Upload
peter-vu
View
222
Download
1
Embed Size (px)
Citation preview
7/24/2019 EGCE 302-(3) Chapter 11 Curvilinear Motion of Particles-C-3
1/18
1
DynamicsEGCE 302
Spring 2016
California State University, Fullerton
Department of Civil and Environmental Engineering
Nagi Abo-Shadi, PhD, PE, SE, PEng
Chapter : Particle Kinematics
Curvilinear Motion of Particles
7/24/2019 EGCE 302-(3) Chapter 11 Curvilinear Motion of Particles-C-3
2/18
2
Chapter 11 Particle Kinematics
Curvilinear Motion of Particles
Curvilinear Motion:Any motion other than straight line
7/24/2019 EGCE 302-(3) Chapter 11 Curvilinear Motion of Particles-C-3
3/18
3
For small r = ds, v is
tangent to the curve
7/24/2019 EGCE 302-(3) Chapter 11 Curvilinear Motion of Particles-C-3
4/18
4
Acceleration
7/24/2019 EGCE 302-(3) Chapter 11 Curvilinear Motion of Particles-C-3
5/18
5
System of Coordinates
Magnitude and direction of velocity and acceleration
are determined with respect reference axis.
This is based on the coordinate system used; such as:
1. Rectangular Coordinates
2. Normal / Tangential Coordinates
3. Radial / Transverse Coordinates
7/24/2019 EGCE 302-(3) Chapter 11 Curvilinear Motion of Particles-C-3
6/18
6
1. Rectangular Coordinates (x, y ,z)
Position vector
Velocity vector
7/24/2019 EGCE 302-(3) Chapter 11 Curvilinear Motion of Particles-C-3
7/18
7
1. Rectangular Coordinates (x, y ,z)
Position vector
Acceleration vector
7/24/2019 EGCE 302-(3) Chapter 11 Curvilinear Motion of Particles-C-3
8/18
8
Example
7/24/2019 EGCE 302-(3) Chapter 11 Curvilinear Motion of Particles-C-3
9/18
9
= cos 25
= sin 25
= 0 = d cos 5= 0 = - d sin 5
= d cos 5=
-ds
in5
7/24/2019 EGCE 302-(3) Chapter 11 Curvilinear Motion of Particles-C-3
10/18
10
cos 25
X-Direction (Uniform motion)
= d cos 5=
-ds
in5
Eq 1
7/24/2019 EGCE 302-(3) Chapter 11 Curvilinear Motion of Particles-C-3
11/18
11
Y-Direction (Uniform acceleration)
- sin 5 = 0 + sin 25 t +
= d cos 5=
-ds
in5
Eq 2
7/24/2019 EGCE 302-(3) Chapter 11 Curvilinear Motion of Particles-C-3
12/18
12
Solve Eq 1 and Eq 2 for (d) and (t)
= 726.06 ft
cos 25
Eq 1
- sin 5 = 0 + sin 25 t +
Eq 2
7/24/2019 EGCE 302-(3) Chapter 11 Curvilinear Motion of Particles-C-3
13/18
13
Relative Motion
Position vector rB/Aof B relative to A
7/24/2019 EGCE 302-(3) Chapter 11 Curvilinear Motion of Particles-C-3
14/18
14
Example
7/24/2019 EGCE 302-(3) Chapter 11 Curvilinear Motion of Particles-C-3
15/18
15
Uniform Rectilinear Motion
As Given Shifted
7/24/2019 EGCE 302-(3) Chapter 11 Curvilinear Motion of Particles-C-3
16/18
16
Two (2) ways of dealing with this problem:
(1)
= 180 25 = 155
+ + = 180 = =/
7/24/2019 EGCE 302-(3) Chapter 11 Curvilinear Motion of Particles-C-3
17/18
17
(2)
= += +
= - += 48 = 109.5 km/hr
7/24/2019 EGCE 302-(3) Chapter 11 Curvilinear Motion of Particles-C-3
18/18
18
(2)
= +
= +
= 0 +
= 48 = 20.3 km/hr
=