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Applying Newton&s Laws
523
SET UP:
The person moves in a circle of radius
3.00 m
(5.00 m)sin30.0
5.50 m
R
=+
=
°
. The freebody diagram
is given in Figure 5.52.
F
!
is the force applied to the seat by the rod.
EXECUTE:
(a)
y
y
Fm
a
=
∑
gives
cos30.0
g
=
°
and
cos30.0
mg
F
=
°
.
x
x
a
=
∑
gives
2
sin30.0
v
R
=
°
.
Combining these two equations gives
2
tan
(5.50 m)(9.80 m/s )tan30.0
5.58 m/s
vR
g
θ
==
=
°
. Then the period
is
22
(
5
.
5
0
m
)
6.19 s
5.58 m/s
R
T
v
ππ
=
.
(b)
The net force is proportional to
m
so in
m
=
∑
F
a
!
!
the mass divides out and the angle for a given rate of
rotation is independent of the mass of the passengers.
EVALUATE:
The person moves in a horizontal circle so the acceleration is horizontal. The net inward force
required for circular motion is produced by a component of the force exerted on the seat by the rod.
Figure 5.52
5.53.
IDENTIFY:
Apply
m
∑
F
=a
!
!
to the composite object of the person plus seat. This object moves in a horizontal
circle and has acceleration
rad
a
, directed toward the center of the circle.
SET UP:
The freebody diagram for the composite object is given in Figure 5.53. Let
x
+
be to the right, in the
direction of
rad
a
!
. Let
y
+
be upward. The radius of the circular path is
7.50 m
R
=
. The total mass is
2
(255 N
825 N)/(9.80 m/s )
110.2 kg
+=
. Since the rotation rate is 32.0 rev/min
0.5333 rev/s
=
, the period
T
is
1
1.875 s
0.5333 rev/s
=
.
EXECUTE:
y
y
a
=
∑
gives
cos40.0
0
A
Tm
g
−
=
°
and
255 N
825 N
1410 N
cos40.0
cos40.0
A
mg
T
+
=
°°
.
x
x
a
=
∑
gives
rad
sin40.0
AB
TT
m
a
°
and
44
(
7
.
5
0
m
)
sin40.0
(110.2 kg)
(1410 N)sin40.0
8370 N
(1.875 s)
BA
R
T
T
=−
=
−
=
.
The tension in the horizontal cable is 8370 N and the tension in the other cable is 1410 N.
EVALUATE:
The weight of the composite object is 1080 N. The tension in cable
A
is larger than this since its
vertical component must equal the weight.
rad
9280 N
ma
=
. The tension in cable
B
is less than this because part of
the required inward force comes from a component of the tension in cable
A
.
Figure 5.53
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Chapter 5
5.54.
IDENTIFY:
Apply
m
∑
F
=a
!
!
to the button. The button moves in a circle, so it has acceleration
rad
a
.
SET UP:
The situation is equivalent to that of Example 5.22.
EXECUTE:
(a)
2
s
v
Rg
μ
=
. Expressing
v
in terms of the period
T
,
2
R
v
T
π
=
so
2
s
2
4
R
Tg
=
. A platform speed of
40.0 rev/min corresponds to a period of 1.50 s, so
2
s
22
4
(0.150 m)
0.269.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics, Force

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