portillo (gdp347) – oldhomework 32 – Turner – (58185)
1
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beFore answering.
001
(part 1 oF 3) 10.0 points
A particle rotates counterclockwise in a circle
oF radius 7
.
4 m with a constant angular speed
oF 7
.
2 rad
/
s
.
At
t
= 0, the particle has an
x
coordinate oF 5
.
9 m and
y >
0
.
x
y
(5
.
9 m
,y
)
Figure:
Not drawn to scale.
radius
7
.
4 m
7
.
2 rad
/
s
Determine the
x
coordinate oF the particle
at
t
= 1
.
81 s
.
Correct answer: 3
.
26647 m.
Explanation:
Let :
x
0
= 5
.
9 m
,
ω
= 7
.
2 rad
/
s
,
t
0
= 0 s
,
and
R
= 7
.
4 m
,
Since the amplitude oF the particle’s mo
tion equals the radius oF the circle and
ω
= 7
.
2 rad
/
s , we have
x
=
A
cos(
ω t
+
φ
)
= (7
.
4 m) cos
b
(7
.
2 rad
/
s)
t
+
φ
B
.
We can fnd
φ
using the initial condition that
x
0
= 5
.
9 m at
t
= 0
(5
.
9 m) = (7
.
4 m) cos(0 +
φ
)
,
which implies
φ
= arccos
x
0
R
= arccos
(5
.
9 m)
(7
.
4 m)
= 37
.
1272
◦
= 0
.
647992 rad
.
ThereFore, at time
t
= 1
.
81 s , the
x
coordinate
oF the particle is
x
=
R
cos
b
ω t
+
φ
B
= (7
.
4 m) cos
b
(7
.
2 rad
/
s) (1
.
81 s)
+ (0
.
647992 rad)
B
=
3
.
26647 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
= 1
.
81 s.
Correct answer:

47
.
8083 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
= 1
.
81 s
,
the argument oF the sine is
φ
2
≡
ω t
+
φ
= (7
.
2 rad
/
s) (1
.
81 s) + (0
.
647992 rad)
= 13
.
68 rad
,
and the
x
component oF the velocity oF the
particle is
v
x
=

ω R
sin(
φ
2
)
=

(7
.
2 rad
/
s) (7
.
4 m) sin(13
.
68 rad)
=

47
.
8083 m
/
s
.
003
(part 3 oF 3) 10.0 points
±ind the
x
component oF the particle’s accel
eration at
t
= 1
.
81 s.
Correct answer:

169
.
334 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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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
= 1
.
81 s
,
the
x
component of the
particle’s acceleration is
a
x
=

ω
2
R
cos
φ
2
=

(7
.
2 rad
/
s)
2
(7
.
4 m) cos(13
.
68 rad)
=

169
.
334 m
/
s
2
.
004
(part 1 of 3) 10.0 points
Consider a uniform rod with a mass
m
and
length
L
pivoted on a frictionless horizontal
bearing at a point
O
p
5
8
L
from the lower
end
P
,
as shown.
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 Spring '10
 McCord
 Simple Harmonic Motion, Correct Answer, Angular frequency, Portillo

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