Rocket Science quiz 4

February 15, 2007

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Question 1.

A 22 lbs spherical payload is dropped from a high building. It falls and reaches terminal velocity. If the object has the following physical characteristics find the terminal velocity of the object.

Weight = W =22 lbs

Cross sectional area opposing the free stream flow = A = 4 ft²

Density of air = ρ = 0.075 lbs/ft³

Drag coefficient = C_d = 7 sec²/ft

Part 1. Draw a free body diagram and label the diagram with forces

Weigh

Part 2. Right down the force equation in the vertical direction. (What are the names of forces acting on the sphere in equation form?)

F _vertical = F_drag –Weight = F_d -W

Part 3. Find the acceleration in the vertical direction at terminal velocity in the down ward direction in ft/sec²

a _vertical = 0 m/s₂

At terminal velocity is v = constant = the change in v is 0

F _vertical = F_drag –Weight = F_d –W = 0

F/m = a _vertical = 0

then F = 0 = F_d –W

this implies à F_d = W

Part 4. Find the drag force at terminal velocity in pounds

F_drag = 22 LBS

Since: F_d = W = 22 lbs

Part 5. Solve for the terminal velocity in ft/sec

V_terminal = 4.577 ft/sec = 3.12 mph

F_d = W = 22 lbs

F_d = ½ ρ C_d A V_term²

W = ½ ρ C_d A V_term²

V_term² = 2W/( ρ C_d A )

V_term = Square root (2 x 22 lbs/((0.075 lbs/ft³) x (7 sec²/ft) x (4 ft²))
Once at terminal velocity the spherical payload encounters a horizontal wind of 20 ft/sec that blows from left to right on the page.

Part 6. Draw a free body diagram of all the forces acting on the shpere and label the diagram with forces

Part 7. Right down the force equation in the horizontal direction. (What are the names of forces acting on the sphere in equation form?)

F_horizontal = Drag _{due to wind} = D_wind-drag

Part 8. Find the force, in lbs, acting on the sphere in the horizontal direction.

F_horizontal = F_wind-drag = ½ ρ C_d A V_wind² = 420 lbs

Show your calculations below

F_wind-drag = ½ ρ C_d A V_wind²

F_wind-drag = ((0.075 lbs/ft³) x (7 sec²/ft) x (4 ft²) x 20 ft/sec)/2

F_wind-drag = 420 lbs

Part 9. Find the acceleration in the horizontal direction. The mass = Weight / 32ft/sec²

a_horizontal = F_wind-drag/ mass

a_horizontal = F_wind-drag / (W/32 ft/sec²) = 32 ft/sec² x F_winddrag/W

a_horizontal = (420 lbs/ 22 lbs) x 32 ft/sec²

a_horizontal = 610 ft/sec²

a_horizontal = 19.1 g’s

Show your calculations below

Part 10. Does the drag force remain constant on the sphere? Explain your answer.

In the vertical direction the force due to gravity and the force due to the vertical drag cancel each other. Thus the drag in the vertical direction remains constant.

In the horizontal direction the drag is initially a maximum. As the rocket gains speed it eventually travels at the wind speed and the relative velocity between the tow becomes zero and so does the drag force in the horizontal direction.

The final total velocity of the rocket will be

V_final = V _vertical + V _horizontal

V _vertical = V _terminal in the downward direction

V _horizontal = V _wind moving to the right