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Phys 205
Chapter 5 Solutions
5.3
The Earth has a radius of 6
.
4
×
10
6
m ad completes one revolution about its axis in 24 h. (a) Find the speed
of a point on the equator. (b) Find the speed of New York City.
This is a kinematics of uniform circular motion problem.
Given:
Radius of the Earth:
r
E
= 6
.
4
×
10
6
m
Period of rotation:
T
= 24
h
= 86
,
400 s
(a) We want to ﬁnd the tangential speed of a point at the equator. Use Equation 5.1.
v
Eq
=
2
πr
E
T
v
Eq
=
2
π
(
6
.
4
×
10
6
m
)
86
,
400 s
= 465 m
/
s
(b) New York City has a latitude of 40.7
◦
so the radius of its orbit will not be the full radius of the Earth (see
the ﬁgure for Problem 5.9 below). The radius of its orbit will be:
r
=
r
E
cos 40
.
7
◦
= 4
.
85
×
10
6
m.
v
NY
=
2
πr
T
v
NY
=
2
π
(
4
.
85
×
10
6
m
)
86
,
400 s
= 353 m
/
s
5.6
Consider the motion of the hand of a mechanical clock. If the minute hand of the clock has a length of 6.0 cm,
what is the centripetal acceleration of a point at the end of the hand?
This is a kinematics of uniform circular motion problem.
Given:
Period
T
= 60 min = 3600 s
Radius:
r
= 0
.
060 m
We want to ﬁnd the centripetal acceleration,
a
c
. Use Equations 5.5 and 5.1.
a
c
=
v
2
r
=
1
r
±
2
πr
T
²
2
→
a
c
=
4
π
2
r
T
2
This last equation is useful to use for problems like this one when you are given the period & radius and want
the centripetal acceleration.
a
c
=
4
π
2
(0
.
060 m)
(3600 s)
2
= 1
.
8
×
10

7
m
/
s
2
5.9
Consider points on the Earth’s surface as sketched in the ﬁgure. Because of the Earth’s
rotation, these points undergo uniform circular motion. Compute the centripetal acceler
ation of (a) a point at the equator, and (b) at latitude of 30
◦
.
This is a kinematics of uniform circular motion problem.
Given:
Radius of the Earth:
r
E
= 6
.
4
×
10
6
m
Period of rotation:
T
= 24
h
= 86
,
400 s
(a) We want to ﬁnd the centripetal acceleration,
a
c
, of a point at the equator. Use the
equation from the previous problem.
!"#
a
c
=
4
π
2
r
T
2
a
c
=
4
π
2
(
6
.
4
×
10
6
m
)
(86
,
400 s)
2
= 0
.
034 m
/
s
2
(b) At a latitude of 30
◦
the radius of orbit will be:
r
=
r
E
cos 30
◦
= 5
.
54
×
10
6
m.
a
c
=
4
π
2
r
T
2
a
c
=
4
π
2
(
5
.
54
×
10
6
m
)
(86
,
400 s)
2
= 0
.
029 m
/
s
2
1
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View Full Document5.12
When a ﬁghter pilot makes a quick turn, he experiences a centripetal acceleration. When this acceleration
is greater than about 8
×
g
, the pilot will usually lose consciousness (“black out”). consider a pilot ﬂying at
a speed of 900 m/s who wants to make a sharp turn. What is the minimum radius of curvature she can take
without blacking out?
This is a kinematics of uniform circular motion problem.
(a) Given:
Maximum centripetal acceleration:
a
c
= 8
×
g
= 78
.
4 m
/
s
2
Tangential speed:
v
= 900 m
/
s
We want to ﬁnd the minimum radius of curvature which corresponds to the maximum centripetal acceleration.
Use Equation 5.5.
a
c
=
v
2
r
→
r
=
v
2
a
c
r
=
(900 m
/
s)
2
78
.
4 m
/
s
2
= 1
.
0
×
10
4
m
The minimum radius is roughly 10 km.
5.14
The Daytona 500 stock car race is held on a track that is approximately 2.5 mi long, and the turns are banked
at an angle of 31
◦
. It is currently possible for cars to travel through the turns at a speed of 180 mi/h. Assuming
these cars are on the verge of slipping into the outer wall of the racetrack (because they are racing!), ﬁnd the
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 Spring '11
 MARKO
 Circular Motion

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