Mechanics Fundamentals

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1. A particle is shot from the ground with an initial speed of 150 m/sec and an elevation angle of 60° with no acceleration in the z direction. (a) Formulate its trajectory, i.e. (horizontal displacement) and y (vertical displacement) vs. time history, for the entire flight period, assuming no air resistance and constant gravitational acceleration g = 9.8 m/s. Find its point of landing. (b) Find the eight alternative points of landing if you use a combination of the following values: 170 m/s and 130 m/s for the initial speed; 65 and 55° for the elevation angle and a constant lateral acceleration = +1 m/s and -1 m/s2 in the z direction. (c) What is the radius of a circle, with its center at the landing point found in part (a), that encloses all those alternative points of landing found part (b)? 2. A 8 lb rocket is launched from the ground with elevation angle of 85° with zero initial speed. The time history of its motor's thrust force is shown below. The thrust force stays at zero for t2 0.4 second. P, Iь 16 0.4 0.2 1, sec Assuming no air resistance, constant gravitational acceleration g = 32.2 fts? a kinetic friction coefficient k = 0.3 between the rocket and launch rail and a particle model for the rocket, what is the trajectory, or x (horizontal displacement) and y (vertical displacement) vs. time history of this rocket if it leaves the launch rail at the end of entire trust period i.e. t = 0.4 second. Use its position at t = 0 as the origin (x= 0 and y = 0). What is the length of the launch rail? 3. Redo problem #2 above if the launch rail length is set at 1 foot. You should try to draw free body diagrams and formulate the equations of motion for different phases of this problem first, assuming constant gravitational acceleration g = 32.2 ftus?, a particle model for the rocket and thrust in the direction of velocity, before you try to develop analytical solution for some of them and comment on the level of difficulty. Notes: This is an individual, not a team, assignment. Each student should submit his/her own solution The submittal deadline is the beginning of class on Tuesday, September 12. Presentation is worth 30% of the assignment points. Staple your submittal in the upper LH corner.
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Explanation & Answer

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1.
Initial velocity: 𝑣0 = 150 π‘š/𝑠𝑒𝑐
Elevation angle: 𝛼 = 60Β°
Initial height: β„Ž = 0 π‘š
Gravitational acceleration: 𝑔 = 9.8 π‘š/𝑠 2
(a) The horizontal displacement: π‘₯ = 𝑣0 𝑑 cos 𝛼 = 150𝑑 cos 60Β° = 75𝑑
𝒙 = πŸ•πŸ“π’•
The vertical displacement:
1
1
𝑦 = β„Ž + 𝑣0 𝑑 sin 𝛼 βˆ’ 𝑔𝑑 2 = 0 + 150𝑑 sin 60Β° βˆ’ Γ— 9.8 Γ— 𝑑 2
2
2
π’š = πŸ•πŸ“βˆšπŸ‘π’• βˆ’ πŸ’. πŸ—π’•πŸ
The particle landed when y =0
∴ 𝑦 = 75√3𝑑 βˆ’ 4.9𝑑 2 = 0 ∴ 75√3 βˆ’ 4.9𝑑 = 0 ∴ 𝑑 =

75√3
= 26.51 𝑠
4.9

Then π‘₯ = 75𝑑 = 75 Γ— 26.51 = 1988.32 β‰ˆ 1988 π‘š
Then, the point of landing: (π‘₯, 𝑦) = (πŸπŸ—πŸ–πŸ–, 𝟎)
1

1

2

2

(b) The landing point is: (π‘₯, 𝑦) = (𝑣0 𝑑 cos 𝛼 , 𝑣0 𝑑 sin 𝛼 βˆ’ 𝑔𝑑 2 , 𝑣0𝑧 𝑑 + π‘Žπ‘§ 𝑑 2 )
Where: initial velocity in z-direction: 𝑣0𝑧 = 0 π‘š/𝑠
π‘Žπ‘§ is the lateral acceleration in z-direction.
So, the landing point when y =0 is:
1

1

2

2

(π‘₯, 𝑦, 𝑧) = (𝑣0 𝑑 cos 𝛼 , 𝑣0 𝑑 sin 𝛼 βˆ’ 𝑔𝑑 2 , βˆ’ π‘Žπ‘§ 𝑑 2 )
1- π’—πŸŽ = πŸπŸ•πŸŽ π’Ž/𝒔 , 𝜢 = πŸ”πŸ“Β°, 𝒂𝒛 = +𝟏 π’Ž/π’”πŸ :
1
2 Γ— 170 sin 65Β°
∴ 𝑦 = 170𝑑 sin 65Β° βˆ’ Γ— 9.8 Γ— 𝑑 2 = 0 ∴ 𝑑 =
= 31.44 𝑠
2
9.8
So, the landing point is:
1
(π‘₯, 𝑦, 𝑧) = (170 Γ— 31.44 Γ— cos 65Β° , 0, Γ— 1 Γ— (31.44)2 )
2
(π‘₯, 𝑦, 𝑧) = (πŸπŸπŸ“πŸ—, 𝟎, πŸ’πŸ—πŸ’)

