918
Chapter 9
(b)
M
M
drum
mass
mg h
K
K
=
+
.
2
1
M
M
drum
2
mv
mg h
K
=
−
and
2
drum
M
M
2
2(250.0 J)
2
2(3.71 m/s
)(13.2 m)
8.04 m/s
15.0 kg
K
v
g h
m
=
−
=
−
=
E
VALUATE
:
We did the calculations without knowing the moment of inertia
I
of the drum, or the mass and radius
of the drum.
9.71.
I
DENTIFY
and
S
ET
U
P
:
All points on the belt move with the same speed. Since the belt doesn°t slip, the speed of
the belt is the same as the speed of a point on the rim of the shaft and on the rim of the wheel, and these speeds are
related to the angular speed of each circular object by
.
v
r
ω
=
E
XECUTE
:
Figure 9.71
(a)
1
1
1
v
r
ω
=
1
(60.0 rev/s)(2
rad/1 rev)
377 rad/s
ω
π
=
=
2
1
1
1
(0.45
10
m)(377 rad/s)
1.70 m/s
v
r
ω
−
=
=
×
=
(b)
1
2
v
v
=
1
1
2
2
r
r
ω
ω
=
2
1
2
1
(
/
)
(0.45 cm/2.00 cm)(377 rad/s)
84.8 rad/s
r
r
ω
ω
=
=
=
E
VALUATE
:
The wheel has a larger radius than the shaft so turns slower to have the same tangential speed for
points on the rim.
9.72.
I
DENTIFY
:
The speed of all points on the belt is the same, so
1
1
2
2
r
r
ω
ω
=
applies to the two pulleys.
S
ET
U
P
:
The second pulley, with half the diameter of the first, must have twice the angular velocity, and this is
the angular velocity of the saw blade.
rad/s
30 rev/min
π
=
.
E
XECUTE
:
(a)
2
rad
s
0.208 m
(2(3450 rev min))
75.1 m s.
30 rev min
2
v
π
⎛
⎞
⎛
⎞
=
=
⎜
⎟⎜
⎟
⎝
⎠
⎝
⎠
(b)
2
2
4
2
rad
rad
s
0.208 m
2(3450 rev min)
5.43
10 m s ,
30
rev min
2
a
r
π
ω
⎛
⎞
⎛
⎞
⎛
⎞
=
=
=
×
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
so the force holding sawdust on the blade would have to be about 5500 times as strong as gravity.
E
VALUATE
:
In
v
r
ω
=
and
2
rad
a
r
ω
=
,
ω
must be in rad/s.
9.73.
I
DENTIFY
and
S
ET
U
P
:
Use Eq.(9.15) to relate
rad
a
to
ω
and then use a constant acceleration equation to
replace
.
ω
E
XECUTE
:
(a)
2
rad
,
a
r
ω
=
2
rad,1
1
,
a
r
ω
=
2
rad,2
2
a
r
ω
=
2
2
rad
rad,2
rad,1
2
1
(
)
a
a
a
r
ω
ω
Δ
=
−
=
−
One of the constant acceleration equations can be written
2
2
2
1
2
1
2
(
),
z
z
ω
ω
α θ
θ
=
+
−
or
2
2
2
1
2
1
2
(
)
z
z
z
ω
ω
α
θ
θ
−
=
−
Thus
rad
2
1
2
1
2
(
)
2
(
),
z
z
a
r
r
α
θ
θ
α
θ
θ
Δ
=
−
=
−
as was to be shown.
(b)
2
2
2
rad
2
1
85.0 m/s
25.0 m/s
8.00 rad/s
2
(
)
2(0.250 m)(15.0 rad)
z
a
r
α
θ
θ
Δ
−
=
=
=
−
Then
2
2
tan
(0.250 m)(8.00 rad/s
)
2.00 m/s
a
r
α
=
=
=
E
VALUATE
:
2
ω
is proportional to
z
α
and
0
(
)
θ
θ
−
so
rad
a
is also proportional to these quantities.
rad
a
increases
while
r
stays fixed,
z
ω
increases, and
z
α
is positive.
I
DENTIFY
and
S
ET
U
P
:
Use Eq.(9.17) to relate
K
and
ω
and then use a constant acceleration equation to replace
.
ω
E
XECUTE
:
(c)
2
1
2
;
K
I
ω
=
2
1
2
2
2
,
K
I
ω
=
2
1
1
1
2
K
I
ω
=
2
2
1
1
2
1
2
1
2
1
2
1
2
2
(
)
(2
(
))
(
),
z
z
K
K
K
I
I
I
ω
ω
α
θ
θ
α
θ
θ
Δ
=
−
=
−
=
−
=
−
as was to be shown.
(d)
2
2
2
1
45.0 J
20.0 J
0.208 kg
m
(
)
(8.00 rad/s
)(15.0 rad)
z
K
I
α
θ
θ
Δ
−
=
=
=
⋅
−
E
VALUATE
:
z
α
is positive,
ω
increases, and
K
increases.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Rotation of Rigid Bodies
919
9.74.
I
DENTIFY
:
wood
lead
.
I
I
I
=
+
m
V
ρ
=
, where
ρ
is the volume density and
m
A
σ
=
, where
σ
is the area density.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '06
 Buchler
 Physics, Energy, Inertia, Kinetic Energy, Mass, Potential Energy, kg

Click to edit the document details