the other stuff in projectile motion · so what are these formulas missing? many things, but one of...

The "other stuff" in Projectile Motion

What do we already know?


Equations that describe the motion of projectiles



d = vit + .5at2

vf = vi + at



d = vit + .5at2

vf = vi + at

Equations that tell us about force and acceleration



d = vit + .5at2

vf = vi + at

Equations that tell us about force and acceleration

F = ma

What do these formulas help describe?

So what are these formulas missing?


Many things, but one of the big ones is the drag force.



The drag force is a force that always opposes motion



The drag force is a force that always opposes motion

Drag is the force of pushing all the fluid (air) out of the way

FD = .5pv2CDA

What are these variables?

FD = .5pv2CDA

p is the density of the fluid

It is measured in kg/m3

FD = .5pv2CDA

p is the density of the fluid

It is measured in kg/m3

Air is about 1.2kg/m3 at 20C

While water is 1000kg/m3

FD = .5pv2CDA

v is the velocity of the object

It is measured in m/s

FD = .5pv2CDA

v is the velocity of the object

It is measured in m/s

This variable tends to change constantly

FD = .5pv2CDA

CD is the coefficient of drag of the object

It is a measured characteristic of the object, and is dependant on the objects shape. A sphere has a coefficient of about .48

FD = .5pv2CDA

CD is the coefficient of drag of the object

It is a measured characteristic of the object, and is dependant on the objects shape. A sphere has a coefficient of about .48

A more aerodynamic object will have a lower coefficient, and vice versa

FD = .5pv2CDAA is the cross sectional area of the object

Area is measured in m2

FD = .5pv2CDAA is the cross sectional area of the object

Area is measured in m2

Think of the cross sectional area as the shadow the object would cast. In the case of a ball, it would be a circle.

FD = .5pv2CDA

What do we end up with?

FD = .5pv2CDA


p = 1.2kg/m3

CD = 0.48

A = pi(.02)2

v is a changing variable

FD = .5pv2CDA


p = 1.2kg/m3

CD = 0.48

A = pi(.02)2

v is a changing variable

FD = 0.000363v2

Terminal Velocity of a ping pong ball

Mass of the ball is 2.7g


Mass of the ball is 2.7gFg = mg


Mass of the ball is 2.7gFg = mg = .0027kg x 9.8m/s2


Mass of the ball is 2.7gFg = mg = .0027kg x 9.8m/s2 = .0265N



So at what velocity does FD = Fg?



So at what velocity does FD = Fg?

FD = .000363v2

for what v is FD = .0265N?


0.000363v2 = .0265


0.000363v2 = .0265

v2 = .0265/0.000363 = 73

v = 8.54m/s


0.000363v2 = .0265

v2 = .0265/0.000363 = 73

v = 8.54m/s

The actual measured terminal velocity is 9.5m/s

Any Questions on the drag force?

Launch Physics

Launch Physics

What is the driving force in the launching of objects?

Launch Physics

What is the driving force in the launching of objects?

You can launch an object a variety of ways.From counterweights to stored energy in something like a rubber band.

More complex solutions include using pressure differentials or even using electromagnetism.

Launching using Air Pressure

We are going to take a look at how an object is launched using air pressure.



This is the mechanism used in well known projectiles such as guns and cannons.



This is the mechanism used in well known projectiles such as guns and cannons.

The formula is quite simple. FP = PA

FP = PA

FP = PA

P is the pressure involved.This is measured in N/m2

FP = PA


Note that the pressure is the DIFFERENCE in pressures. We are all under 1 atmosphere of pressure (101 kPa or 14.7 psi) just from all the air above us pressing down.

FP = PA


Note that the pressure is the DIFFERENCE in pressures. We are all under 1 atmosphere of pressure (101 kPa or 14.7 psi) just from all the air above us pressing down.So when we say something has 1psi, what we mean is that it is under 15.7 psi, but compared to the pressure just outside of it, it has 1 psi.

FP = PA

A is the area involvedThis is measured in m2

While it is not always the case, we can use the cross sectional area once again here.

What about the other forces?


