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Chapter E  Problems
Blinn College  Physics 2425  Terry Honan
Problem E.1
(a) What is the centripetal (radial) acceleration of a point on the earth's equator?
(b) Give an expression for the centripetal acceleration as a function of the latitude angle,
q
L
. What is this at the latitude of Bryan,
Texas, at
q
L
=
30.7 °?
(c) What are the speeds of a point at the equator and at Bryan, Texas due to the eaarth's rotation?
(d) The earth's rotation is slowing at a rate of 2.2 s every 100,000 years.
.
„
T
„
t
=
D
T
D
t
=
2.3 s
100,000 years
What is the (very small) tangential component of the acceleration of a point on the earth's equator? Hint: use the chain rule.
„
„
t
2
p
r
T
= 
2
p
r
T
2
„
T
„
t
Solution to E.1
a
c
=
v
2
r
and
v
=
2
p
r
T
ï
a
c
=
J
2
p
T
N
2
r
For the earth:
T
=
1 day
=
24
ÿ
3600
s
(a) At the equator:
r
=
R
E
ï
a
c
=
K
2
p
24
ÿ
3600
O
2
6.37
μ
10
6
=
0.0337
m
s
2
Comment: Compared to the gravitational acceleration of g, this is small but not negligible.
(b) r is the radius of the circle but something on the rotating earth moves along a path that traces a latitude line. At some point other
that the equator this is:
r
=
R
E
cos
q
L
ï
a
c
=
0.0337
m
s
2
μ
cos
q
L
q
L
=
30.7 °
ï
a
c
=
0.0290
m
s
2
(c) The speed is found form the period:
v
=
2
p
r
T
.
At equator:
v
=
2
p
24
ÿ
3600
6.37
μ
10
6
=
463
m
s
At Bryan:
v
=
2
p
24
ÿ
3600
6.37
μ
10
6
cos 30.7 °
=
398
m
s
(d) To find the tangential acceleration use
a
t
= „
v
ê
„
t
and the hints.
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t
=
„
v
„
t
=
„
„
t
2
p
r
T
= 
2
p
r
T
2
„
T
„
t
= 
2
p
r
T
2
D
T
D
t
= 
2
p
R
E
H
1 day
L
2
2.3 s
100,000 years
= 
2
p
R
E
H
1 day
L
2
2.3
s
100,000 year
= 
2
p
6.37
μ
10
6
H
24
μ
3600
L
2
2.3
100,000
μ
H
365.24
μ
24
μ
3600
L
=
3.91
μ
10

15
m
s
2
Problem E.2
A car drives around a 200 m radius circle with a speed that decreases uniformly from 30 m
ê
s to 20 m
ê
s in 8 s. At the instant the
speed is 25 m
ê
s then:
(a) what is the centripetal acceleration,
(b) what is the tangential acceleration and
(c) what is the magnitude of the acceleration?
Solution to E.2
(a)
a
c
=
v
2
r
=
25
2
200
=
3.125
m
s
2
(b)
a
t
=
„
v
„
t
=
D
v
D
t
=
20

30
8
= 
1.25
m
s
2
(c)
a
=
a
c
2
+
a
t
2
=
3.37
m
s
2
Problem E.3
While moving in a circle with a 12 m radius the speed of a particle varies with time by
v
H
t
L
=
2
+
10
t

4
t
2
in SI units. At
t
=
2 s
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 Spring '09
 Acceleration

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