CHAPTER 12
ROTATIONAL MOTION
ActivPhysics
can help with these problems:
Activities 7.7, 7.8, 7.10, 7.12–7.15
Section 121:
Angular Velocity and
Acceleration
Problem
1. Determine the angular speed, in rad/s, of (a) Earth
about its axis; (b) the minute hand of a clock;
(c) the hour hand of a clock; (d) an egg beater
turning at 300 rpm.
Solution
The angular speed is
ω
= Δ
θ/
Δ
t.
(a)
ω
E
= 1 rev
÷
1 d = 2
π/
86
,
400 s = 7
.
27
×
10
−
5
s
−
1
.
(b)
ω
min
=
1 rev
/
1 h = 2
π/
3600 s = 1
.
75
×
10
−
3
s
−
1
.
(c)
ω
hr
=
1 rev
/
12 h =
1
12
ω
min
= 1
.
45
×
10
−
4
s
−
1
.
(d)
ω
=
300 rev
/
min = 300
×
2
π/
60 s = 31
.
4 s
−
1
.
(Note:
Radians are a dimensionless angular measure, i.e.,
pure numbers; therefore angular speed can be
expressed in units of inverse seconds.)
Problem
2. What is the linear speed (a) of a point on Earth’s
equator? (b) At your latitude?
Solution
(a) On the equator,
v
=
ω
E
R
E
= (7
.
27
×
10
−
5
s
−
1
)
×
(6
.
37
×
10
6
m) = 463 m
/
s; see Problem 1(a). (b) At
latitude
θ, r
=
R
E
cos
θ,
so
v
=
ω
E
r
= (463 m
/
s) cos
θ.
Problem
3. Express each of the following in radians per
second: (a) 720 rpm; (b) 50
◦
/
h; (c) 1000 rev/s;
(d) 1 rev/year (which is the angular speed of
Earth in its orbit).
Solution
(a) (720 rev
/
min)(2
π/
rev)(min
/
60 s) = 24
π
s
−
1
=
75
.
4 s
−
1
.
(b) (50
◦
/h
)(
π/
180
◦
)(h
/
3600 s) = 2
.
42
×
10
−
4
s
−
1
.
(c) (1000 rev
/
s)(2
π/
rev) = 2000
π
s
−
1
≈
6
.
28
×
10
3
s
−
1
.
(d) (1 rev
/
y)
≈
2
π/
(
π
×
10
7
s) =
2
×
10
−
7
s
−
1
.
(See note in solution to Problem 1. The
approximate value for 1 y used in part (d) is often
handy for estimates, and is fairly accurate; see
Chapter 1, Problem 12.)
Problem
4. A 25cmdiameter circular saw blade spins at
3500 rpm. How fast would you have to push a
straight hand saw to have the teeth move through
the wood at the same rate as the circular saw teeth?
Solution
The linear speed of the saw teeth is
v
=
ωr
=
(
π/
30)(s
−
1
/
rpm)(3500 rpm)(0
.
25 m
/
2) = 45
.
8 m
/
s
(over 100 mi/h).
Problem
5. A wheel turns through 2.0 revolutions while being
accelerated from rest at 18 rpm/s. (a) What is the
final angular speed? (b) How long does it take to
turn the 2.0 revolutions?
Solution
For constant angular acceleration, (a) Equation 1211
gives
ω
f
=
radicalbig
ω
2
0
+ 2
α
(
θ
f
−
θ
0
) =
radicalBig
0 + 2(18 rev
×
60
/
min
2
)(2 rev) = 65
.
7 rpm
,
and
(b) Equation 1210 gives
θ
f
−
θ
0
= 0 +
1
2
αt
2
,
or
t
=
radicalbig
2(2 rev)
/
(18 rev
/
60 s
2
) = 3
.
65 s
.
Problem
6. You switch a food blender from its high to its low
setting; the blade speed drops from 3600 rpm to
1800 rpm in 1.4 s. How many revolutions does it
make during this time?
Solution
At constant angular acceleration, Equations 121 and
8 give Δ
θ
=
1
2
(
ω
0
+
ω
f
) Δ
t
=
1
2
(3600 + 1800)
×
(rev
/
60 s)(1
.
4 s) = 63.0 rev.
Problem
7. A compact disc (CD) player varies the rotation rate
of the disc in order to keep the part of the disc from
which information is being read moving at a
constant linear speed of 1
.
30 m
/
s
.
Compare the
rotation rates of a 12.0cmdiameter CD when
information is being read from (a) its outer edge
and (b) a point 3.75 cm from the center. Give your
answers in rad/s and rpm.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
CHAPTER 12
171
Solution
Equation 124 gives the relation between linear speed
and angular speed,
ω
=
v/r,
where
r
is the distance
from the center of rotation. With
v
a constant
130 cm
/
s
,
(a)
ω
= (130 cm
/
s)
/
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '09
 Huang
 Energy, Kinetic Energy, Moment Of Inertia, Icm

Click to edit the document details