Saturday, October 31, 2015

Episode 076: Halloween 2015 pt 3, 9 Biggest Unsolved Mysteries in Physics!





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I came across this incredible article from livescience.com entitled "The 9 Biggest Mysteries in Physics"(click for link), as I was infosnacking on Twitter. It's a great article and really embodies where Physics is headed in the next 25 years, if we can even answer one of these questions with any degree of certainty. In this episode we discuss #3-1 in some detail, since as budding physicists, this are the monster questions you have to grapple with. I also realize that maybe the order that I've been passively heckling them about for the last 2 episodes may actually be decent. Good luck!



If you're feeling like supporting the show, grab the Exploring Physics: Force app by clicking below. It has a step-by-step for doing any Free Body Diagram problem, examples, flash cards, and much more!



Friday, October 30, 2015

Episode 075: Halloween 2015 pt 2, 9 Biggest Unsolved Mysteries in Physics!





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I came across this incredible article from livescience.com entitled "The 9 Biggest Mysteries in Physics"(click for link), as I was infosnacking on Twitter. It's a great article and really embodies where Physics is headed in the next 25 years, if we can even answer one of these questions with any degree of certainty. In this episode we discuss #6-4 in some detail, since as budding physicists, this are the monster questions you have to grapple with. Good luck!



If you're feeling like supporting the show, grab the Exploring Physics: Force app by clicking below. It has a step-by-step for doing any Free Body Diagram problem, examples, flash cards, and much more!



Episode 074: Halloween 2015 pt 1, 9 Biggest Unsolved Mysteries in Physics!





Download this episode (right click and save)







I came across this incredible article from livescience.com entitled "The 9 Biggest Mysteries in Physics"(click for link), as I was infosnacking on Twitter. It's a great article and really embodies where Physics is headed in the next 25 years, if we can even answer one of these questions with any degree of certainty. In this episode we discuss #9-7 in some detail, since as budding physicists, this are the monster questions you have to grapple with. Good luck!



If you're feeling like supporting the show, grab the Exploring Physics: Force app by clicking below. It has a step-by-step for doing any Free Body Diagram problem, examples, flash cards, and much more!



Wednesday, October 28, 2015

Episode 073: Dot And Cross Product Examples Pt. 1





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Question: A strapping young man exerted a force of [10N,10N,10N] moving a box from position (1,2,3) to position (4,5,6) on a 3-D cartesian axis, where all positions are defined in meters. What is the total work done? Answer: Work is defined as follows:

W = F ⋅ d

where W is the work (scalar), F is the force vector, and d is the displacement vector. In the problem, the displacement is not exactly given to us. For that, we need to know the following relationship:

d = xf - xi

where xf is the final position and xi is the initial position. First, since we know the initial and final displacements, let’s calculate the displacement vector:

d = (4,5,6) - (1,2,3)

d = [4-1,5-2,6-3]

d = [3,3,3]

So, our displacement vector is [3,3,3]. Notice in the vector subtraction that each component of the position was subtracted individually, and all three directions combined then give the final displacement vector. With both vectors in hand, we can now calculate the work:

W = F ⋅ d

= [10N,10N,10N] ⋅ [3m,3m,3m] (1)

= 10N*3m + 10N*3m + 10N*3m (3)

W = 30J + 30J + 30J (4)

W = 90J (5)

Where J stands for Joule, which is defined to be a N*m. Notice that the work could be the exact same if say the x-component of the dot product was 10 Joules more and the z-component 10 J less. Were you able to catch the steps in there? Each equation has a number in parentheses to the far right. These are the steps of dot product calculation. Notice that step (2) is missing. In the problem, we know that the coordinates are 3-dimensional. As you grow more comfortable with dot products, step (2) becomes more of a mental note than something that needs to be officially addressed.

