I'm Done! (+ Roller Coaster Tips)

Well this semester has come to an end...I survived all 7 (?) units and even the roller coaster and exam. Didn't do too badly either. I must say I will miss this class...but tomorrow is a whole new semester.

Tips to future students:
Ask questions! Think a little beyond what you hear in the class and ask why...don't accept what they tell you without question. Good luck future semesters!

Concerning the roller coaster, here are our top 10 tips. Please take our word for it: if you keep these things in mind, your life will be SO much easier.

1. Use gentle back-and-forth motions when bending the I beam (to avoid kinking)
2. Roll your cylindrical object against the I beam when bending to keep the loops round (seriously, make your loops as round as possible. Pointy = ball no likey)
3. ***Make one length of I beam, and immediately begin attaching it to your frame. That way you will get a sense of what will work, what you’ll need to do to make the rest of the I beams work and how many I beams you’ll be able to fit in so that you don’t waste money buying and looping more (like we did)***
4. Get at least part of your major work figured out early, such as your theme, whether you’re focusing on artistic or technical merit, which materials you want to use, where you’re going to build
5. Check the trash bins of your home hardware store; you might find free useful materials that could come in handy later during building (like us...the store people were amused at how happy we were to get free stuff)
6. Wear gloves when you are bending the I beam, and use a hammer as a lever to press the I beam into shape when your hands alone don’t have enough strength
7. Bend loops in opposite directions to save space in your frame
8. Look at images of previous Wondercoasters for inspiration when you’re stuck on something
9. Don’t leave decorating to the last minute, because your decorations may alter how your track works
10. Start as soon as possible when the project is assigned (you’ll regret it if you don’t) (this sounds cheesy but unfortunately it's true)

Also, here are challenges we faced in construction and how we overcame them (there were lots of these):

Bending the I beams

• We kinked the beam many times because we bent it too fast, wasting a lot of I beam
• We often bent an angle instead of a round loop
• These problems meant that the marble’s ride would be bumpy or difficult, which really made attaching to the frame frustrating

How we overcame these challenges

• Kinking the I beams could be prevented by not bending the beam too fast, or by not unbending the beam after it was bent
• The kinks could be “cured” somewhat by pressing the protruding sides together using big pliers
• When bending the I beam into loops, use slow back and forth motions with the beam and roll the cylinder you’re using against the beam

Attaching the I beams to the frame

• Since the I beams were badly bent in the first place, they caused a lot of trouble when we tried to attach them to the frame
• They had to be in exactly the right position in order for the marble to complete them
• Finding this position, and keeping it in place, especially when we weren’t constantly holding it, was extremely difficult
• Also, the angle and amount of space each length of track took up in order to work was much higher than expected, so we ended up only using half of the lengths of track we made

How we overcame these challenges

• Patience! And a lot of it
• Using various materials to make “Band-Aids” to smooth over the bumps in the I beam
• The glue gun helped significantly in securing the lengths in place once we found that perfect position
• Creative use of the materials at hand: wire; hooks from the curtain rod; corrugated cardboard; duct, electrical and Scotch tape

Polytubing

• Finding a way to smoothly attach polytubing track to the I beam track without the marble bouncing off the edges of the I beam
• Finding a way to smoothly transition polytubing to the cardboard support and turn
• Making the polytubing lie straight instead of staying in its natural curve

How we overcame challenges

• Making a smooth transition between tubing and I beam was something that eluded us…we did our best by at first slitting the tubing and sticking it over the edges of the I beam. Later we tried using the glue gun to secure the two pieces of tubing underneath the middle of the I beam
• We made two “fangs” with the end of each tube, stuck them into the support, and secured them with glue gun glue
• We pushed the many nails sticking out of the polytubing track through a flat, straight piece of compressed cardboard on either side of the track to keep the polytubing straight

20 Sound Notes

And by notes, no, I do not mean musical notes, I mean taking notes or points in writing.

Basically the notes are about waves and the speed of sound.

