Class 3
MAE 242: DYNAMICS
Dr. Samir N. Shoukry
Professor, MAE & CEE
assistants
K. Mc Bride
Dhananjy Rao
ony Oomen
Praveen
© Samir N. Shoukry, 2004, Dynamics MAE 242
Quiz 2 (5 minutes)
A car accelerates according to the relation a=0.02s m/s2.
Determine its velocity when s=100 m if s=v=0 when t=0.
© Samir N. Shoukry, 2004, Dynamics MAE 242
How Position May Be Described?
Morgantown
M
rM/W
New York
N
North
rN/W
East
West
W
South
Point of reference that may be placed
anywhere …, say Charleston, WV
Position is defined relative to a point on which we set a coordinate system
© Samir N. Shoukry, 2004, Dynamics MAE 242
OR
Morgantown
N
Morganown
W
Path
E
S
Position
Vector
New York
rM/W
North
rN/M
Morganown
East
West
New York
South
Path
rM/N
Position
Vector
W
rN/W
N
New York
E
S
Position is defined by a vector “r”
that originates from a reference point
Remember: A vector has a length and direction
© Samir N. Shoukry, 2004, Dynamics MAE 242
Rectangular Coordinates
Position is defined by the distance from
three perpendicular lines from an origin O.
Y
y
Z
r
Y

X
x
X
r=xi+yj
Length of
r=xi+yj+zk
Z
r
 γβ
Length of r
r
r x y
2
2
Direction of r
y
  tan 1 ( )
x
r x y z
Direction of r
2
cos 
|r|
x
2
,
2
cos  
© Samir N. Shoukry, 2004, Dynamics MAE 242
y
z
x
X
|r|
y
,
cos  
|r|
z
Y
Rectangular Coordinates:
Velocity & Acceleration
Z
Y
X
r=xi+yj+zk
v=
dr d (xi)
dt = dt
+ ddt (yj) + ddt(zk)
di dx
dx
d
i
+
x
dt = dt i
dt (xi)= dt
v = vx i + vy j + v z k
v  vx2  v y2  vz2
a = ax i + ay j + az k
a  a x2  a y2  a z2
© Samir N. Shoukry, 2004, Dynamics MAE 242
=vxi
Example: Given r  t 2i  2t 3 j  k , Calculate the magnitude and
direction of the velocity and acceleration vectors. Calculate the magnitude
of the position vector.
dr
v
 2ti  6t 2 j
dt
 vx  2t
,
2 2
y
6t 2
2
6t
v  tan
2t
1
v  (2t )  (6t )
2
v y  6t
&
2
v
2t
a
dv
 2i  12tj
dt
a  (2) 2  (12t ) 2
rx  t 2
,
ax  2
,
a  tan 1
ry  23t 3
&
a y  12t
12t
2
&
rz  1
r  (t 2 )2  (2t 3 )2  (1)2
© Samir N. Shoukry, 2004, Dynamics MAE 242
Motion of A Projectile
v  v0  a t
ax i  0
ay j  gj
vx  v0 x
1
s  s0  v0 t  a t 2
2
v 2  v02  2 a ( s  s0 )
vx  v0 x
v0
v0 y

vy
v0 x
vx  v0 x
v
x  x0  vx 0 t
yf
xup
xdown
y0
R
x0
1
x  x0  vx 0 t  a x t 2
2
vx2  v02x  2 a ( x  x0 )
If ax  0

ymax
In x Direction :
vx  v0 x  ax t
In y Direction :
v y  v0 y  g t
1 2
gt
2
v y2  v02 y  2 g ( y  y0 )
y  y0  v y 0 t 
x
© Samir N. Shoukry, 2004, Dynamics MAE 242
Class 4
MAE 242: DYNAMICS
Instructor
Dr. Samir N. Shoukry
Professor, MAE & CEE
assistants
Dr. G. William, CEE
Mr. K. Mc Bride, MAE
© Samir N. Shoukry, 2004, Dynamics MAE 242
Quiz 3 (5 minutes)
A particle moves according to the relation r=5t3 i +2t2 j
Calculate the magnitude and direction of its acceleration
when t =0.5 second.
© Samir N. Shoukry, 2004, Dynamics MAE 242
Example: A particle accelerates according to the relation: a  x i  vy j  5tk
Given the initial conditions at t=0: x, y, z =0 and vx=vy=1 & vz=0, calculate the
particle velocity & position.
dvx dx dvx dx dvx
dv


 vx x
dt dx dt
dt dx
dx
ax  x 
vx
x
x 2 vx2 1
0 x dx  1 vx dx  2  2  2
vx  x  1
2
ay 
t
dv y
dt
vy
 dt  
0
az 
vy
1

 v y  dt 
dv y
dx
vx 
,
dt
t
x
0
0
 dt  
dx
x2  1
, t  sinh 1 x  x  sinh t
dv y
vy
y
t
0
0
t
t
dy

e
dt

y

e
1
 
 t  ln v y  v y  e ,
t
dvz
 5t  dvz  5t dt
dt
vz
t
0
0
 dvz   5t dt
 vz  2.5t 2 ,
z
t
0
0
2
 dz   2.5t
5
 z  t3
6
© Samir N. Shoukry, 2004, Dynamics MAE 242
Motion of A Projectile
v  v0  a t
ax i  0
ay j  gj
vx  v0 x
1
s  s0  v0 t  a t 2
2
v 2  v02  2 a ( s  s0 )
vx  v0 x
v0
v0 y