There are definitely other forces involved in launching the projectile. There is the force of gravity pushing down, as well as the friction force, the drag force, and the normal force.


There are definitely other forces involved in launching the projectile. There is the force of gravity pushing down, as well as the friction force, the drag force, and the normal force.

However, as we will see in a few slides, these will be miniscule compared to the pressure force, so we can ignore them.

Now to the Cannon

Now to the Cannon

We are going to be shooting a 40mm diameter ping pong ball that weighs about 2.7g. We will be shooting through a tube that is about 25cm long at a pressure of 25 psi.

Now to the Cannon


So what can we predict from this?

Now to the Cannon


So what can we predict from this?

We can start by figuring out the force on the ball due to pressure!

So how fast is this thing gonna go?


We can use FP = PA


We can use FP = PAWe know P = 25 psi = 172 368 pascalsWe can find A using pi x r2 pi x (.02)2 = .00126m2



So FP = 172 368 x .00126 = 217N

What can we do next?



So FP = 172 368 x .00126 = 217N

What can we do next?We can get the acceleration using F=ma


F=ma


F=ma ---------> a = F/m


F=ma ---------> a = F/mWe know F = 217NWe know m = .0027 kgso a = 217/.0027 = 80 370m/s2


F=ma ---------> a = F/mWe know F = 217NWe know m = .0027 kgso a = 217/.0027 = 80 370m/s2

Thats a lot of acceleration!Well, all thats left now is to find how fast it leaves the barrel.


We know the acceleration and the distance traveled, so we can use this to find the time it takes to move through the barrel using



d = vit + .5at2

.5at2 = d - vit



d = vit + .5at2

.5at2 = d - vit

t2 = 2(d - vit)/a = 2(.25 - 0)/(80 370) = .000 006t = .002 sNow we can use this time to find the velocity


We will use vf = vi + at



vf = 0 + 80 370 x .002 = 160m/s



vf = 0 + 80 370 x .002 = 160m/s

That seems a bit fastLets look at the high speed shot, and get an idea of what the speed is

http://www.youtube.com/watch?v=fzv-6jE2u9k&feature=youtu.be


When played frame by frame, we can find that the ball travels roughly the diameter of itself every 3 frames.


When played frame by frame, we can find that the ball travels roughly the diameter of itself every 3 frames.The video was shot at 1200 frames per secondThis means that the ball moved 40 mm in 3 1200ths of a secondSince v = d/t = (.04)/(3/1200) = 16 m/s


When played frame by frame, we can find that the ball travels roughly the diameter of itself every 3 frames.The video was shot at 1200 frames per secondThis means that the ball moved 40 mm in 3 1200ths of a secondSince v = d/t = (.04)/(3/1200) = 16 m/s

So our guess of ~160 m/s was wrong. Why?


The fit isn't perfect, only the ~5mm by the o-ring will actually have the full pressure force applied.

The pressure will also drop to about half of what we measured in the tank


The fit isn't perfect, only the ~5mm by the o-ring will actually have the full pressure force applied.

The pressure will also drop to about half of what we measured in the tank

Putting this together, and redoing the calculations, we get a final velocity of about 20m/s, which is much closer to the measured value

Now to put it all together

But first, any questions about what we have gone over?


So we know the exit velocity will be around 16m/s


So we know the exit velocity will be around 16m/sAnd lets say we aim the cannon at a 30 degree angle


So we know the exit velocity will be around 16m/sAnd lets say we aim the cannon at a 30 degree angleCan we guess where the ball will land?


So we know the exit velocity will be around 16m/sAnd lets say we aim the cannon at a 30 degree angleCan we guess where the ball will land?

First, lets see what our old formulas tell us (when we ignore air resistance)

Ignoring Air Resistance

vi = 16m/s at a 30 degree anglebreak vi into viy and vix



viy = 16 x sin(30) = 8m/svix = 16 x cos(30) = 13.9m/s



viy = 16 x sin(30) = 8m/svix = 16 x cos(30) = 13.9m/s

using viy, and that the launch point is about 30cm up, we find the time it takes to hit the ground.0 = .3 + 8t + .5(-9.8)t2


Finding the roots give us t = 1.67s


Finding the roots give us t = 1.67sUsing that t, and vix we find the distance traveled to bed = vixt = 13.9 x 1.67 = 23.2m

So how will considering air resistance affect the calculation?

