Muraj, Hamza – Quiz 3 – Due: Apr 5 2006, 10:00 pm – Inst: Florin
1
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printout
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have
20
questions.
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before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
The tub of a washer goes into its spindry
cycle, starting from rest and reaching an an
gular speed of 3
.
4 rev
/
s in 5 s
.
At this point
the person doing the laundry opens the lid,
and a safety switch turns off the washer. The
tub slows to rest in 11
.
5 s
.
Through how many revolutions does the
tub turn?
Assume
Constant
angular
acceleration
while it is starting and stopping.
Correct answer: 28
.
05 rev.
Explanation:
We will break the motion into two stages:
(1) an acceleration period and
(2) a deceleration period.
While speeding up,
θ
1
=
ω t
=
0 +
ω
2
t
1
=
1
2
ω t
1
=
1
2
(3
.
4 rev
/
s) (5 s)
= 8
.
5 rev
.
While slowing down,
θ
2
=
ω t
=
ω
+ 0
2
t
2
=
1
2
ω t
2
=
1
2
(3
.
4 rev
/
s) (11
.
5 s)
= 19
.
55 rev
.
So
θ
=
θ
1
+
θ
2
= (8
.
5 rev) + (19
.
55 rev)
=
28
.
05 rev
.
002
(part 1 of 1) 10 points
Assume:
When the disk lands on the surface
it does not bounce.
The disk has mass
m
and outer radius
R
with a nonuniform radial mass distribution
so that its moment of inertia
I
=
5
6
m R
2
.
A disk is given a hard kick along a hori
zontal surface.
The kicking force acts along
a horizontal line through the disk’s center, so
the disk acquires a linear velocity
v
0
but no
angular velocity.
The kinetic friction force between the sur
face and the disk slows down its linear motion
while at the same time making the disk spin
on its axis at an accelerating rate. Eventually,
the disk’s rotation catches up with its linear
motion, and the disk begins to roll without
slipping on the surface.
The acceleration of gravity is
g .
I
=
5
6
m R
2
R ,
radius
m
v
0
μ
How long Δ
t
does it take for the ball to roll
without slipping?
1.
Δ
t
=
2
5
v
0
μ g
2.
Δ
t
=
4
11
v
0
μ g
3.
Δ
t
=
7
16
v
0
μ g
4.
Δ
t
=
8
17
v
0
μ g
5.
Δ
t
=
1
3
v
0
μ g
6.
Δ
t
=
3
7
v
0
μ g
7.
Δ
t
=
5
11
v
0
μ g
correct
8.
Δ
t
=
1
2
v
0
μ g
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Muraj, Hamza – Quiz 3 – Due: Apr 5 2006, 10:00 pm – Inst: Florin
2
9.
Δ
t
=
3
8
v
0
μ g
10.
Δ
t
=
1
4
v
0
μ g
Explanation:
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
=
v
0

a t
=
v
0

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
; the angular acceleration is
X
τ
=
I α ,
so
α
=
τ
I
=
μ m g R
5
6
m R
2
=
6
5
μ g
R
.
(2)
The time dependence of
ω
is
ω
=
α t
=
6
5
μ g
R
t ,
so
v
disk
≡
R ω
=
6
5
μ g 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
v
disk
=
v
surface
6
5
μ g t
=
v
0

μ g t ,
or
11
5
μ g t
=
v
0
,
so
t
=
5
11
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 Angular Momentum, Kinetic Energy, Mass, Moment Of Inertia, Rotation

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