• A vibrating source moves with simple oscillating motion 
• A transverse wave is when the direction of wave travel is perpendicular to the motion of the source (for example, light)
• A longitudinal wave is when the direction of wave travel is parallel to the motion of the source (for example, sound)
• A cycle is a complete sequence of motion
• Wavelength, represented by 
  l 
  , is the length in metres of one cycle
• Period, represented by T, is the time it takes to complete one cycle in seconds
• Frequency (f): the number of cycles / time, measured in Hertz (Hz)
• Amplitude: the maximum displacement (height) of the wave from 0
• The wave equation is v = 
  l 
  f, which is derived from the kinematics formula v =
  D
  d / 
  D
  t

• Dense areas of sound waves are compressions
• Less dense areas of sound waves are rarefactions
• Factors that affect the speed of sound: the producer, the temperature and density of its medium
• Stiffer materials result in a faster speed of sound
• vs = 332 m/s + (0.6 m/s 
  ÷
    °C) (T °C)
• Speeds close to the speed of sounds are measured in Mach numbers
• < Mach 1 is subsonic
• > Mach 1 is supersonic
• Examples of planes/aircraft that have approached and surpassed the speed of sound include Concorde, while Boeing, although fast, is still subsonic
• The sound barrier is when pressure is built up when something approaches the speed of sound and catches up to its own sound waves, creating a "wall"
• A plane breaking the sound barrier creates cone-shaped shock waves that hit the Earth's surface, where the sound is heard as a "boom". This is a sonic boom. 

Newton and His Problems

Poor Sir Isaac Newton. After he got hit on the head by the infamous gravity-demonstrating apple, he began having problems--four of them to be exact. Since therapists and such had not been invented yet, Newton realized he had to solve his own problems...but he could only do this by making certain assumptions. Even though Isaac Newton has been dead for a long time, his problems still exist (they outlived him) and they still have to be solved, only now by us poor grade 11s. Here is a list of the assumptions Newton had to make and that are necessary to apply to the solutions of these problems.

Problem #1: Equilibrium
Assumptions:

• Acceleration: since ax = 0 and ay = 0, overall a = 0 (there is no acceleration)
• Gravity: g = 9.81 m/s^2 
• Positive Axes: a set of positive axes must be set

Example Problem:

For some unimaginable reason, Marco decides that he wants to slide down a frictionless rope across Niagara Falls. However, halfway through he realizes that the maximum tension the rope can support is 500 N. If the angle between the rope and the horizontal is 15 º, help Marco figure out what his mass must be to safely cross the waterfall (hint: right now Marco is hanging on the rope). 

Solution: 

Given: T_{1 max} = 500 N     g = 9.81 m/s²    θ = 15 º               Required: m = ?

Analysis: ∑ F = ∑ F_x = 0

∑ F_x = Fg – 2Tsin15 º

Fg = 2Tsin15 º

mg = 2Tsin15 º

Substitution: m_max = 2Tsin15 º / g = 2(500)sin15 º / 9.81 = 26.38 kg

Therefore, he must weigh 26.38 kg or under in order to cross safely. 

Problem #2: Inclined Planes

Assumptions (Static or no movement):

• Acceleration: since ax = 0 and ay = 0, there is no acceleration or a = 0
• Axes: the x-axis is now parallel to the direction of friction or to the surface, and the y-axis is now parallel to the normal force or perpendicular to the surface
• Positive X-Axis: the positive x-axis is in the direction of potential motion
• Coefficient of Friction: µ = tanθ

Example Problem: If Marco has a mass of 37.6 kg and is sitting without moving on an incline with a coefficient of friction of 0.42, what is the force of friction that he experiences?

Solution:

Given: µ = 0.42     m = 37.6 kg           g = 9.81 m/s²         Required: f = ?

Analysis: ∑F = 0    ∑F_x = ∑F_y = 0       X: ∑F_x = 0   F_gx – f = 0    f = F_gx = mgsinθ

                                                            Y: ∑F_y = 0    F_N – F_gy = 0 F_N = F_gy = mgcosθ

                                                             µ = tanθ

                                                             f = µF_N = mgcosθtanθ

Substitution: θ = tan^-1 µ = tan^-1 (0.42) = 22.78 º

       Either   f = mgsinθ = (37.6)(9.81)sin(22.78º) = 142.83 N

      Or        f = mgcosθtanθ = µmgcosθ = (0.42)(37.6)(9.81)cos(22.78º) = 142.83 N

Therefore, he experiences a frictional force of 142.83 N.  

Assumptions (Kinetic or with movement):

• Kinetic friction: f_k = µ_k x F_N
• _{Axes: the x-axis is parallel to the direction of friction or the surface, and the y-axis is parallel to the normal force or perpendicular to the surface}
• _{Positive X-Axis: the positive x-axis is in the direction of motion}
• Forces: ∑ F = ∑Fx + ∑Fy , ∑Fy = 0 and ∑Fx ≠ 0

Example Problem: Marco wants to feel the wind in his new hair, so he decides to go sliding. The slide is at an angle of 11º and Marco is travelling with a force of 20.5 N despite 54 N of friction. What is the coefficient of friction of the slide?