Using Air Resistance

First, it will add an extra force in both "components". This force will always be against the direction of travel.

Second, it will be constantly changing. It is dependant on the velocity of the object, and as the total velocity changes, so too does the drag force.


We know the equation for finding the drag forceFD = .000363v2



Lets look at the drag force as the ball exits the barrel.FD = .000363(16)2 = .093N



Lets look at the drag force as the ball exits the barrel.FD = .000363(16)2 = .093NThis force will be antiparallel to the direction of travel. Now to apply it accordingly


FNetx = FDx

FNety = FDy + Fg

So the x and y components will be found the same way we find the x and y components of velocity: by multiplying the drag force by the cos and sin of the current angle, which we know initially to be 30 degrees.


FNetx = FDx = -.081NFNety = FDy + Fg = -.047N + (-.0265N) = -.0735N


FNetx = FDx = -.081NFNety = FDy + Fg = -.047N + (-.0265N) = -.0735N

We can then turn those forces into accelerations using F = ma

ax = -.081/.0027 = -30m/s2

ay = -.0735/.0027 = -27.2m/s2


ax = -30m/s2 ay = -27.2m/s2

Using these accelerations, we can find the change in position as well as the change in velocities.


ax = -30m/s2 ay = -27.2m/s2

Using these accelerations, we can find the change in position as well as the change in velocities.

However, this time we are going to choose the time, we call this a time step. We can choose any size time step, but for now, lets use 0.1s.


ax = -30m/s2 vx = 13.9m/sTo find the change in position, we use dx = vxt + .5at2


ax = -30m/s2 vx = 13.9m/sTo find the change in position, we use dx = vxt + .5at2 = 13.9 x 0.1 + .5 x (-30) x (.1)2

dx = 1.24m



dx = 1.24m

To find the change in velocity we usevxf = vxi + at = 13.9 + (-30) x (.1) = 10.9m/s



dx = 1.24m

To find the change in velocity we usevxf = vxi + at = 13.9 + (-30) x (.1) = 10.9m/s

We then do the same for the y component


ay = -27.2m/s2 vy = 8m/sTo find the change in position, we use dy = vyt + .5at2 = 8 x 0.1 + .5 x (-27.2) x (.1)2

dy = .664m


ay = -27.2m/s2 vy = 8m/sTo find the change in position, we use dy = vyt + .5at2 = 8 x 0.1 + .5 x (-27.2) x (.1)2

dy = .664m

To find the change in velocity we usevyf = vyi + at = 8 + (-27.2) x (.1) = 5.28m/s


So, we have now seen how the ball will change over that time step. It will have moved .664m up and 1.24m across. We add these to our previous (x,y) coordinates of (0, .3) to get (1.24, .964)


So, we have now seen how the ball will change over that time step. It will have moved .664m up and 1.24m across. We add these to our previous (x,y) coordinates of (0, .3) to get (1.24, .964)

We update the velocity as well. It now is moving at 10.9m/s in the x direction and 5.28m/s in the y direction. This gives it a new total velocity of 12.1m/s.


We can use that information to repeat all those calculations over again.

The values of vx and vy will give us the angle of travel. While v will tell us the value of FD. We will keep track of the (x,y) coordinates, and we know to stop when the y value is less than or equal to 0.

Can we shoot the cannon yet?


Almost


Almost

The last point is that those calculations can get quite annoying. So what we can do is have a computer do them for us.


Almost

The last point is that those calculations can get quite annoying. So what we can do is have a computer do them for us.

By letting a computer do the work (and it does it billions of times faster than us) we can choose a smaller step size to get a more accurate estimate.


Using such a program, I have a few estimates

time step Distance.2 6.45m.1 7.87m.05 8.18m.01 8.47m.001 8.53m

Initial no drag guess: 23.2m


Ok, we can shoot it now

And feel free to ask any questions you may have

the other stuff in projectile motion · so what are these formulas missing? many things, but one of...

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