oldmidterm 03 – FIERRO, JEFFREY – Due: Apr 1 2008, 11:00 pm
1
Question 1, chap 12, sect 2.
part 1 of 1
10 points
The tub of a washer goes into its spin
dry cycle, starting from rest and reaching an
angular speed of 3
.
3 rev
/
s in 4
.
6 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 6
.
9 s
.
Through how many revolutions does the
tub turn?
Assume
Constant
angular
acceleration
while it is starting and stopping.
Correct answer:
18
.
975
rev (tolerance
±
1
%).
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
.
3 rev
/
s) (4
.
6 s)
= 7
.
59 rev
.
While slowing down,
θ
2
=
ω t
=
ω
+ 0
2
t
2
=
1
2
ω t
2
=
1
2
(3
.
3 rev
/
s) (6
.
9 s)
= 11
.
385 rev
.
So
θ
=
θ
1
+
θ
2
= (7
.
59 rev) + (11
.
385 rev)
=
18
.
975 rev
.
Question 2, chap 13, sect 2.
part 1 of 1
10 points
Consider a wheel of mass
m
, outer radius
R
, unknown but radially symmetric mass dis
tribution, and moment of inertia
3
7
mR
2
.
The wheel is given a spin around its axis,
and then it’s carefully lowered on a horizontal
surface.
(The wheel does not bounce.)
At
the moment of contact, the wheel has angular
velocity
ω
0
but zero linear velocity,
v
0
= 0, so
it slips against the surface.
The kinetic friction force between the sur
face and the wheel slows down the rotation of
the wheel and at the same time gives it a hor
izontal acceleration.
Eventually, the wheel’s
linear motion catches up with its rotation,
and the wheel begins to roll without slipping.
I
=
3
7
m R
2
R ,
radius
m
ω
0
μ
Once the wheel rolls without slipping, what
is its angular speed?
1.
ω
rot
=
1
3
ω
0
2.
ω
rot
=
2
5
ω
0
3.
ω
rot
=
5
12
ω
0
4.
ω
rot
=
3
11
ω
0
5.
ω
rot
=
2
9
ω
0
6.
ω
rot
=
3
8
ω
0
7.
ω
rot
=
7
16
ω
0
8.
ω
rot
=
3
10
ω
0
correct
9.
ω
rot
=
3
7
ω
0
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oldmidterm 03 – FIERRO, JEFFREY – Due: Apr 1 2008, 11:00 pm
2
10.
ω
rot
=
6
13
ω
0
Explanation:
Once the wheel lays on the surface, there
are 3 forces acting on it: the weight
mg
, the
normal force
N
, and the kinetic friction force
f
.
The weight and the normal force acts
along the vertical line through the wheel’s
center (which is where the center of mass is),
so they create no torque around the center of
mass.
On the other hand, the friction force
acts perpendicularly to the wheel’s radius at
the distance
R
from the center, so it creates a
torque
τ
=
R
×
f.
(1)
Consequently, the wheel’s rotation
decelerates
at the rate
α
=
τ
I
=
Rf
I
,
(2)
ω
(
t
) =
ω
0
−
αt.
(3)
At the same time, the friction force gives the
wheel a horizontal linear acceleration
a
=
f
m
,
(4)
so at time
t
the wheel’s linear speed is
v
=
at
=
f
m
×
t.
(5)
The wheel’s linear motion catches up with
its rotations when
v
(
t
) =
R
×
ω
(
t
)
.
(6)
According to eqs. (3) and (5), this happens
when
at
=
Rω
0
−
Rαt.
(7)
Solving this equation, we find
t
=
Rω
0
a
+
Rα
,
(8)
at which time the wheel rotates at angular
velocity
ω
(
t
) =
ω
0
−
α
×
Rω
0
a
+
Rα
=
ω
0
×
a
a
+
αR
.
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 Spring '08
 Turner
 Angular Momentum, Kinetic Energy, Moment Of Inertia, Rotation, kg

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