Page | 1

2- π’—πŸŽ = πŸπŸ•πŸŽ π’Ž/𝒔 , 𝜢 = πŸ”πŸ“Β°, 𝒂𝒛 = βˆ’πŸ π’Ž/π’”πŸ :
1
2 Γ— 170 sin 65Β°
∴ 𝑦 = 170𝑑 sin 65Β° βˆ’ Γ— 9.8 Γ— 𝑑 2 = 0 ∴ 𝑑 =
= 31.44 𝑠
2
9.8
So, the landing point is:
1
(π‘₯, 𝑦, 𝑧) = (170 Γ— 31.44 Γ— cos 65Β° , 0, Γ— (βˆ’1) Γ— (31.44)2 )
2
(π‘₯, 𝑦, 𝑧) = (πŸπŸπŸ“πŸ—, 𝟎, βˆ’ πŸ’πŸ—πŸ’)
3- π’—πŸŽ = πŸπŸ•πŸŽ π’Ž/𝒔 , 𝜢 = πŸ“πŸ“Β°, 𝒂𝒛 = +𝟏 π’Ž/π’”πŸ :
1
2 Γ— 170 sin 55Β°
∴ 𝑦 = 170𝑑 sin 55Β° βˆ’ Γ— 9.8 Γ— 𝑑 2 = 0 ∴ 𝑑 =
= 28.42 𝑠
2
9.8
So, the landing point is:
1
(π‘₯, 𝑦, 𝑧) = (170 Γ— 28.42 Γ— cos 65Β° , 0, Γ— 1 Γ— (28.42)2 )
2
(π‘₯, 𝑦, 𝑧) = (πŸπŸŽπŸ’πŸ, 𝟎, πŸ’πŸŽπŸ’)
4- π’—πŸŽ = πŸπŸ•πŸŽ π’Ž/𝒔 , 𝜢 = πŸ“πŸ“Β°, 𝒂𝒛 = βˆ’πŸ π’Ž/π’”πŸ :
1
2 Γ— 170 sin 55Β°
∴ 𝑦 = 170𝑑 sin 55Β° βˆ’ Γ— 9.8 Γ— 𝑑 2 = 0 ∴ 𝑑 =
= 28.42 𝑠
2
9.8
So, the landing point is:
1
(π‘₯, 𝑦, 𝑧) = (170 Γ— 28.42 Γ— cos 65Β° , 0, Γ— (βˆ’1) Γ— (28.42)2 )
2
(π‘₯, 𝑦, 𝑧) = (πŸπŸŽπŸ’πŸ, 𝟎, βˆ’πŸ’πŸŽπŸ’)
5- π’—πŸŽ = πŸπŸ‘πŸŽ π’Ž/𝒔 , 𝜢 = πŸ”πŸ“Β°, 𝒂𝒛 = +𝟏 π’Ž/π’”πŸ :
1
2 Γ— 130 sin 65Β°
∴ 𝑦 = 130𝑑 sin 65Β° βˆ’ Γ— 9.8 Γ— 𝑑 2 = 0 ∴ 𝑑 =
= 24.04 𝑠
2
9.8
So, the landing point is:
1
(π‘₯, 𝑦, 𝑧) = (130 Γ— 24.04 Γ— cos 65Β° , 0, Γ— 1 Γ— (24.04)2 )
2
(π‘₯, 𝑦, 𝑧) = (πŸπŸ‘πŸπŸ, 𝟎, πŸπŸ–πŸ—)
π’Ž

6- π’—πŸŽ = πŸπŸ‘πŸŽ , 𝜢 = πŸ”πŸ“Β°, 𝒂𝒛 = βˆ’πŸ π’Ž/π’”πŸ :
𝒔

1
2 Γ— 130 sin 65Β°
∴ 𝑦 = 130𝑑 sin 65Β° βˆ’ Γ— 9.8 Γ— 𝑑 2 = 0 ∴ 𝑑 =
= 24.04 𝑠
2
9.8
Page | 2

