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

[Benjamin_harsh]

Homework Statement:

Find the Moment of inertia of the shaded region w.r.t the line AB.

Relevant Equations:

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

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

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

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

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

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

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

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

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: $\frac {π}{64}d^{4}$

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

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

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

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

$I_{AB} = 341.33 + 100.79 – 12.56 = 429.55 cm^{4}$

 

[Chestermiller]

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.

 

[Benjamin_harsh]

So centroid of any semicircular section is always located at $\frac{4R}{3\pi}$?

 

[Doc Al]

List of centroids

 

[Benjamin_harsh]

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?

 

[Doc Al]

did they use both $I_{XX}$ and $I_{YY}$ values in it?

No. Why would they do that?

 

[Benjamin_harsh]

No. Why would they do that?

So what values have they substitute in this formulae: $I_{G1x}$?

 

[Doc Al]

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 $mD^2$.

 

[Benjamin_harsh]

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.

 

[Doc Al]

Why they took zero as distance between centroid and parallel axis ?

The center of the circle is on the axis.

 

[Benjamin_harsh]

The center of the circle is on the axis.

how can you tell?

 

[Doc Al]

how can you tell?

Just by looking at the diagram.

 

[Benjamin_harsh]

Just by looking at the diagram.

In circles, axis always lies on the centroid?

 

[Benjamin_harsh]

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?

 

[haruspex]

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.
I_XX is the MoI about a line through the centroid and parallel to the X axis. Likewise, I_YY is about a line through the centroid and parallel to the Y axis.
For a lamina, you can find I_ZZ simply by adding these together. This is the perpendicular axis theorem.
In the question in this thread you are only asked for I_XX, so you do not need to find I_YY.

 

[Benjamin_harsh]

In the question in this thread you are only asked for I_XX, so you do not need to find I_YY?

How can you tell that no need to find I_YY?

 

[Doc Al]

How can you tell that no need to find I_YY?

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.

 

[Benjamin_harsh]

So all moments of inertia needed for this problem will be parallel to that axis.

So we need to neglect $I_{YY}$ and consider $I_{XX}$?

 

[haruspex]

So we need to neglect $I_{YY}$ and consider $I_{XX}$?

It's not a matter of neglecting I_YY. 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 I_XX and apply the parallel axis theorem; you do not need to know I_YY.

 

[Benjamin_harsh]

So When should we to need calculate both $I_{XX}$ and $I_{YY}$?

 

[haruspex]

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 I_XX and I_YY and add them together to get I_ZZ.

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 I_ZZ; the result is the same.
E.g. if the centroid is at (x,y) but you want the MoI about the Z axis (I_Z0) then you can do I_X0=I_XX+My², I_Y0=I_YY+Mx², I_Z0=I_X0+I_Y0
or
I_ZZ=I_XX+I_YY, I_Z0=I_ZZ+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 I_ZZ then finding I_XX and I_YY 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 I_XX.

In this thread it definitely only asks for I_XX, so you do not need I_YY.

 

[Benjamin_harsh]

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 I_XX and I_YY and add them together to get I_ZZ.

How can I tell lamina is in the XY plane?

 

[haruspex]

How can I tell lamina is in the XY plane?

I am defining the XY plane as the plane containing the lamina.

 

[Benjamin_harsh]

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 $M$?

 

[haruspex]

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 I_ZZ for the MoI about a line through the centroid in the Z direction and I_Z0 for the MoI about a line through the origin in the Z direction, etc.
M is the mass of the lamina.