Overton, Mays – Homework 23 – Due: Mar 28 2005, 4:00 am – Inst: Turner
1
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printout
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have
12
questions.
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before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
A girl throws a long stick with a length of
0
.
92 m and a mass of 0
.
93 kg into the air
in such a way that the center of mass rises
vertically.
At the moment it leaves her hand, the stick
is horizontal and the speed of the end of the
stick nearest to her is zero.
When the center of mass of the stick reaches
its highest point, the stick is horizontal, and
it has made 27 complete revolutions.
Assume:
The stick’s cross sectional area
and mass is uniform.
The acceleration of
gravity is 9
.
8 m
/
s
2
.
0
.
46 m
0
.
92 m
v
cm
0
0
.
93 kg
v
end
0
How long did it take for the center of mass
to reach its highest point?
Correct answer: 2
.
82187 s.
Explanation:
From kinematics
v
0
=
g t
(1)
From definition of angular velocity
v
0
=
r ω
=
‘
2
ω .
(2)
From kinematics
θ
=
ω t
⇒
ω
=
θ
t
.
(3)
Substituting
ω
from Eq. 3 into Eq. 2 gives
v
0
=
‘
2
θ
t
.
(4)
Substituting
v
0
from Eq. 1 into Eq. 4 gives
‘
2
θ
t
=
g t .
(5)
Since,
θ
=
n
(2
π
)
,
we have
t
=
s
n π ‘
g
(6)
=
s
(27)
π
(0
.
92 m)
(9
.
8 m
/
s
2
)
= 2
.
82187 s
.
002
(part 1 of 2) 10 points
Assume:
When the disk lands on the surface
it does not bounce.
A disk (with radius 50 cm
,
mass 3 kg
,
and
moment of inertia
m R
2
/
2) rotates at angular
speed 9 rad
/
s around its axis.
The disk is
carefully lowered onto a horizontal surface
and released with zero initial linear velocity
along the surface.
The kinetic friction force between the sur
face and the disk slows down the rotation of
the disk and at the same time gives it a hor
izontal acceleration.
Eventually, the disk’s
linear motion catches up with its rotation,
and the disk begins to roll without slipping on
the surface.
The acceleration of gravity is 9
.
8 m
/
s
2
.
50 cm
,
radius
3 kg
9 rad
/
s
μ
= 0
.
04
Determine the time it takes the disk to start
rolling without slipping.
Correct answer: 3
.
82653 s.
Explanation:
Let :
r
= 50 cm = 0
.
5 m
,
ω
0
= 9 rad
/
s
,
m
= 3 kg
,
and
μ
= 0
.
04
.
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Overton, Mays – Homework 23 – Due: Mar 28 2005, 4:00 am – Inst: Turner
2
From the perspective of the surface, let the
speed of the center of the disk be
v
surface
.
Using the frictional force
f
, we can determine
the acceleration
X
F
s
=
m a ,
and
f
=
μ m g ,
so
a
=
μ g .
Then
v
surface
=
a t
=
μ g t .
(1)
After pure rolling begins there is no longer any
frictional force and consequently no longer
any acceleration. From the perspective of the
center of the disk, let the tangential velocity
of the rim of the disk be
v
disk
, the angular
velocity be
ω
, and the angular acceleration is
X
τ
=
I α ,
so
α
=
τ
I
=
μ m g R
m R
2
2
=
2
μ g
R
(2)
=
2 (0
.
04) (9
.
8 m
/
s
2
)
(0
.
5 m)
= 1
.
568 rad
/
s
2
.
The time dependence of
ω
is
ω
=
ω
0

α t
=
ω
0

2
μ g
R
t ,
so
v
disk
≡
R ω
=
R
ω
0

2
μ g
R
t
¶
.
(3)
When the disk reaches pure rolling, the veloc
ity from the perspective of the surface will be
the same as the velocity from the perspective
of the center of the disk; that is, there will
be no slipping. Setting the velocity
v
disk
from
Eq. 3 equal to
v
surface
from Eq. 1 gives
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 Spring '08
 Turner
 Physics, Friction, Mass, Work, Moment Of Inertia, Overton

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