Uni Physics stuff-self directed and independent learning

Hello everyone,

This is just an extra page about physics stuff, and I won't be making a new page every time I update the physics section.

*CHANGE-TO TEACHERS-THIS IS MY SECTION ON SELF DIRECTED AND INDEPENDENT RESEARCH LEARNING. I SHALL BE DOING THIS PART ON PHYSICS INSTEAD OF SEARCHING FOR MANY IRRELEVANT TOPICS AND INSTEAD FOCUS ON ONE PARTICULAR TOPIC. PLEASE MARK WITH YOUR RUBRICS BASED ON THIS PAGE.

Just some clarification: Don't think anything big when you hear uni textbook; its not really that hard (some of the things are actually done in Sec 3, in fact the first chapter is devoted to estimation and the number of important digits, as well as exponentials, which is just a fancy word for expressing things in powers of 10: (calculator format), and I haven't gone that all that far yet.

My father is always telling me: many people think that physics is basically memorising a bunch of formulas. However, my father says that to do physics well one needs to understand the concepts of physics well. Though I do not see the importance of this now, I hope to in the future.

A few apologies, as this is all done in blogging format, I will be unable to insert some symbols and might need words. If you feel something is wrong with the equations or you don't understand some things, just post it in the comments.

I will probably be arranging my things in chapters.

Kinematics in one direction(=speed and acceleration)
14/1/14:
When I started my physics textbook, this is the first chapter with any real math in it. So it was quite daunting at first. The first stuff is basically the rate and speed stuff you do in primary school. The only new thing is that velocity instead of speed is used. This means that it also includes the direction and at least one of the directions must be stated as positive and negative (which makes the other the opposite).

Basically, 1D kinematics is movement in one dimension (forward and backward). There are some important equations for 1D kinematics:

v=v0+at
x=x0+v0t+1/2a(t)(square)
vsquare=(v0)square+2a(x-x0)
v(average)=(v+v0)/2

All these equations require a to be constant. For each of these equations, v is velocity, x is distance, a is acceleration and t is time. (The 0s are meant to be subscripts but I can't do that in Blogger). v without a subscript basically means that it is the velocity of an object at a designated point of time (e.g. at t1, the velocity of the object is v1 and the distance it has travelled is x1-x0 where x0 is the starting point). Normally the subscript 1 is not used unless there are two designated points (subscript 2) where there is a greater need to differentiate the two. For t however, when there is no 0, it refers to a duration of time. A specified point in time will usually always have a subscript.

I will try to put these equations into layman terms/explain them. I probably won't do very well, but I'll try my best.

Eq 1: v is v1. The time for the object to accelerate from v0 to v1 is t. The acceleration of the object is in m/s(square), which basically means that the m is divided by s twice. For example:

A car was going at 45m/s. It accelerated constantly for 6s to 87m/s. Find its acceleration.
87=45+6a
a=7

Eq 2: t is the time interval from the time the object is at x0 to the time where it is reaches x. The logic behind it is that v0t is the distance the object would have travelled if it had not accelerated, and 1/2a(t)(square) is the extra distance it goes through acceleration. The reason why v0t is the distance the object would have travelled if it had not accelerated is pretty straightforward: it is the original velocity multiplied by the time it took. But why is 1/2a(t)(square) the extra distance the object goes through acceleration?

In the first equation we see that a times t is the difference between v and v0. In equation 4, we see that the average velocity is 1/2(v+v0). x-x0=v(average)t, thus x-x0=1/2(v+v0)t=1/2(v0+at+v0)t =v0t+a(t)(square)
Thus is equation 2.

Eq. 4: As it has been used in the previous equation, I don't think it needs a lot more explanation. The average velocity of an object, assuming it accelerates at a constant rate, is the average of its starting and ending velocity.

Eq 3: This is HARD!!! I will try to explain this, but it needs quite a bit of algebra and some things that are rather hard to explain in this manner. Thankfully, this equation is mostly only needed for certain proof and is not usually used in problems.

