hernandez (ejh742) – oldmidterm 03 – Turner – (58120)
1
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printout
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have
24
questions.
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before answering.
001
10.0 points
A rotating bicycle wheel has an angular
speed of 40
◦
/
s at 3
.
6 s and a constant angular
acceleration of 37
◦
/
s
2
. With the center of the
wheel at the origin, the valve stem is on the
positive
x
axis (horizontal) at
t
0
= 1
.
3 s.
Through what angle has the valve stem
turned between 1
.
3 s and 7
.
1 s?
Correct answer: 360
.
76
◦
.
Explanation:
Let :
ω
1
= 40
◦
/
s
,
t
1
= 3
.
6 s
,
α
= 37
◦
/
s
2
,
θ
0
= 0
◦
,
t
0
= 1
.
3 s
,
and
t
= 7
.
1 s
.
The initial angular velocity can be deter
mined from
ω
1
=
ω
0
+
α
(
t
1
−
t
0
)
ω
0
=
ω
1
−
α
(
t
1
−
t
0
)
= 40
◦
/
s
−
(37
◦
/
s
2
) (3
.
6 s
−
1
.
3 s)
=
−
45
.
1
◦
/
s
.
This
θ
2
=
ω
0
(
t
2
−
t
0
) +
1
2
α
(
t
2
−
t
0
)
2
= (
−
45
.
1
◦
/
s) (7
.
1 s
−
1
.
3 s)
+
1
2
(37
◦
/
s
2
) (7
.
1 s
−
1
.
3 s)
2
=
360
.
76
◦
.
002
(part 1 of 3) 10.0 points
A beetle takes a joy ride on a pendulum. The
string supporting the mass of the pendulum
is 160 cm long.
If the beetle rides through a swing of 37
◦
,
how far has he traveled along the path of the
pendulum?
Correct answer: 103
.
324 cm.
Explanation:
Let :
r
= 160 cm
,
and
θ
= 37
◦
.
Arc length is defined as
s
=
r θ
= (160 cm)(37
◦
)
·
parenleftBig
π
180
◦
parenrightBig
=
103
.
324 cm
.
003
(part 2 of 3) 10.0 points
What is the displacement experienced by the
beetle while moving theough the same angle
37
◦
?
Correct answer: 101
.
538 cm.
Explanation:
Using the law of cosines, we have
bardbl
vector
R
bardbl
=
radicalbig
r
2
+
r
2
−
2
r r
cos
θ
=
r
radicalbig
2 (1
−
cos
θ
)
= (160 cm)
radicalbig
2(1
−
cos 37
◦
)
=
101
.
538 cm
.
Alternative Solution:
Divide the isosce
les triangle in half. Then
bardbl
vector
R
bardbl
= 2
.
0
r
sin
parenleftbigg
θ
2
parenrightbigg
= 2
.
0 (160 cm) sin 18
.
5
◦
= 101
.
538 cm
.
004
(part 3 of 3) 10.0 points
If the pendulum at some instant is swinging
at 2
.
5 rad
/
s, how fast is the beetle traveling?
Correct answer: 400 cm
/
s.
Explanation:
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hernandez (ejh742) – oldmidterm 03 – Turner – (58120)
2
Let :
ω
= 2
.
5 rad
/
s
.
Linear and angular velocity are related by
v
=
r ω
= (160 cm)(2
.
5 rad
/
s)
=
400 cm
/
s
.
keywords:
005
10.0 points
At
t
= 0, a wheel rotating about a fixed axis at
a constant angular deceleration of 0
.
56 rad
/
s
2
has an angular velocity of 2
.
5 rad
/
s and an
angular position of 6
.
2 rad.
What is the angular position of the wheel
after 3 s?
Correct answer: 11
.
18 rad.
Explanation:
Let :
α
=
−
0
.
56 rad
/
s
2
,
ω
0
= 2
.
5 rad
/
s
,
θ
0
= 6
.
2 rad
,
and
t
= 3 s
.
The angular position is
θ
f
=
θ
0
+
ω
0
t
+
1
2
α t
2
= 6
.
2 rad + (2
.
5 rad
/
s) (3 s)
+
1
2
(
−
0
.
56 rad
/
s
2
) (3 s)
2
=
11
.
18 rad
.
006
10.0 points
A massless rod of length
L
has a small mass
m
fastened at its center and another mass
m
fastened at one end.
On the opposite end
from the mass
m
, the rod is hinged with a
frictionless hinge.
The rod is released from
rest from an initial horizontal position; then
it swings down.
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 Spring '08
 Turner
 Physics, Angular Momentum, Kinetic Energy, Mass, Moment Of Inertia, Rotation

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