obennoskey (blo242) – hw8 – Demkov – (59910)
1
This printout should have 47 questions.
Multiplechoice questions may continue on
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beFore answering.
001
(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 170 cm long.
IF the beetle rides through a swing oF 35
◦
,
how Far has he traveled along the path oF the
pendulum?
Correct answer: 103
.
847 cm.
Explanation:
Arc length is defned as
s
=
r θ
= (170 cm)(35
◦
)
·
p
π
180
◦
P
= 103
.
847 cm
,
002
(part 2 oF 3) 10.0 points
What is the displacement experienced by the
beetle while moving theough the same angle
35
◦
?
Correct answer: 102
.
24 cm.
Explanation:
Using the law oF cosines, we have
b
v
R
b
=
r
r
2
+
r
2
−
2
r r
cos
θ
=
r
r
2 (1
−
cos
θ
)
= (170 cm)
r
2(1
−
cos 35
◦
)
= 102
.
24 cm
.
Alternative Solution:
Divide the isosce
les triangle in halF. Then
b
v
R
b
= 2
.
0
r
sin
±
θ
2
²
= 2
.
0 (170 cm)(sin 17
.
5
◦
)
= 102
.
24 cm
.
003
(part 3 oF 3) 10.0 points
IF the pendulum at some instant is swinging
at 3
.
2 rad
/
s, how Fast is the beetle traveling?
Correct answer: 544 cm
/
s.
Explanation:
Linear and angular velocity are related by
V
=
r ω
= (170 cm)(3
.
2 rad
/
s)
= 544 cm
/
s
.
keywords:
004
10.0 points
A large wheel is coupled to a wheel with halF
the diameter as shown.
r
2
How does the rotational speed oF the
smaller wheel compare with that oF the larger
wheel?
How do the tangential speeds at
the rims compare (assuming the belt doesn’t
slip)?
1.
The smaller wheel has Four times the ro
tational speed and the same tangential speed
as the larger wheel.
2.
The smaller wheel has twice the rotational
speed and the same tangential speed as the
larger wheel.
correct
3.
The smaller wheel has halF the rotational
speed and halF the tangential speed as the
larger wheel.
4.
The smaller wheel has twice the rotational
speed and twice the tangential speed as the
larger wheel.
Explanation:
V
=
r ω
The tangential speeds are equal, since the
rims are in contact with the belt and have the
same linear speed as the belt.
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View Full Documentobennoskey (blo242) – hw8 – Demkov – (59910)
2
The smaller wheel (with half the radius)
rotates twice as fast:
p
1
2
r
P
(2
ω
) =
r ω
=
v
005
10.0 points
A phonograph record has an initial angular
speed of 10 rev/min. The record slows to 1.8
rev/min in 33 s.
What is the record’s average angular accel
eration during this time interval?
Correct answer:
−
2
.
82743 rad
/
s
2
.
Explanation:
Let :
ω
o
= 10 rev
/
min
,
ω
f
= 1
.
8 rev
/
min
,
and
Δ
t
= 33 s
.
α
avg
=
ω
f
−
ω
o
Δ
t
=
10 rev
/
min
−
37 rev
/
min
1 s
×
2
π
rad
1 rev
·
1 min
60 s
=
−
2
.
82743 rad
/
s
2
.
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 Spring '08
 DEMOVK
 Angular Momentum, Kinetic Energy, Mass, Moment Of Inertia, Correct Answer

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