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Chapter 10 1
Page 1
CHAPTER 10  Rotational Motion About a Fixed Axis
1. (
a
) 30° = (30°)(
π
rad/180°) =
p/6 rad = 0.524 rad
;
(
b
) 57° = (57°)(p rad/180°) =
19p/60 = 0.995 rad
;
(
c
) 90° = (90°)(p rad/180°) =
p/2 = 1.571 rad
;
(
d
) 360° = (360°)(p rad/180°) =
2p = 6.283 rad
;
(
e
) 420° = (420°)(p rad/180°) =
7p/3 = 7.330 rad
.
2.
The subtended angle in radians is the size of the object divided by the distance to the object:
θ
= 2
r
Sun
/
r
;
(0.5°)(p rad/180°) = 2
r
Sun
/(150
×
10
6
km), which gives
r
Sun
˜ 6.5
×
10
5
km
.
3.
We find the distance from
=
h
/
r
;
(7.5°)(p rad/180°) = (300 m)/
r
; which gives
r
=
2.3
×
10
3
m
.
4.
From the definition of angular acceleration, we have
α
= ?
ω
/?
t
= [(20,000 rev/min)(2p rad/rev)/(60 s/min) – 0]/(5.0 min)(60 s/min)
=
7.0 rad/s
2
.
5.
From the definition of angular velocity, we have
= ∆
/
∆
t
, and we use the time for each hand to turn through a complete circle, 2p rad.
(
a
)
second
= ∆
/
∆
t
= (2p rad)/(60 s) =
0.105 rad/s
.
(
b
)
minute
= ∆
/
∆
t
= (2p rad)/(60 min)(60 s/min) =
1.75
×
10
–3
rad/s
.
(
c
)
hour
= ∆
/
∆
t
= (2p rad)/(12 h)(60 min/h)(60 s/min) =
1.45
×
10
–4
rad/s
.
(
d
) For each case, the angular velocity is constant, so the angular acceleration is
zero
.
6. (
a
) The Earth moves one revolution around the Sun in one year, so we have
orbit
= ∆
/
∆
t
= (2p rad)/(1 yr)(3.16
×
10
7
s/yr) =
1.99
×
10
–7
rad/s
.
(
b
) The Earth rotates one revolution in one day, so we have
rotation
= ∆
/
∆
t
= (2p rad)/(1 day)(24 h/day)(3600 s/h) =
7.27
×
10
–5
rad/s
.
7.
All points will have the angular speed of the Earth:
=
∆
/
∆
t
= (2p rad)/(1 day)(24 h/day)(3600 s/h) = 7.27
×
10
–5
rad/s.
Their linear speed will depend on the distance from the rotation axis.
(
a
) On the equator we have
v
=
r
Earth
= (6.38
×
10
6
m)(7.27
×
10
–5
rad/s) =
464 m/s
.
(
b
) At a latitude of 66.5° the distance is
r
Earth
cos 66.5°, so we have
v
=
r
Earth
cos 66.5°
= (6.38
×
10
6
m)(cos 66.5°)(7.27
×
10
–5
rad/s) =
185 m/s
.
(
c
) At a latitude of 40.0° the distance is
r
Earth
cos 40.0°, so we have
v
=
r
Earth
cos 40.0°
= (6.38
×
10
6
m)(cos 40.0°)(7.27
×
10
–5
rad/s) =
355 m/s
.
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View Full DocumentChapter 10 2
Page 2
8.
The subtended angle in radians is the size of the object divided by the distance to the object.
A pencil
with a diameter of 6 mm will block out the Moon if it is held about 60 cm from the eye.
For the angle
subtended we have
θ
Moon
=
D
pencil
/
r
pencil
˜ (0.6 cm)/(60 cm)
˜ 0.01 rad
.
We estimate the diameter of the Moon from
Moon
=
D
Moon
/
r
Moon
;
0.01 rad =
D
Moon
/(3.8
×
10
5
km), which gives
D
Moon
˜ 4
×
10
3
km
.
9. (
a
)
ω
= (2500 rev/min)(2p rad/rev)/(60 s/min) =
262 rad/s
.
(
b
) The linear speed of the point on the edge is the tangential speed:
v
=
r
= (0.175 m)(262 rad/s) =
46 m/s
.
Because the speed is constant, the tangential acceleration is zero.
There will be a radial
acceleration:
a
R
=
2
R
= (262 rad/s)
2
(0.175 m) =
1.2
×
10
4
m/s
2
radial
.
10. (
a
) The angular speed of the merrygoround is
= (1 rev)(2p rad/rev)/(4.0 s) = 1.57 rad/s.
The linear speed of the child is the tangential speed:
v
=
r
= (1.2 m)(1.57 rad/s) =
1.9 m/s
.
(
b
) The child will have a radial acceleration:
a
R
=
2
R
= (1.57 rad/s)
2
(1.2 m) =
3.0 m/s
2
radial
.
11. In each revolution the ball rolls a distance equal to its circumference, so we have
L
=
N
(p
D
);
3.5 m = (15.0)p
D
, which gives
D
= 0.074 m =
7.4 cm
.
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 Spring '09
 Fielding
 Physics

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