plane.
IXX is the MoI about a line through the centroid and parallel to the X axis. Likewise, IYY is about a line through the centroid and parallel to the Y axis.
For a lamina, you can find IZZ simply by adding these together. This is the perpendicular axis theorem.
In the question in this thread you are only asked for IXX, so you do not need to find IYY.
 
  • #16
haruspex said:
In the question in this thread you are only asked for IXX, so you do not need to find IYY?

How can you tell that no need to find IYY?
 
  • #17
Benjamin_harsh said:
How can you tell that no need to find IYY?
You are asked to find the moment of inertia about the line AB, which is parallel to the x-axis. So all moments of inertia needed for this problem will be parallel to that axis.
 
  • #18
Doc Al said:
So all moments of inertia needed for this problem will be parallel to that axis.

So we need to neglect ##I_{YY}## and consider ##I_{XX}##?
 
  • #19
Benjamin_harsh said:
So we need to neglect ##I_{YY}## and consider ##I_{XX}##?
It's not a matter of neglecting IYY. You are only asked to find the moment about AB. That is a line parallel to the X axis, so you only need to find IXX and apply the parallel axis theorem; you do not need to know IYY.
 
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  • #20
So When should we to need calculate both ##I_{XX}## and ##I_{YY}##?
 
  • #21
Benjamin_harsh said:
So When should we to need calculate both ##I_{XX}## and ##I_{YY}##?
I already explained that in post #15 above. If you have a lamina in the XY plane and you need to find the MoI about an axis in the Z direction then one way to do that is to find IXX and IYY and add them together to get IZZ.

If the axis in the Z direction is not through the centroid then you can use the parallel axis theorem either separately for the XX and YY directions before adding or apply it after finding IZZ; the result is the same.
E.g. if the centroid is at (x,y) but you want the MoI about the Z axis (IZ0) then you can do IX0=IXX+My2, IY0=IYY+Mx2, IZ0=IX0+IY0
or
IZZ=IXX+IYY, IZ0=IZZ+M(x2+y2)

In post #14 above you linked to a problem which, according to your post #1 in that thread, asked for the MoI about "the centroidal axis". As I mentioned in post #12 in that thread that is not well defined. A centroidal axis is any axis through the centroid, not necessarily normal to the plane. If that question does intend IZZ then finding IXX and IYY is a good way. But as I noted, the diagram there does not provide any X coordinates and does not look symmetric, so I still think it may only have been seeking IXX.

In this thread it definitely only asks for IXX, so you do not need IYY.
 
  • #22
haruspex said:
If you have a lamina in the XY plane and you need to find the MoI about an axis in the Z direction then one way to do that is to find IXX and IYY and add them together to get IZZ.

How can I tell lamina is in the XY plane?
 
  • #23
Benjamin_harsh said:
How can I tell lamina is in the XY plane?
I am defining the XY plane as the plane containing the lamina.
 
  • #24
What does ##X## and ##Y## indicates in this equation: ##I_{Z0} = I_{zz} + M(x^{2} + y^{2})##? Are there distance between centroid and the axis?

What value should I put in ##M##?
 
  • #25
Benjamin_harsh said:
What does ##X## and ##Y## indicates in this equation: ##I_{Z0} = I_{zz} + M(x^{2} + y^{2})##?

Are there distance between centroid and the axis? What value should I put in ##M##?
As I wrote, (x,y) are the coordinates of the centroid.
I am using IZZ for the MoI about a line through the centroid in the Z direction and IZ0 for the MoI about a line through the origin in the Z direction, etc.
M is the mass of the lamina.
 

1. Where does the value of 4R/3π come from in scientific calculations?

The value of 4R/3π is derived from the equation for the volume of a sphere, which is V = (4/3)πr^3. In this equation, 4/3 is equivalent to 4R/3π, where R is the radius of the sphere.

2. Why is 4R/3π used in calculations instead of a simpler value?

The value of 4R/3π is used in calculations because it is the most accurate representation of the volume of a sphere. Using a simpler value, such as 1 or 2, would result in less accurate calculations.

3. Can you explain the significance of the number 4R/3π in scientific equations?

The number 4R/3π is significant in scientific equations because it represents the proportion of the volume of a sphere to its radius. This value is crucial in many calculations involving spheres, such as in physics, chemistry, and engineering.

4. How is the value of 4R/3π related to the shape of a sphere?

The value of 4R/3π is directly related to the shape of a sphere, as it represents the volume of a sphere. This value is derived from the fundamental properties of a sphere, such as its radius and the constant pi (π).

5. Is there any other context in which the value of 4R/3π is used?

Aside from its use in calculations involving spheres, the value of 4R/3π also appears in various scientific equations and concepts, such as the ideal gas law and the concept of surface area to volume ratio in biology. It is a fundamental value that has wide-ranging applications in different fields of science.

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