So, the landing point is:
1
(π‘₯, 𝑦, 𝑧) = (130 Γ— 24.04 Γ— cos 65Β° , 0, Γ— (βˆ’1) Γ— (24.04)2 )
2
(π‘₯, 𝑦, 𝑧) = (πŸπŸ‘πŸπŸ, 𝟎, βˆ’πŸπŸ–πŸ—)
7- π’—πŸŽ = πŸπŸ‘πŸŽ π’Ž/𝒔 , 𝜢 = πŸ“πŸ“Β°, 𝒂𝒛 = +𝟏 π’Ž/π’”πŸ :
1
2 Γ— 130 sin 55Β°
∴ 𝑦 = 130𝑑 sin 55Β° βˆ’ Γ— 9.8 Γ— 𝑑 2 = 0 ∴ 𝑑 =
= 21.73 𝑠
2
9.8
So, the landing point is:
1
(π‘₯, 𝑦, 𝑧) = (130 Γ— 21.73 Γ— cos 55Β° , 0, Γ— 1 Γ— (21.73)2 )
2
(π‘₯, 𝑦, 𝑧) = (πŸπŸ”πŸπŸŽ, 𝟎, πŸπŸ‘πŸ”)
8- π’—πŸŽ = πŸπŸ‘πŸŽ π’Ž/𝒔 , 𝜢 = πŸ“πŸ“Β°, 𝒂𝒛 = βˆ’πŸ π’Ž/π’”πŸ :
1
2 Γ— 130 sin 55Β°
∴ 𝑦 = 130𝑑 sin 55Β° βˆ’ Γ— 9.8 Γ— 𝑑 2 = 0 ∴ 𝑑 =
= 21.73 𝑠
2
9.8
So, the landing point is:
1
(π‘₯, 𝑦, 𝑧) = (130 Γ— 21.73 Γ— cos 55Β° , 0, Γ— (βˆ’1) Γ— (21.73)2 )
2
(π‘₯, 𝑦, 𝑧) = (πŸπŸ”πŸπŸŽ, 𝟎, βˆ’πŸπŸ‘πŸ”)
(c) The landing point in part (a) is: (1988, 0,0)
The farthest points from the center point (1988, 0,0) are: (1321, 0, βˆ’289)
and (1321, 0, 289) and they have equal distance from the center.
The radius of the circle is the distance between the points (1988, 0,0)
and(1321, 0, 289):
π‘Ÿ = √(1321 βˆ’ 1988)2 + (0 βˆ’ 0)2 + (289 βˆ’ 0)2 = πŸ•πŸπŸ• π’Ž

Page | 3

2.

x
πΉπ‘‘β„Ž
πΉπ‘‘β„Ž
y
𝐹𝑁

π‘Š cos πœƒ

πœ‡π‘˜ 𝐹𝑁
π‘Š sin πœƒ

π‘Š

85Β°

𝐹𝑁 = π‘Š cos πœƒ = 8 cos 85Β° = 0.6972 lb
- For 𝟎 < 𝒕 ≀ 𝟎. 𝟐
πΉπ‘‘β„Ž βˆ’ πœ‡π‘˜ 𝐹𝑁 βˆ’ π‘Š sin πœƒ =
βˆ΄π‘Ž=

π‘Š
8
π‘Ž ∴ 16 βˆ’ 0.3 Γ— 0.6972 βˆ’ 8 sin 85Β° =
π‘Ž
𝑔
32.2

32.2
Γ— (16 βˆ’ 0.3 Γ— 0.6972 βˆ’ 8 sin 85Β°) = 31.48 𝑓𝑑/𝑠 2
8

To get the displacement:
1
1
𝑑1 = 𝑣0 𝑑 + π‘Žπ‘‘ 2 = 0 Γ— 0.2 + Γ— 31.48 Γ— 𝑑 2 = πŸπŸ“. πŸ•πŸ’ π’•πŸ 𝒇𝒕
2
2
At t= 0.2 s, 𝑑1 = 15.74 Γ— 0.22 = 0.6296 𝑓𝑑
The horizontal displacement: π‘₯1 = 𝑑1 π‘π‘œπ‘ πœƒ = 15.74 𝑑 2 cos 85Β° = 𝟏. πŸ‘πŸ• π’•πŸ 𝒇𝒕
The vertical displacement: 𝑦1 = 𝑑1 sin πœƒ = 15.74 𝑑 2 sin 85Β° = πŸπŸ“. πŸ”πŸ– π’•πŸ 𝒇𝒕
- To get the final velocity at the first interval which is equal to initial velocity of the
seco...


Anonymous
Really useful study material!

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