hernandez (ejh742) – oldhomework 30 – Turner – (58120)
1
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001
(part 1 of 3) 10.0 points
A particle rotates counterclockwise in a circle
of radius 4
.
8 m with a constant angular speed
of 13 rad
/
s
.
At
t
= 0, the particle has an
x
coordinate of 1
.
4 m and
y >
0
.
x
y
(1
.
4 m
, y
)
Figure:
Not drawn to scale.
radius
4
.
8 m
13 rad
/
s
Determine the
x
coordinate of the particle
at
t
= 1
.
44 s
.
Correct answer: 1
.
98143 m.
Explanation:
Let :
x
0
= 1
.
4 m
,
ω
= 13 rad
/
s
,
t
0
= 0 s
,
and
R
= 4
.
8 m
,
Since the amplitude of the particle’s mo
tion
equals
the
radius
of
the
circle
and
ω
= 13 rad
/
s , we have
x
=
A
cos(
ω t
+
φ
)
= (4
.
8 m) cos
bracketleftBig
(13 rad
/
s)
t
+
φ
bracketrightBig
.
We can find
φ
using the initial condition that
x
0
= 1
.
4 m at
t
= 0
(1
.
4 m) = (4
.
8 m) cos(0 +
φ
)
,
which implies
φ
= arccos
x
0
R
= arccos
(1
.
4 m)
(4
.
8 m)
= 73
.
0423
◦
= 1
.
27483 rad
.
Therefore, at time
t
= 1
.
44 s , the
x
coordinate
of the particle is
x
=
R
cos
bracketleftBig
ω t
+
φ
bracketrightBig
= (4
.
8 m) cos
bracketleftBig
(13 rad
/
s) (1
.
44 s)
+ (1
.
27483 rad)
bracketrightBig
=
1
.
98143 m
.
Note:
The angles in the cosine are in radians.
002
(part 2 of 3) 10.0 points
Find the
x
component of the particle’s veloc
ity at
t
= 1
.
44 s.
Correct answer:

56
.
8353 m
/
s.
Explanation:
Differentiating the function
x
(
t
) with re
spect to
t
, we find the
x
component of the
particle’s velocity at any time
t
v
x
=
d x
dt
=

ω A
sin(
ω t
+
φ
)
,
so at
t
= 1
.
44 s
,
the argument of the sine is
φ
2
≡
ω t
+
φ
= (13 rad
/
s) (1
.
44 s) + (1
.
27483 rad)
= 19
.
9948 rad
,
and the
x
component of the velocity of the
particle is
v
x
=

ω R
sin(
φ
2
)
=

(13 rad
/
s) (4
.
8 m) sin(19
.
9948 rad)
=

56
.
8353 m
/
s
.
003
(part 3 of 3) 10.0 points
Find the
x
component of the particle’s accel
eration at
t
= 1
.
44 s.
Correct answer:

334
.
862 m
/
s
2
.
Explanation:
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hernandez (ejh742) – oldhomework 30 – Turner – (58120)
2
Differentiating the
x
component of the par
ticle’s velocity with respect to
t
, we find the
x
component of the particle’s acceleration at
any time
t
a
x
=
d v
x
dt
=

ω
2
A
cos(
ω t
+
φ
)
,
so at
t
= 1
.
44 s
,
the
x
component of the
particle’s acceleration is
a
x
=

ω
2
R
cos
φ
2
=

(13 rad
/
s)
2
(4
.
8 m) cos(19
.
9948 rad)
=

334
.
862 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
parenleftbigg
5
8
L
from the lower
end
parenrightbigg
,
as shown.
5
8
L
L
O
θ
What is the moment of inertia of the rod
about the pivot point
O
?
The moment of
inertia of a uniform rod about its center of
mass is
1
12
M L
2
.
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 Spring '08
 Turner
 Physics, Simple Harmonic Motion, Work, Correct Answer, Angular frequency, D2

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