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- Thread starter ChiralSuperfields
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- #36

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- #37

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##a^2 + a^2 = R^2##

##2a^2 = R^2##

##\implies a = \frac{R}{\sqrt{2}} = \frac{\sqrt{2}}{2} R##

Just carless algebra...late at night for me.

- #38

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No worries! We all make algebra mistakes! Thank you for your reply @erobz!##x^2 + x^2 = R^2##

##2x^2 = R^2##

##\implies x = \frac{R}{\sqrt{2}} = \frac{\sqrt{2}}{2} R##

Just carless algebra...late at night for me.

So, I'm just trying to understand how you got ## y\hat j = x\hat j##. I have drawn a grey square below.

I'm trying not to use circular reasoning, but did you get ## y\hat j = x\hat j## because ##\vec r## will always form a right isosceles triangle since it is a square so ##\theta = 45##?

Many thanks!

- #39

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- #40

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Thank you for your reply @haruspex!

I was trying to consider cases where ##\vec r ≠ R## for when the position vector is pointing somewhere along to the mass element on the sides not to the corners.

When ##\vec r = R##, I can see that since it square then each side of the right triangle will have equal sides of length ##x##.

I think the range of x for that case would be from ##0## at the midpoint to ##\frac {R}{\sqrt{2}}## at the centre of one side.

Dose this special case hold for ##\vec r ≠ R##?

Many thanks!

- #41

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Why do you want to solve this using vectors? It's straightforward using scalars and the parallel axis theorem.Thank you for your reply @haruspex!

I was trying to consider cases where ##\vec r ≠ R## for when the position vector is pointing somewhere along to the mass element on the sides not to the corners.

When ##\vec r = R##, I can see that since it square then each side of the right triangle will have equal sides of length ##x##.

I think the range of x for that case would be from ##0## at the midpoint to ##\frac {R}{\sqrt{2}}## at the centre of one side.

Dose this special case hold for ##\vec r ≠ R##?

Many thanks!

Edit: rereading your post, I don’t understand it. How are you defining ##\vec r## and x?

- #42

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- #43

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The vector ## \vec{r}## varies, in both magnitude and direction, but for that horizontal segment It will always have the form:No worries! We all make algebra mistakes! Thank you for your reply @erobz!

So, I'm just trying to understand how you got ## y\hat j = x\hat j##. I have drawn a grey square below.

View attachment 322331

I'm trying not to use circular reasoning, but did you get ## y\hat j = x\hat j## because ##\vec r## will always form a right isosceles triangle since it is a square so ##\theta = 45##?

Many thanks!

$$ \vec{r} = x ~{\hat i} + \frac{\sqrt{2}}{2} R ~{\hat j}$$

- #44

- #45

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Thank you for your reply @haruspex !Why do you want to solve this using vectors? It's straightforward using scalars and the parallel axis theorem.

Edit: rereading your post, I don’t understand it. How are you defining ##\vec r## and x?

Sorry, let me try to explain it better.

##\vec r_1 = \sqrt {x^2 + y_1^2} ## since it has a ##x\hat i## and ##y_1\hat j## component. However since ##y_1 = x## in this special case then ##\vec r_1 = \sqrt {2x^2} = R ##

For this other vector their y-component is not ##x##

##\vec r_2 = \sqrt {x^2 + y_2^2} ≠ R## (cannot equal R since ##\vec r_2 < \vec r_1 ##)

##\vec r_3 = \sqrt {x^2 + y_3^2} ≠ R##

So what I was saying is that we should consider the general case where,

##\vec r = \sqrt {x^2 + y^2} ##

Where x is a constant equal to ##\frac {\sqrt {2}}{2}R## and y is a variable position vector component in the y-direction.

However, if we choose a different coordinate system, such as what @erobz has done in post #43

Then now y a constant equal to ##\frac {\sqrt {2}}{2}R## and x is variable position vector component in the y-direction.

I think the confusing thing about this problem is choosing the right side to find the moment of inertia for and defining the right coordinate system. What do you think?

How would you solve this using the parallel axis theorem out of curiosity?

Many thanks!

Last edited:

- #46

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- #47

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Thank you @erobz!

I think I now understand where you get that constant from now. So basically, you assume that both sides of the triangle are equal which means that angle must be 45 degrees.

So ##\sin45 = x/R## which gives your result.

Is that how you did it?

Many thanks!

- #48

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Thank you for that @erobz!The vector ## \vec{r}## varies, in both magnitude and direction, but for that horizontal segment It will always have the form:

$$ \vec{r} = x ~{\hat i} + \frac{\sqrt{2}}{2} R ~{\hat j}$$

- #49

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Yes, that will always be the case for a square.Thank you @erobz!

I think I now understand where you get that constant from now. So basically, you assume that both sides of the triangle are equal which means that angle must be 45 degrees.

View attachment 322347

So ##\sin45 = x/R## which gives your result.

Is that how you did it?

Many thanks!

- #50

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For one side of the square, the moment of inertia relative to the center is just $$\int_{-L/2}^{L/2} [(L/2)^2+x^2]\lambda dx =\lambda L (L/2)^2+\lambda \frac{x^3}{3}\Big|_{-L/2}^{L/2} =\lambda L (L/2)^2+\lambda \frac{2L^3}{3\times 8}$$

I considered the x axis parallel to the side I am looking at.

Using ##M=\lambda L## you get

$$I_{side} =m\frac{L^2}{4}+m \frac{L^2}{12}$$

The last term is the moment of inertia for an axis going through the center of the bar (look it up in a table) and the first term is what you add by using the parallel axis theorem. You could write this right away, without any integral.

Then the monemt of the square is just ##I_{square}=4I_{side}##.

Only now is time to see what happens if ##L=R\sqrt{2}##.

- #51

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Thank you for your reply @erobz !Yes, that will always be the case for a square.

- #52

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Thank you very much for that @nasu!

For one side of the square, the moment of inertia relative to the center is just $$\int_{-L/2}^{L/2} [(L/2)^2+x^2]\lambda dx =\lambda L (L/2)^2+\lambda \frac{x^3}{3}\Big|_{-L/2}^{L/2} =\lambda L (L/2)^2+\lambda \frac{2L^3}{3\times 8}$$

I considered the x axis parallel to the side I am looking at.

Using ##M=\lambda L## you get

$$I_{side} =m\frac{L^2}{4}+m \frac{L^2}{12}$$

The last term is the moment of inertia for an axis going through the center of the bar (look it up in a table) and the first term is what you add by using the parallel axis theorem. You could write this right away, without any integral.

Then the monemt of the square is just ##I_{square}=4I_{side}##.

Only now is time to see what happens if ##L=R\sqrt{2}##.

- #53

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##I_{side}= \frac {2mR^2}{4} + \frac{2mR^2}{12} ##

##I_{side} = \frac {mR^2}{2} + \frac{mR^2}{6} ##

##I_{square} = 4[\frac {mR^2}{2} + \frac{mR^2}{6}]##

##I_{square} = 2mR^2 + \frac{2mR^2}{3}##

##I_{square} = \frac{8mR^2}{3}##

Many thanks!

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