vy
v0 x
vx  v0 x
v
x  x0  vx 0 t
yf
xup
xdown
y0
R
x0
1
x  x0  vx 0 t  a x t 2
2
vx2  v02x  2 a ( x  x0 )
If ax  0

ymax
In x Direction :
vx  v0 x  ax t
In y Direction :
v y  v0 y  g t
1 2
gt
2
v y2  v02 y  2 g ( y  y0 )
y  y0  v y 0 t 
x
© Samir N. Shoukry, 2004, Dynamics MAE 242
A skier leaves the ramp A at angle  A  25 with the horizontal.
He strikes the ground at point B, Determine his initial speed
vA and the time of flight from A to B.
vA sin  A
vA cos  A
-64
3
100( )  60
5
Inetial and final positions are given. Thus let us use the
1
position relation S=S0  v0t  at 2
2
In X direction : 80  0  (v A cos 25)t  0
InY direction :  64  0  (v A sin 25)t 
Solve Eq. 1& 2  v A  19.41 m / s
..........(1)
1 2
gt ...........(2)
2
AND t  4.547 s
© Samir N. Shoukry, 2004, Dynamics MAE 242
4
100( )  80
5
15sin 60  12.99 ft / s
15cos 60  7.5 ft / s
y
x
Initial position and velocity are given. The relation between
x and y components of he final position is given.
1
Apply s  s0  v0t  at 2 in x and y directions.
2
In x direction : x  0  7.5 t
............(1)
1
In y direction : y  0  12.99t  gt 2 ...........(2)
2
Also we have : y  0.05 x 2
............(3)
Solve for x & y
 x  5.15 ft
and
y  1.33 ft
© Samir N. Shoukry, 2004, Dynamics MAE 242
14
ft / s
5
y
5 5
10
28
14

5 5
5
5m
Initial velocity and position are given, also given x  coordinate of the final position.
1
Apply : s  s0  v0t  at 2
2
28
In x direction : 10  0 
t  t  0.7986 s
5
14
1
In y direction : y  0 
t  gt 2  y  1.8718 m  h  5  1.8718  3.1282 m
2
5
Another Methode : Determine the distance the squirrel drops in time t.
1
h  gt 2  .5*9.81*.7986 2  3.1282 m
2
© Samir N. Shoukry, 2004, Dynamics MAE 242
The boy at A attempts to throw a ball over the roof
of a barn such that it is launched at an angle of 40
degrees. Determine the speed vA at which he must
throw the ball so that it reaches its maximum height
at C. Also, find the distance d where the boy must
stand so that he can make the throw.
Given: Y0 = 1m, At max. height, vY = 0, Y = 8m and v0x = v0cos400; v0y = v0sin400
(1) X = v0xt;
(2) vx = v0x
(constant velocity in x-direction);
(3) Y = Y0 1/2gt2 + v0Yt
(4) vY = v0Y gt (constant acceleration -g in y-direction)
At the top of the trajectory vy = 0 and from (4) we get: v0y = gt
Substituting the above into (3) we get 7 = ½ g t2 and solving the above to get the
time to reach maximum elevation.
t = 1.195 seconds
Substituting for v0Y at the top of the trajectory, we get v0sin40o = 9.81(1.195)
which is solved for the initial velocity.
v0 = 18.24 m/sec
Substituting into (1) we get the total distance X.
X = 18.24(cos40)(1.195) = 16.7 m
Which gives d = X 4 = 12.70m
© Samir N. Shoukry, 2004, Dynamics MAE 242
The ball is thrown from the tower with a velocity of 20 ft/s as
shown. Determine the x and y coordinates to where the ball
strikes the slope. Also, determine the speed at which the ball hits
the ground.
Initial Conditions:
v0 = 20 ft/s v0x = 20(3/5) = 12 ft/s v0y = 20(4/5) = 16 ft/s
theta = 26.60 , X0 = 0
Y0 = 80 ft
Motion in the x-direction at a constant velocity.
(1) vx = v0x = 12 ft/s
(2) X = vxt = 12t
Motion in the y-direction at a constant acceleration a = -g. Integrating for velocity
(3) vy = v0y - gt = 16 - 32.2(t)
and the y-displacement is obtained by integrating the latter, which gives
Y = -1/2gt2 + v0yt + Y0
or (4) Y = -16.1t2 +16t + 80
The x and y coordinates of the ball when it hits the ground can be written as.
; (6) y = Rsin26.6o
(5) x = 20 + Rcos26.6o
Equations (2), (4) (5) and (6) are four equations in four unknowns. They can be solved
for t - the time when the ball hits the ground. Thus t = 2.7 sec.
Substituting into (2) and (4) provides the x and y coordinates of the ball when it hits the
ground.
x = 32.4 ft;
y = 6.56 ft
To solve for the speed, we substitute t = 2.7s into equation (3) which gives.
vx = 12 ft/s
and vy = 70.9 ft/s
The total speed is found from v = (vx2 + vy2)1/2 ; v = 71.9ft/s
© Samir N. Shoukry, 2004, Dynamics MAE 242