Cannons

So hard to find any background on this topic, so I did not find much helpful information beside the obvious...
There are few factors that I think will determine the effectiveness of the cannon.
First factor: Angle for the maximum range
In a hypothetical situation, suppose all other factors are controlled, including friction, air resistance, velocity, etc. And the only variables that are being altered is the launching angle of the cannon, how will the range of the cannon ball launched be affected? Or otherwise, what angle will give the biggest maximum range for this projectile motion of cannon?

Imagine launching at 30, 45 and 60.

It seems that the launch angle of 45 will result the farthest distance travelled. From what we learned in kinematics, range of a projectile motion will be determined by the product of hang time and Vx. For angle of 30, it had the biggest value for Vx out of the 3 cases, but it's maximum range was limited by its hangtime due to lack of Vy. For angle of 60, it had the biggest value for Vy. Therefore, its hangtime was the greatest out of all three, however, its maximum range was then limited by its Vx, the lowest of all 3. The angle of 45, which did not have the highest value in any categories, had the biggest maximum range because it had a decent value for both the hangtime and also Vx.


Tight Sealing:
For the duct tape that are used to stick together the pieces of pop can, it ensures that first of all, fuel does not leak when we shake the cannon, and secondly to make sure the no force/energy caused by the explosion "escape" from the launch. This is because we want the force caused by the explosion to focus on the "bullet", which would be the paper cup.

Mass of the base:
The mass should not be too light, because the force that causes the cannon ball to fly on its path will cause an equal force on the base of the cannon (opposite force, newton's 3rd law).This is why tanks and cannons are made out of metal as its for resisting the opposite force after launching.

In summary, what will make a better cannon? A combination of high hang time, initial horizontal velocity, little friction and built almost flawlessly.(Completely enclosed taping, a strong base and other factors)

Dynamics

Equilibrium
This is achieved when the forces of both x and y components are balanced. Therefore, we get an overall acceleration of 0, hence the object is at a state of rest.
Assumptions:

• No Friction
• No Acceleration

The forces in the y and x components of the object being acted on are balanced. Hence, we must find the x and y component of force that is acting on the object.This can be accomplished by using vectors component, when we know the force in N at a given angle.
For example:

Object B has a force of 9.2N acting on it at a angle of 70 degrees with the horizontal, therefore
Bx = 9.2 cos 70
By = 9.2 sin 70
With multiple forces acting on an object, add my x directions forces and y direction forces together to get Fnetx and Fnety.
The other forces acting on the object would be gravity. Since the forces are balanced and the object is not moving up and down, we can conclude that:

Fg = F net y, furthermore, we know that Fg = ma.
We already know that a is equal to 9.81m/s^2, therefore the mass of the object can be calculated by:
M = F nety / acceleration.
We can always check if our answer is right by calculating and see if the x direction forces are balanced as they're suppose to in an equilibrium.

Static Inclines 

Assumptions:

• No acceleration
• + tive axes in the direction acceleration
• No air resistance
• fs (static friction) = µ * Fn
• Fn is perpendicular to the surface

Looking at diagram, we already know that there is no acceleration at all, therefore:
Fgx=fs, where Fgx = Mg sin θ => Mg sin θ = M Fn
Fn=Fgy, where Fgy = Mg cos θ => Mg cos θ = Fn (sub into first equation)
Mg sin θ = µ Mg cos θ
 µ = sinθ/cosθ
 µ = tanθ

Kinetic Incline
 In this case, the incline is almost like a static incline, except the object is moving and accelerating in the x direction. Therefore, ax does not = 0
Other assumptions:

• No acceleration in y direction
• + tive axes in the direction acceleration
• No air resistance
• fk (kinetic friction) = µk * Fn
• Fn is perpendicular to the surface

In the x direction: 
Fgx - fk = ma, Mg sin θ- µFn = Ma (1)
In the y direction:
Fn - Fgy = 0 (since acceleration is still 0), Mg cos θ = Fn (2)
Sub (2) into (1)
Mg sin θ- µ Mg cos θ= Ma
a = Mg sin θ- µ Mg cos θ / M

Pulleys
Assumptions:

• No friction from pulley, rope or any other materials in the system
• no air resistance
• Remember to draw multiple FBDs depending on situation
• Always make acceleration the positive axes in all FBDs
• T1=T2
• acceleration of the system is the same

Notice in this sample pulley problem, the positive axes is different in both diagrams because acceleration is opposite in both due to Newton's 3rd law.