2-D Cross Product Question: For the following vectors c = [2,5] and d = [-4,9] calculate c x d and calculate d x c. First, let’s calculate c x d:

Step 1: Identify Two Vectors in Coordinate System as [ax,ay] & [bx,by]- These are given in the problem as [2,5] and [-4,9]. For later reference, let’s identify each component:

ax = 2

ay = 5

bx = -4

by = 9

Step 2: Multiply ax and by- Here we can easily reference step #1 and perform the following operation:

ax * by = 2 * 9 = 18

Step 3: Multiply ay and bx- Here again we can easily reference step #1 and perform the following operation:

ay * bx = 5 * -4 = -20

Step 4: Subtract step 2 from step 3, i.e. ax*by-ay*bx:

18 - -20 = 38

Step 5: Place a k-hat at the end and box your answer, you’re done! So our k-hat represents that our vector is pointing entirely in the z-direction. So our answer could be written 38*k-hat. Alternatively, we can write this

c x d = [0,0,38].

Next, let’s calculate the opposite, d x c.

Step 1: Identify Two Vectors in Coordinate System as [ax,ay] & [bx,by]- These are given in the problem as [-4,9] and [2,5]. Notice that these are OPPOSITE from last time. For later reference, let’s identify each component:

ax = -4

ay = 9

bx = 2

by = 5

Step 2: Multiply ax and by- Here we can easily reference step #1 and perform the following operation:

ax * by = -4 * 5 = -20

Step 3: Multiply ay and bx- Here again we can easily reference step #1 and perform the following operation:

ay * bx =9 * 2 = 18

Step 4: Subtract step 2 from step 3, i.e. ax*by-ay*bx:

-20 - 18 = -38

Step 5: Place a k-hat at the end and box your answer, you’re done! So our k-hat represents that our vector is pointing entirely in the z-direction. So our answer could be written -38*k-hat. Alternatively, we can write this

d x c = [0,0,-38].

What do you notice between these two different operations?

What is different?

The vector is pointing in the opposite direction (change of sign)

What is similar?

The magnitude.

The entire vector is directed in the z-direction (i.e. x and y components are 0)

What can we draw from this?

One thing that we can notice, is that

c x d = -(d x c)

Is it possible to prove this?

Wednesday, October 21, 2015

Episode 072: The Right Hand Rule.





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Before we dive into the examples, there is one last item. Throughout the coordinate systems and vector properties/components sections, have you wondered how the x, y, and z-coordinates are related? They are related by something known as the right hand rule. With your right hand, point your fingers flat out. If you were to point your fingers towards the along the x-axis, and then curl your fingers towards the y-axis, your thumb will point in the direction of the z-axis. This is something that you will see all the time if you study electricity and magnetism. Current, electric, and magnetic fields all relate in direction by the right hand rule.

Cross products follow the right hand rule: i-hat cross j-hat is k-hat. Of course, i-hat is the unit vector pointing in the x-direction, j-hat aligns with the y-direction and k-hat the z-direction. Following the axes, j-hat cross k-hat is i-hat. You could use your right hand rule to prove this.

Here is a trick: Think of the letters ijkijk. If you run to the right, any combination of three letters yields a positive cross product, i x j = k, j x k = i, k x i = j. If you run the letters to the left, any combination of three letters yields a negative cross product: k x j = -i, j x i = -k, and so on. This can give you a general idea of which direction your cross product vector is pointing. Now, on with the examples.

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Episode 071: The Cross Product in 2 & 3 Dimensions!





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Like the dot product, the vector cross product is another means of vector multiplication. The difference with the cross product being that the operation yields a vector quantity. To be honest, I had some hesitation including this in the bootcamp. It’s one of those things I honestly don’t think you’ll use incredibly often. However, if needed, it’s a very handy tool, and if you’re in a class where you do need it, it’s best to have seen it at least once before. So, without further ado, here are the 5 steps to performing the cross product operation in 2 Dimensions:

1) Identify Two Vectors in Coordinate System as [ax,ay] & [bx,by]

2) Multiply ax and by

3) Multiply ay and bx

4) Subtract step 2 from step 3, i.e. ax*by-ay*bx

5) Place a k-hat at the end and box your answer, you’re done!

