m
kg
2000
8
cm
0
.
8
s
60
min
1
min
rev
800
1
m
s
N
10
1.8
9
2
5
3
2
2
2
5
sediment
≈
=
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
⋅
⎟
⎠
⎞
⎜
⎝
⎛
⋅
×
=
Δ
−
−
t
(
b
) In Problem 72 it was shown that
the rate of fall of the particles in air
is 2.42 cm/s. Find the time required
to fall 8.0 cm in air under the
influence of
gravity:
s
31
.
3
cm/s
42
.
2
cm
8.0
Δ
Δ
air
=
=
=
v
x
t
Find the ratio of the two times:
86
ms
38.47
s
31
.
3
Δ
Δ
sediment
air
≈
=
t
t
With the drag force in Problem 72 it takes about 86 times longer than it does
using the centrifuge.

Additional Applications of Newton’s Laws
461
Motion Along a Curved Path
74
•
A rigid rod with a 0.050-kg ball at one end rotates about the other end
so the ball travels at constant speed in a vertical circle with a radius of 0.20 m.
What is the maximum speed of the ball so that the force of the rod on the ball
does not exceed 10 N?
Picture the Problem
The force of
the rod on the ball is a maximum
when the ball is at the bottom of the
vertical circle. We can use Newton’s
second law to relate the force of the
rod on the ball to the mass
m
of the
ball, the radius
r
of its path, and the
constant speed of the ball along its
circular path. The diagram to the
right shows the forces acting on the
ball when it is at the bottom of the
vertical circle.
g
m
F
r
r
=
g
max
T
r
r
Apply
radial
radial
ma
F
=
∑
to the ball:
r
v
m
mg
T
2
max
max
=
−
Solving for
v
max
yields:
r
g
m
T
v
⎟
⎠
⎞
⎜
⎝
⎛
−
=
max
max
Substitute numerical values and evaluate
v
max
:
()
m/s
2
.
6
m
20
.
0
m/s
81
.
9
kg
050
.
0
N
10
2
max
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
=
v
75
•
[SSM]
A 95-g stone is whirled in a horizontal circle on the end of an
85-cm-long string. The stone takes 1.2 s to make one complete revolution.
Determine the angle that the string makes with the horizontal.

Chapter 5
462
Picture the Problem
The only forces
acting on the stone are the tension in
the string and the gravitational force.
The centripetal force required to
maintain the circular motion is a
component of the tension. We’ll solve
the problem for the general case in
which the angle with the horizontal is
θ
by applying Newton’s second law of
motion to the forces acting on the
stone.
Apply
∑
=
a
F
r
r
m
to the stone:
r
v
m
ma
T
F
x
2
c
cos
=
=
=
∑
(1)
and
0
sin
=
−
=
∑
mg
T
F
y
(2)
Use the right triangle in the diagram
to relate
r
,
L
, and
:
cos
L
r
=
(3)
Eliminate
T
and
r
between equations
(1), (2) and (3) and solve for
v
2
:
cos
cot
2
gL
v
=
(4)
Express the speed of the stone in
terms of its period:
rev
1
2
t
r
v
π
=
(5)
Eliminate
v
between equations (4)
and (5) and solve for
:
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
−
L
gt
2
2
rev
1
1
4
sin
Substitute numerical values and
evaluate
:
( )( )
()
°
=
⎥
⎦
⎤
⎢
⎣
⎡
=
−
25
m
0.85
4
s
1.2
m/s
9.81
sin
2
2
2
1
π
76
••
A 0.20-kg stone is whirled in a horizontal circle on the end of an 0.80-
m-long string. The string makes an angle of 20º with the horizontal. Determine
the speed of the stone.

Additional Applications of Newton’s Laws
463
Picture the Problem
The only forces
acting on the stone as it moves in a

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