ROTATIONAL VECTORS AND
ANGULAR MOMENTUM
E
XERCISES
Section 11.1 Angular Velocity and Acceleration Vectors
12.
If we assume that the wheels are rolling without slipping (see Section 10.5), the magnitude of the angular velocity
is
1
cm
v
/
(70 m/3.6 s)/(0.31 m)
62.7 s
.
r
ϖ

=
=
=
With the car going north, the axis of rotation of the wheels is
eastwest. Since the top of a wheel is going in the same direction as the car, the righthand rule gives the direction
of
ϖ
as west.
13.
I
NTERPRET
The problem asks about the angular acceleration of the wheels as the car traveling north with a speed
of 70 km/h makes a
90
°
left turn that lasts for 25 s.
D
EVELOP
The speed of the car is
cm
70 km/h
19.4 m/s.
v
=
=
Assuming that the wheels are rolling without
slipping, the magnitude of the initial angular velocity is
cm
19.4 m/s
62.7 rad/s
0.31 m
v
r
ϖ
=
=
=
With the car going north, the axis of rotation of the wheels is eastwest. Since the top of a wheel is going in the
same direction as the car, the righthand rule gives the direction of
i
ϖ
r
as west. In unitvector notation, we write
ˆ
.
i
i
ϖ
ϖ
= 
r
After making a left turn, the angular speed remains unchanged, but the direction of
f
ϖ
r
is now south (see sketch).
In unitvector notation, we write
ˆ
.
f
j
ϖ
ϖ
= 
r
E
VALUATE
Using Equation 10.4, we find the angular acceleration to be
av
ˆ
ˆ
(
)
ˆ
ˆ
(
)
f
i
j
i
i
j
t
t
t
t
ϖ
ϖ
ϖ
ϖ
ϖ
ϖ
α

∆

 
=
=
=
=

∆
∆
∆
∆
r
r
r
r
The magnitude of
av
α
r
is
2
av
2
2(62.7 rad/s)


3.55 rad/s
25 s
t
ϖ
α
=
=
=
∆
r
and
av
α
r
points in the southeast direction (in the direction of the vector
ˆ
ˆ
).
i
j

A
SSESS
Angular acceleration
av
α
r
points in the same direction as
.
ϖ
∆
r
14.
Suppose that the
x
axis is horizontal in the direction of the final angular velocity
ˆ
(
(60 rpm) )
f
i
=
ϖ
and the
y
axis is
vertical in the direction of the initial angular velocity
ˆ
ˆ
(
(45 rpm)
).
j
=
i
ϖ
Equation 11.1 implies that
av
(
)/
f
i
t
=

∆ =
α
ϖ
ϖ
ˆ
ˆ
ˆ
ˆ
(60
45
) rpm/15 s
(4
3
) rpm/s.
i
j
i
j
=


Its magnitude is
2
2
av
(4)
(
3)
rpm/s
α
=
+ 
=
2
2
5 rpm/s
5(
/30) s
0.524 s
,
π


=
=
at an angle
1
3
4
tan
(
)
36.9
θ

=

= 
°
to the
x
axis (i.e., below the horizontal).
11.1
11
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11.2
Chapter 11
15.
I
NTERPRET
The problem asks about the angular velocity of the wheels after an angular acceleration has been
applied within a time interval.
D
EVELOP
Take the
x
axis east and the
y
axis north, with positive angles measured CCW from the
x
axis. In unit
vector notation, the initial angular velocity
i
ϖ
r
and the angular acceleration
α
r
can be expressed as
2
2
2
ˆ
ˆ
(140 rad/s)
ˆ
ˆ
ˆ
ˆ
(cos
sin
)
(35 rad/s )[cos(90
68 )
sin(90
68 ) ]
ˆ
ˆ
( 32.45 rad/s )
(13.11 rad/s )
i
i
i
i
i
j
i
j
i
j
α
α
ϖ
ϖ
α
α
θ
θ
=
=
=
+
=
° +
°
+
° +
°
= 
+
r
r
The final angular velocity can be found by using Equation 10.8.
E
VALUATE
Using Equation 10.8, the angular velocity at
5.0 s
t
=
is
2
2
ˆ
ˆ
ˆ
(140 rad/s)
[( 32.45 rad/s )
(13.11 rad/s ) ](5.0 s)
ˆ
ˆ
( 22.3 rad/s)
(65.6 rad/s)
f
i
t
i
i
j
i
j
ϖ
ϖ
α
=
+
=
+

+
= 
+
r
r
r
The magnitude and direction of
f
ϖ
r
are
2
2


( 22.3 rad/s)
(65.6 rad/s)
69.2 rad/s
f
f
ϖ
ϖ
=
=

+
=
r
and
,
1
1
,
22.3 rad/s
tan
tan
18.8
65.6 rad/s
f y
f
f x
ϖ
θ
ϖ



=
=
= 
°
or 18.8
°
west of north.
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 Spring '09
 Prof.Kim
 Physics, Acceleration, Angular Momentum, Momentum, Rotation, li

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