For m1:

F = ma

 

Fₓ = ma

Fₓ = 0

Fy = m1ay

m1g - T = m1a

T = m1g - m1a (1)

For m2:

F = ma

Fₓ = ma

Fₓ = 0

Fy = m2ay

T - m2g = m2a

T = m2a + m2g (2)

By comparing equation (1) and (2)

set (1)=(2)

m1g - m1a = m2a + m2g

m1g - m2g = m2a + m1a

m1g - m2g = a (m2 + m1)

(m1g - m2g) / (m2 + m1) = a

To Find T

Sub the value of a into equation 1 or 2 to get value for T

*sometimes pulley will not necessarily be parallel to each other, they could be perpendicular or at any angle

Trains:

Assumptions:

• 1 FBD for acceleration of the whole system, where a is constant throughout the system
• multiple FBDs depending for the number of tensions there are
• ay = 0
• no air resistance
• weightless cables
• + axes in direction of a

For system acceleration:

F = ma

Fₓ = maₓ

Fₐ - f = maₓ 

Fₐ - µmg = maₓ

aₓ = (Fₐ - µmg) / m

This acceleration will be same for all parts of the train. There's no acceleration in y component since we do not see the train going up and down. This is the same for all components of the train

For m1:
F = ma

Fₓ = m1aₓ

Fₐ - T1 - f1 = m1aₓ
Fₐ - T1 - µm1g = m1aₓ   

T1 = Fₐ - µmg - m1aₓ

Now there's a choice of using m3 or m2, but I think Mr. Chung said using m3 is suppose to be less likely to make a mistake. So for the sake of grade, we should stick with m3

For m3:

F = ma

Fy = m3ay

Fy = 0

Fn3 - m3g = 0

Fn3 = m3g

 

Fₓ = m3aₓ

T2 - f3 = m3aₓ

T2 = m3aₓ + f3

T2 = m3aₓ + µm3g

Projectile Motion

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Projectile Motion

A general summary of solving projectile motion problems, finding the usual things (Range, Height, hang time), without regard of air resistance.

Since all these problems occur on Earth, the acceleration of all the problems will be the gravity (-9.8m/s²)

However, gravity will only affect the vertical-y component, therefore the acceleration of horizontal speed is 0 - constant velocity (v1x = v2x = vx)

Components of original velocity:

Using basic trignometry, we can calculate x and y components when given the inital velocity and angle.

X component of original velocity: v_x1 = v cos(θ)
Y component of original velocity: v_y1 = v sin(θ)

Height:

At its maximum height, Vy becomes 0

Given v_y2 = 0, v_y1 = V₁ sin θ, a_y = -9.8m/s²

v₂²=v₁²+2ad

v₂²=v₁²+2aH

   0=v₁²-2gH

   H=v₁²/2g

Hang Time: After finding out  V_1x and V_1y with trignometry, we know that the displacment vertically is 0 after the total time.

Given : dy = 0, ay = -g

 d = v₁t + 0.5at²

 0 = v_1yt + 0.5(-g)t²

 v_1yt -0.5gt²=0

 t(v_1y-0.5gt)=0

Therefore, t = 0 or T = 2v_1y/g

Range:

This is the distance the object travelled horizontally, since the horizontal velocity is constant, we can find out the distance from equation : 

Dx = (Vx)(T)

With reference to the time of the projectile, we can then easily calculate the range. 

(I prefer doing them one by one)

Different Types of Projectile Motion

The Physics behind Roller Coasters

Many people do not realize the force driven behind rolle coasters is not from electric energy, but potential and kinetic energy. The mechanism of roller coasters is all about the physics behind the energy from its motions.

Think about when you lift a textbook above the table, what is happening to the book? It's actually gaining potential energy because when the book is dropped at some point, the energy can be observed and heard when the book hits the table. This is all because another force - gravity, is acting on the object. Relate this observation to rollercoasters, the potential energy of the rollercoaster will be highest when it's at the highest peak of the rail, because more kinetic energy will be produced when it swoshes down the rail. Therefore, as the coaster travels lower and lower, it converts all of its potential energy into kinetic energy. So the potential energy is the lowest when its at a low altitude, but inversely, kinetic energy is the highest.

Modern roller coasters usually have a launcher, which gives it an extra push to begin with, building up more kinetic energy.We know that energy cannot be created or destroyed, rollercoasters work on the principles of the conversion of one form of energy into another. However, in theory, the train should be able to climb to its original height after the first drop (energy conversion). But dissipative forces such as friction and air resistance denies this possiblity. This is why the first peak for a rollercoast is always the highest in order for it to gain more potential energy.