NAME:
PAGE 1
Problem 1 (5p)
One-Dimensional Motion: Velocity
A particle is released from rest in the x direction from initial position x0 = 5m with acceleration
a = 9x m/s2 , where x is in meters. Determine the velocity of the particle as a function of x.
Solution:
a(x)dx = vdv
Integrating we get:
9
Z x
xdx =
Z
vdv
x0
2
9(x − x20 ) = v 2
Answer:
v = 3(x2 − 25)1/2
MAE242 EXAM March 6, 2008
NAME:
Problem 2 (5p)
PAGE 2
Relative Motion.
Two planes, A and B are flying at the same altitude. If their velocities are vA and vB , such that the angle
between their straight-line courses is θ, determine the magnitude of the relative velocity vAB of plane B
with respect to plane A, and the angle, α (in degrees), the velocity vector makes with the direction of
plane B.
Given: vA , vB , α
Directions: Select the X axis in the opposite direction of vB vector (see the figure), and Y axis in the
vertical direction.
A.
Solution:
Relative velocity:
~vAB = ~vB − ~vA
Projecting on the horizontal and vertical axes, we get:
(vAB )x = vB − (−vA cos θ) = vB + vA cos θ
(vAB )y = vA sin θ
After we found the components we can compute both the magnitude of velocity, vB and its angle with the
horizontal direction, α:
vAB =
α=
q
(vAB )2x + (vAB )2y =
q
(vB + vA cos θ)2 + (vA sin θ)2
180o
(aAB )y
arctan
π
(aAB )x
MAE242 EXAM March 6, 2008
(1)
NAME:
Problem 3 (5p)
PAGE 3
Mine Car.
The mine car of mass m is hoisted up the incline using the cable and motor M. For a short time, the force
in the cable is F = Ct2 . If the car has an initial velocity v at s = 0 and t = 0, determine the distance it
moves up the plane as a function of t. Given: m, C, t, v.
A.
Solution:
Write the balance of forces on the direction of motion:
F − mg sin α = ma
where α is the angle of the incline with the horizontal direction. From the figure:
sin α =
8
17
Thus the acceleration is:
a(t) =
8
F (t)
−g
m
17
Using
a(t) =
dv
dt
separating variables and integrating, we get:
Integrating:
Z t
a(t)dt =
Z v
0
v1
dv = v − v1
(2)
where
Z t
a(t)dt =
Z t
C
0
0
8
t −g
dt
m
17
2
C t3 8gt
=
−
m3
17
(3)
Combining (2) and (3), we get:
v = v1 −
8gt Ct3
+
17
3m
MAE242 EXAM March 6, 2008
NAME:
And using: v =
PAGE 4
ds
,
dt
we get:
Z s
0
ds =
Z t
vdt =
0
Z t
0
8gt Ct3
v1 −
+
17
3m
4gt2
Ct4
= v1 t −
+
17
12m
Thus
s = v1 t −
4gt2
Ct4
+
17
12m
MAE242 EXAM March 6, 2008
!
NAME:
Problem 4 (5p)
PAGE 5
A Centrifuge
A man lies against the cushion for which the coefficient of static friction is µ. If he rotates about the z
axis on a radius R with a constant speed v, determine the smallest angle α of the cushion, at which he
will begin to slip off. Given: R, µ, v.
Solution
See P.13-66
v2
R
−µ N sin α + N cos α = mg
µ N cos α + N sin α = m
Dividing the two equations, we get:
v2
µ + tan α
=
1 − µ tan α
R
tan α =
v 2 /R − µ
1 + µ v 2 /R
MAE242 EXAM March 6, 2008