1st step: simultaneous equation
v=v0+at
x=x0+v0t+1/2at(square)
t=(v-v0)/a
Substituting into second equation:
x=x0+v0(v-v0)/a+a(v-v0)(square)/2a
Cancel a:
x=x0+v0(v-v0)/a+(v-v0)(square)/2
Multiplying everything by 2a:
2ax=2ax0+2v0(v-v0)+a(v-v0)(square)
2a(x-x0)=2v0v-2v0(square)+v(square)-vv0-vv0+v0(square)
2a(x-x0)=v(square)+v0(square)

Thus is equation 3. This is normally only used in situations where t is not mentioned and is unobtainable using any of the other equations (all of which contain t)

15/1/14

Here is one of the questions of the book that uses some (but certainly not all) of these equations. I will change it a bit or else it would be too complicated to explain without a diagram.

A car decelerates from 14m/s at 6.0 m/s^2. How far has it travelled during this time period?

Solution:
v/a=t
14/6=7/3=(about)2.333
(average)v=(v+v0)/2
(average)vt=x
(14+0)/2*2.333=(around)16
Ans: 16m

In 1D Kinematics, air resistance is ignored, thus all objects accelerate when falling at a constant rate of 9.8m/s^2.

Another problem:

A tower is 70m high. How far will a ball dropped from the top have travelled after (a) 1 second, (b) 2 seconds and (c) 3 seconds?

Solution:
(a)
In 1 second, the ball will have accelerated to 9.8 m/s since 9.8m/s^2 times 1s=9.8 m/s. The average velocity of the ball will be the average of the starting speed and the ending speed, namely 0m/s and 9.8m/s (note: this is only true for constant acceleration of the ball), which is (0 m/s+9.8m/s)/2 which is 4.9 m/s. You can then multiply this by the time to get 4.9m since 4.9 m/s times 1s=4.9m.
(b)
I shall not explain this one in words; if you don't know why I do certain steps refer to the example above.

9.8*2=19.6
(0+19.6)/2=9.8
9.8*2=19.6
Ans: 19.6m

(c)
Try this one on your own! The answer is 44.1m.

Note: the values are in the ratio 1:4:9. This is basically 1^2:2^2:3^2.
4.9/1=4.9
19.6/4=4.9
44.1/4=4.9

4.9=9.8/2
     =a/2

Thus we derive the equation 1/2a*t^2=x (recap: x is total distance travelled) which I have proved earlier (you can refer there for the proof). Try this equation for different gravities and the equation will still hold true. Actually, it is not just gravity-you can use this equation for any type of acceleration!

I think that will be it for 1D Kinematics. Now for 2D Kinematics:
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2D Kinematics

2nd March 2014

2D Kinematics and 1D Kinematics are quite similiar except that now there is motion in 2 directions instead of 1. The main concept for 2D Kinematics is that you need to treat the motion as two separate motions in 2 different directions-not as one! For example, for a force exerted at a certain angle, you need to use trigonometry to calculate the force exerted in each direction, then calculate the resultant motion before finally using trigonometry again to convert both motions on the x and y axis into a movement at a certain angle in a certain direction.

For 2D Kinematics, you need some very basic understanding of sine, cosine, tangent, arcsine, arccosine and arctangent. Don't worry, most of the time you just need to know which function to use for what setting-let the calculator do the hard work for you!

One more thing-for physics, don't be afraid to use your calculator-for physics you need precision and accuracy, and calculators are there for a reason-to shorten the time you take to do the problem. Even if you can do the sum with some triple digit multiplication, just do it with your calculator. Our goal here is not to give you superb mental calculation, but instead it is to ensure that you understand the basic concepts and stuff-the actual calculation is not as important. Also for trigonomic functions you either use tables or a calculator, whereby the latter is much faster and more accurate, so try to use your calculator.

A basic introduction about a bit of the trigonomic function (note: what I will be stating will not be enough to equip you with sufficient trigonomic knowledge to solve the questions! If you want to do some reading up on your own)

First and foremost, sine, cosine and tangent only work for right-angled triangles!
However, when angles are involved, a vertical line can be drawn connecting the two lines that will form a right-angled triangle. From there sine, cosine and tangent can be used.

Triangles are made up of the hypotenuse, adjacent and opposite lines. Which line is which type all depends on the placement of the angle, theta. The theta sign looks like an italicised 0 that has a line in the middle.

Yup, that's the one.

Coming back to triangles.

The hypotenuse is the longest line in the triangle. The adjacent is the other line that is touching angle theta yet not the longest line in the triangle.

