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obennoskey (blo242) – hw12 – Demkov – (59910)
1
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001
(part 1 oF 3) 10.0 points
A particle rotates counterclockwise in a circle
oF radius 6
.
4 m with a constant angular speed
oF 17 rad
/
s
.
At
t
= 0, the particle has an
x
coordinate oF 0
.
85 m and
y >
0
.
x
y
(0
.
85 m
,y
)
Figure:
Not drawn to scale.
radius
6
.
4 m
17 rad
/
s
Determine the
x
coordinate oF the particle
at
t
= 2
.
87 s
.
Correct answer: 6
.
39541 m.
Explanation:
Let :
x
0
= 0
.
85 m
,
ω
= 17 rad
/
s
,
t
0
= 0 s
,
and
R
= 6
.
4 m
,
Since the amplitude oF the particle’s mo
tion equals the radius oF the circle and
ω
= 17 rad
/
s , we have
x
=
A
cos(
ω t
+
φ
)
= (6
.
4 m) cos
b
(17 rad
/
s)
t
+
φ
B
.
We can fnd
φ
using the initial condition that
x
0
= 0
.
85 m at
t
= 0
(0
.
85 m) = (6
.
4 m) cos(0 +
φ
)
,
which implies
φ
= arccos
x
0
R
= arccos
(0
.
85 m)
(6
.
4 m)
= 82
.
3679
◦
= 1
.
43759 rad
.
ThereFore, at time
t
= 2
.
87 s , the
x
coordinate
oF the particle is
x
=
R
cos
b
ω t
+
φ
B
= (6
.
4 m) cos
b
(17 rad
/
s) (2
.
87 s)
+ (1
.
43759 rad)
B
=
6
.
39541 m
.
Note:
The angles in the cosine are in radians.
002
(part 2 oF 3) 10.0 points
±ind the
x
component oF the particle’s veloc
ity at
t
= 2
.
87 s.
Correct answer: 4
.
12169 m
/
s.
Explanation:
Di²erentiating the Function
x
(
t
) with re
spect to
t
, we fnd the
x
component oF the
particle’s velocity at any time
t
v
x
=
dx
dt
=

ω A
sin(
ω t
+
φ
)
,
so at
t
= 2
.
87 s
,
the argument oF the sine is
φ
2
≡
ω t
+
φ
= (17 rad
/
s) (2
.
87 s) + (1
.
43759 rad)
= 50
.
2276 rad
,
and the
x
component oF the velocity oF the
particle is
v
x
=

ω R
sin(
φ
2
)
=

(17 rad
/
s) (6
.
4 m) sin(50
.
2276 rad)
=
4
.
12169 m
/
s
.
003
(part 3 oF 3) 10.0 points
±ind the
x
component oF the particle’s accel
eration at
t
= 2
.
87 s.
Correct answer:

1848
.
27 m
/
s
2
.
Explanation:
Di²erentiating the
x
component oF the par
ticle’s velocity with respect to
t
, we fnd the
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View Full Documentobennoskey (blo242) – hw12 – Demkov – (59910)
2
x
component of the particle’s acceleration at
any time
t
a
x
=
dv
x
dt
=

ω
2
A
cos(
ω t
+
φ
)
,
so at
t
= 2
.
87 s
,
the
x
component of the
particle’s acceleration is
a
x
=

ω
2
R
cos
φ
2
=

(17 rad
/
s)
2
(6
.
4 m) cos(50
.
2276 rad)
=

1848
.
27 m
/
s
2
.
004
10.0 points
When a body executes simple harmonic mo
tion, its period is
1.
proportional to its acceleration.
2.
independent of its amplitude.
correct
3.
the reciprocal of its speed.
4.
inversely proportional to its accelera
tion.
5.
proportional to its amplitude.
Explanation:
The period of simple harmonic motion is
the same regardless of the amplitude.
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 Spring '08
 DEMOVK

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