10
o
v
g
d
Starting with sketching the coordinate vectors, ˆ
e
t
is parallel to the velocity
pointing to the right and slightly up, and ˆ
e
n
points down and slightly to the
right. Based on our kinematic expression, we can write the acceleration vector
as the following.
a
= ˙
v
ˆ
e
t
+
1000
2
ρ
ˆ
e
n
(2.55)
From the given information, we know that the acceleration vector is composed
of two physical components.
The drag component acts opposite the velocity
vector, in the
−
ˆ
e
t
direction. The gravity component acts straight down with
components in both directions.
a
=
−
1160ˆ
e
t
+ 9
.
81 (cos 10
◦
ˆ
e
n
−
sin 10
◦
ˆ
e
t
)
(2.56)
From these two expressions for the acceleration vector, we extract the scalar
equations.
˙
v
=
−
1160
−
9
.
81 sin10
◦
(2.57)
1000
2
ρ
= 9
.
81 cos10
◦
(2.58)
Solving for the unknowns, ˙
v
=
−
1162 m/s
2
and
ρ
= 103
,
500 m = 103
.
5 km.

32
CHAPTER 3.
PARTICLE KINETICS
3.3
Example Problems
Example 3.1
Consider a fuzzy die hanging on a cord from the rearview mirror of a car. The
car is decelerating at 0
.
2
g
. Find the steady-state angle the cord makes with the
vertical.
θ
i
^
j
^
1. Kinematics - Let’s use the coordinate vectors indicated in the diagram. In the
steady-state condition, the die’s acceleration is the same as the car:
a
= 0
.
2
g
ˆ
i
.
2. FBD - In sketching your own FBD, the only forces acting on the die are
gravity,
mg
, and the tension in the cord,
T
. The resulting force expression is
Σ
F
=
T
sin
θ
ˆ
i
+ (
T
cos
θ
−
mg
)
ˆ
j
.
3. N2L - Substituting into Newton’s second law:
T
sin
θ
ˆ
i
+ (
T
cos
θ
−
mg
)
ˆ
j
= 0
.
2
mg
ˆ
i
(3.5)
Extracting scalar equations of motion:
T
sin
θ
= 0
.
2
mg
(3.6)
T
cos
θ
−
mg
= 0
From these equations, solve for
θ
.
tan
θ
= 0
.
2
⇒
θ
= 11
.
31
◦
(3.7)

34
CHAPTER 3.
PARTICLE KINETICS
Example 3.3
The rod rotates about
B
in the horizontal plane with a constant angular rate
˙
θ
= 0
.
05 rad/s. Collar
A
has a mass of 0.3 kg and slides along the rod without
friction. At the current instant, the collar is at a radius of
r
= 0
.
15 m and is
sliding along the rod with ˙
r
= 0
.
2 m/s. Find the value of ¨
r
and the force exerted
on the collar by the bar.
0.15 m
θ
= 0.05 rad/s
.
A
B
1. Kinematics - Choosing polar coordinate vectors, ˆerpoints along the rod, andˆeθpoints up and to the left.Using the given information, we can write anexpression for the acceleration vector.
2. FBD - In sketching your own FBD, the rod exerts a force
N
on the collar.
The force is perpendicular to the rod, but we’re not sure if it’s in the positive or
negative ˆ
e
θ
direction. So for now, let’s just guess it’s in the positive ˆ
e
θ
direction.
Σ
F
=
N
ˆ
e
θ
(3.14)
Of course, gravity is also acting on the collar. But since we’re focused on the
motion in the horizontal plane, we can neglect gravity.
3. N2L - Note that the collar has a mass of 0.3 kg.
N
ˆ
e
θ
= 0
.
3
(
¨
r
−
3
.
75
·
10
−
4
)
ˆ
e
r
+ 0
.
02ˆ
e
θ
(3.15)
Extracting scalar equations of motion:
0 = 0
.
3
(
¨
r
−
3
.
75
·
10
−
4
)
(3.16)
N
= 0
.
3
·
0
.
02
From these equations, ¨
r
= 3
.
75
·
10
−
4
m/s
2
and
N
= 0
.
006 N. We had guessed
that
N
acts in the positive ˆ
e
θ
direction. Since we got a positive answer for
N
,
this guess must have been correct.

48
CHAPTER 3.
PARTICLE KINETICS
Example 3.11
Block
A
has a mass of 50 kg, and block

#### You've reached the end of your free preview.

Want to read all 25 pages?

- Spring '08
- josephmansour
- Acceleration, Force, Friction, Angular velocity