Trigonometry is based on one concept: when you have two lines, the ratio between them for a certain angle theta will always be constant. For example, when you have the opposite and hypotenuse, if you know the angle between them, and you know one of the measurements, you can use sine to find the relationship between the two measurements (a coefficient), then you can find the other measurement.

For example, I have a triangle. Theta=36 degrees. It is between the opposite and the hypotenuse. Thus here I can use sine.

If the hypotenuse is 25m:
sin(25)=opposite/25m
sin(25)=0.4226182617
0.4226182617=opposite/25m
0.4226182617*25m=opposite
Thus opposite=10.56545654.

Sine formula: sin(theta)=opposite/hypotenuse

The other two formulas work similiarly.

Cosine formula: cos(theta)=adjacent/hypotenuse

Tangent formula: tan(theta)=opposite/adjacent

To memorise the formulas for each trigonomic function, remember SOHCAHTOA. S, C and T are sine, cosine and tangent, while O, H and A are opposite, hypotenuse and adjacent.

There are three other trigonomic functions also involving these three sides but you will most likely not be using them for this section.




Now on to the physics...
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27/4/14

In 2D Kinematics, you will be using a lot of VECTORS.

Vector~a quantity having direction as well as magnitude, especially as determining the position of one point in space relative to another

Vector diagrams are diagrams using arrows to specify the direction and distance travelled.

Here's a vector question.

John walked 27.8 degrees north from east for 65.7m. He then walked another 13.6 degrees south from east for 156.2m. How far is he from the starting point.

These questions are usually best visualised using a vector diagram but being unable to insert proper vector diagrams in here I shall try my best to explain the solution. If you do not understand just leave a comment in the comment box. (and your email?)

Remember what I said at the start of this chapter-the key is to see each motion as two separate motions in two separate directions. First you separate the two motions into 4 motions, 2 on each of the x and y planes. Find the resultant motion for both x and y before then combining them into a single motion with an angle.

(You will need trigonometry for this question)

First, 27.8 degrees north from east.
sin(27.8)*65.7=30.6416 (for physics questions I like to round to 4 decimal places, you can use 3)
cos(27.8)*65.7=58.1169

Since the vector arrow for this will be pointing more east than north, the motion will be largely east. Thus we can see that the first motion can be separated into two motions of 30.6416m north and 58.1169m east.

Next, 13.6 degrees south from east.
sin(13.6)*156.2=36.7292
cos(13.6)*156.2=151.8203

Since the vector arrow again will be pointing more east than south, the motion will be largely east. Thus the second motion can be separated into 36.7292m south and 151.8203m east.

Thus

30.6416-36.7292=-6.0876
58.1169+151.8203=209.9372

John travelled 6.0876m south and 209.9372m east.

To visualise combining them into one vector, we need a vector diagram.

--------->
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             V
Not drawn to scale (obviously)

At the meeting point between the two vectors, there is a 90 degrees angle. Here, we can use Pythagoras theorem. (You know what that is right)

For those who don't,



(Square root both sides to find c which will be the square root of the sum of a^2+b^2)

Thus using your calculator:

The square root of 6.0876^2+209.9372^2=210.0284.

To find the angle, you need to use tangent. Tangent theta is opposite/adjacent and in this case the downwards line is opposite angle theta marked by the brackets (sorry I don't have a better substitute) and the adjacent  is the horizontal line.

--------->
()            |
              |
              |
              |
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              |
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             V
tan(theta)=209.9372/6.0876
               =34.4860

To find what theta is, we use arc tangent (a.k.a tan^-1) to get theta=88.3390

Thus the resultant motion of John is 210.0284 metres, 88.339 degrees south of east.

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A bit of clarification on how to know which value (sin or cos) is what value (horizontal or vertical)|:

Some times it can be blatantly obvious which direction corresponds to which value but at other times it can get rather confusing resulting in careless mistakes (which I make a lot of sad to say)

sin-opposite/hypotenuse
cos-adjacent/hypotenuse

--------->
              |
              |
              |
              |
              |
              |
              |
             V

Remember that the horizontal line is the adjacent and the vertical is the opposite (for this case only---always remember the definition of sin cos and tan using SOHCAHTOA in case you forget-it is a common careless mistake)

For a vector of m degrees, y direction above/below x direction with original vector distance k,

sin(m)*k=distance in x
cos(m)*k=distance in y

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