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Projectile Motion (Two Dimensional)

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Projectile Motion(Two Dimensional)

Accounting for Drag

- Know the equation to compute the drag force on an object due to air friction
- Apply Newton's Second Law and the relationship between acceleration, velocity and position to solve a two-dimensional projectile problem, including the affects of drag.
- Prepare an Excel spreadsheet to implement solution to two-dimensional projectile with drag.

V0

y

Position:

q

x

Velocity: Acceleration:

Vx = Vocos(q) ax = 0

Vy = Vosin(q) - g t ay = -g

- All projectiles are subject to the effects of drag.
- Drag caused by air is significant.
- Drag is a function of the properties of the air (viscosity, density), projectile shape and projectile velocity.

- The drag FORCE acting on the projectile causes it to decelerate according to Newton's Law:
aD = FD/m

where: FD = drag force

m = mass of projectile

- The drag force due to wind (air) acting on an object can be found by:
FD = 0.00256 CDV2A

where: FD = drag force (lbf)

CD = drag coefficient (no units)

V = velocity of object (mph)

A = projected area (ft2)

- As a pair, take 3 minutes to convert the proportionality factor in the drag force equation on the previous slide if the
- units of velocity are ft/s, and
- the units of area are in2

- The drag coefficient is a function of the shape of the object (see Table 10.4).
- For a spherical shape the drag coefficient ranges from 0.1 to 300, depending upon Reynolds Number (see next slide).
- For the projectile velocities studied in this course, drag coefficients from 0.6 to 1.2 are reasonable.

q

- Consider the projectile, weighing W, and travelling at velocity V, at an angle q.

- The drag force acts opposite
to the velocity vector, V.

q

q

- The three forces acting on the projectile are:
- the weight of the projectile
- the drag force in the x-direction
- the drag force in the y-direction

- The total drag force can be computed by:
FD = 8.264 x 10-6 (CDV2 A)

where:

|V2|= Vx2 + Vy2

- The X and Y components of the drag force can be computed by:
FDx = -FD cos(q)

FDy = -FD sin(q)

where: q = arctan(Vy/Vx)

- Derive equations for ax and ay from FDx and FDy.
- Assuming ax and ay are constant during a brief instant of time, derive equations for Vx and Vy at time ti knowing Vx and Vy at time ti-1 .
- Assuming Vx and Vy are constant during a brief instant of time, derive equations for x and y at time ti knowing x and y at time ti-1 .

- Develop an Excel spreadsheet that describes the motion of a softball projectile:
1) neglecting drag and

2) including drag

More

- Plot the trajectory of the softball (Y vs. X)
- assuming no drag
- assuming drag

- Answer the following for each case:
- max. height of ball
- horizontal distance at impact with the ground

More

- Assume the projectile is a softball with the following parameters:
- W = 0.400 lbf
- m = 0.400 lbm
- Diameter = 3.80 in
- Initial Velocity = 100 ft/s at 30o
- CD = 0.6
- g = 32.174 ft/s2 (yes, assume you are on planet Earth)

More

- Reminder for the AES:
F = ma/gc

where gc = 32.174 (lbm ft)/(lbf s2)

- The equations of acceleration for this problem are:
ax = (FDx )gc/m

ay = (FDy -W)gc/m

More

- What is a reasonable Dt ?
- What happens to the direction of the drag force after the projectile reaches maximum height?

More