Solution:

Given: g = 9.81 m/s²   θ = 11º             f = 54 N          F = 20.5 N

Required: µ = ? (m = ?)

Analysis: ∑F = 20.5N    ∑F_x + ∑F_y = 20.5N    ∑F_y = 0        ∑F_x = 20.5 N

              X: ∑F_x = 20.5 N   Fgx – f = 20.5

              Y: ∑F_{y = 0          Fgy = FN =} mgcosθ         

              f = µFN                µ = f/FN  = f/mgcosθ         

              m = ?    Fgx – f = 20.5     mgsinθ – f = F           mgsinθ = F + f             m = (F + f)/gsinθ

Substitution: m = (F + f)/gsinθ = (20.5 + 54)/(9.81)sin11º = 39.8 kg

                      µ = f/mgcosθ = 54/(39.8)(9.81)cos11º = 0.14

Therefore, the coefficient of friction on the slide was 0.14. 

Problem #3: Pulleys

Assumptions:

• Air resistance: there is none
• Friction: there is no friction when using the pulley
• Multiple Systems: each side of the pulley can be broken into multiple systems
• Tension: the tension in both systems is equal and opposite
• Positive Axis: for each system, the positive axis is in the direction of motion or potential motion
• Acceleration: the acceleration is consistent for both systems

Example Problem: Marco is hanging off one end of a frictionless pulley while Marco’s friend Tony is hanging off the other end. If Tony has twice as much mass as Marco and the tension in the pulley is 78 N, what is Marco’s mass?

Solution: 

Given: m₁ = Marco     m₂ = Tony      m₂ = 2m₁           T = 78 N      g = 9.81 m/s²

Required: m₁ = ?

Analysis: System 1: ∑ F₁ = m₁a = T - m₁g                        a = (T - m₁g)/ m₁

              System 2: ∑ F₂ = m₂a = m₂g – T              a = (m₂g – T)/ m₂

                  a = a

(T - m₁g)/m₁ = (m₂g – T)/m₂

(T - m₁g)/m₁ = (2m₁g – T)/2m₁

(T - m₁g)(2m₁) = (2m₁g – T)(m₁)

(T - m₁g)(2m₁) = (2m₁g – T)(m1)

2T – 2m₁g = 2m₁g – T

3T = 4m₁g

3T/ (4g) = m₁

Substitution: m₁ = 3(78) / (4)(9.81) = 5.96 kg

Therefore, Marco has a mass of 5.96 kg. 

Problem #4: Trains

Assumptions:

• Multiple Systems: each car of the train can be separated into its own system
• Air Resistance: there is none
• Positive Axes: the positive axis (generally it is the x-axis that involves motion) is in the direction of motion
• Acceleration: the acceleration is consistent for all the system. For this reason, the train can be treated as one entity in order to find acceleration

Example Problem: In Marco's train (shown below), what is the amount of force opposing the motion of the middle car (hint: the amount of force in the opposite direction of motion for the middle car).

Solution:

Given: m₁ = 20 kg       m₂ = 30 kg       m₃ = 40 kg       µ = 0.16           F_A = 450 N      g = 9.81 m/s²

Required: T₁ – m₂a = x = ? (a = ? T₁ = ?)

Analysis: and Substitution: Total System: ∑F_T = m_Ta           

∑F_T = ∑Fx + ∑Fy 

∑Fy = 0 

∑Fx = F_A – f  

        = mTa

a = (F_A – f)/m_T

   = (F_A - µF_N)/m_T

   = (F_A - µm_Tg)/m_T

   = [450 – (0.16)(90)(9.81)]/90

   = 3.4304 m/s²

System 1: ∑F₁ = m₁a    

∑F₁ = ∑F_x + ∑F_y    

∑F_y = 0 

∑F_x = F_A – T₁ – f     

m₁a = F_A – T₁ – f

       = F_A – T₁ - µF_N

       = F_A – T₁ - µm₁g  

T₁ = F_A - µm₁g – m₁a

     = 450 – (0.16)(20)(9.81) – (20)(3.4304)

     = 350 N

System 2: ∑F₂ = m₂a    

∑F₂ = ∑F_x + ∑F_y    

∑F_y = 0 

∑F_x = T₁ – T2 – f

       = T₁ – x

x = T₁ – m₂a

  = 350 N – (30)(3.4304)

  = 247.088 N

Therefore, 247.088 N is the amount of force pulling the middle car back. 


Thank you for having the patience to look through this long and incredibly time-consuming to make blog!