What the heck k-hat? In 2-dimensions, the unit vectors which point in the “x” and “y” directions are called i-hat and j-hat. k-hat points along the z-direction in a 3-D coordinate system and has a length of 1 unit. This gives the vector its direction. When given two vectors, the cross product vector will always point perpendicular to both vectors. More on this shortly. But, it is possible to write your cross product vector as [0,0,ax*by-ay*bx]. So, the cross product kind of forces you to be in 3 dimensions whether you like it or not.

So if that’s the case, how does it work in 3 dimensions? Different steps, and 6 steps this time:

1) Identify Two Vectors in Coordinate System as [ax,ay,az] & [bx,by,bz]

2) Calculate [ay*bz-by*az]*i-hat

3) Calculate [az*bx-ax*bz]*j-hat

4) Calculate [ax*by-ay*bx]*k-hat

5) Write your vector as a vector consisting of the last 3 steps: [step2,step3,step4]

6) Box your answer, you’re done!

So, these are the steps to follow when calculating a cross product in 2 or 3 dimensions. Actually performing examples will make this process clearer and more fluid.

Monday, October 19, 2015

Episode 070: The Determinant.





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This week, I decided to go back to a topic that was in-progress and sort of went on hold: vector operations. On the heels of the dot product, we're at the point where it's time to tackle the big one: the cross product. But first, there is a small bit of back story that is worth telling in order for the cross product to make sense: The Determinant.

So without going into too much painful detail, there is a wonderful subsection of mathematics called Elementary Linear Algebra, which is pretty much entirely dedicated to matrix operations. Fascinating course. So a matrix, since we haven't already covered it in any detail here on the blog/podcast you can think of as a square or rectangular array of numbers. Below is an example of a 2x2 matrix (pronounced 2 by 2):

You can put anything in these slots 1,2,3,4. Numbers, variables, trig functions, anything mathematical that you can think of. However, our focus right now is how to calculate the determinant. You can do this by multiplying the items in slots 1 and 4, and subtracting the product of slots 3 and 2.

Now, you can in the same way visualize a 3x3 matrix by thinking of each entry as one of the numbers on your phone.

In slots 1,2, &3, we will be putting our i-hat, j-hat, and k-hat unit vectors. To perform the first determinant, which will be for the i-hat vector, we will think of our lowest right-hand corner 2x2 block to be a 2x2 matrix, for which we will be calculating the determinant:

When you calculate the i-hat component of your determinant, we will draw one line horizontal, and one line vertically which will "cancel" out or show us which values we will not be using to calculate our determinant. This leaves (5*9-8*7)i-hat.

For the k-hat component, we perform roughly the same operation. The two pink lines show which components of the matrix we will not be using, and we calculate the determinant of the 2x2 matrix which remains: (4*8-7*5)k-hat.

Wait! You forgot about j-hat! What gives?? i-hat and k-hat are very similar in that they give "nice" 2x2 matrices to calculate determinants of. When we draw our pink lines again for this example:

We have a small problem. We have to cross over the pink lines! There is not an easy 2x2 matrix to calculate the determinant of! Well, if you were to move reproduce the i-hat column to the right of the k-hat column, there would be a nice 2x2 matrix to calculate. However, so as not to move the matrix around, there is a rule: when you cross the pink line for j-hat, you need to multiply the whole thing by -1: this gives -(4*9-7*6)j-hat, or (7*6-4*9) j-hat.

Altogether, we now have: (5*9-8*7)i-hat+(7*6-4*9)j-hat+(4*8-7*5)k-hat. Written as a vector where i-hat, j-hat and k-hat correspond to x,y,z coordinates we have [(5*9-8*7),(7*6-4*9),(4*8-7*5)]. Bear in mind, that these are not the numbers 1,2,3,4,5,6,7,8 & 9. They correspond to whatever variables, constants, or trig functions that rest in these spaces. More on this later. This whole operation may be marginally clumsy and a little confusing for newcomers, but stay tuned for the cross product step-by-step. The steps will be very similar, and this step in the journey will make the cross product more familiar, and seriously clarify its origins.