kuruvila (lk5992) – HW 9 – opyrchal – (11113)
1
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printout
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have
11
questions.
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before answering.
001
10.0 points
The angular position of an object that rotates
about a fixed axis is given by
θ
(
t
) =
θ
0
e
βt
,
where
β
= 4 s

1
,
θ
0
= 1
.
3 rad, and
t
is in
seconds.
What is the magnitude of the total linear
acceleration at
t
= 0 of a point on the object
that is 2
.
5 cm from the axis?
Correct answer: 85
.
2863 cm
/
s
2
.
Explanation:
Let :
r
= 2
.
5 cm
.
We can find the angular velocity and accel
eration:
θ
(
t
) =
θ
0
e
βt
ω
(
t
) =
d θ
(
t
)
dt
=
β θ
0
e
βt
and
α
(
t
) =
d ω
(
t
)
dt
=
β
2
θ
0
e
βt
.
At
t
= 0, since
e
0
= 1,
ω
=
β θ
0
and
α
=
β
2
θ
0
,
so the tangential and radial accelerations are
a
t
=
α r
=
β
2
θ
0
r
and
a
r
=
ω
2
r
=
β
2
θ
2
0
r ,
and the magnitude of the total linear acceler
ation is
a
=
radicalBig
a
2
t
+
a
2
r
=
radicalBig
β
4
θ
2
0
r
2
+
β
4
θ
4
0
r
2
=
β
2
θ
0
r
radicalBig
1 +
θ
2
0
= (4 s

1
)
2
(1
.
3 rad) (2
.
5 cm)
radicalBig
1 + (1
.
3 rad)
2
=
85
.
2863 cm
/
s
2
.
002 (part 1 of 3) 10.0 points
What is the tangential acceleration of a bug
on the rim of a 78 rpm record of diameter
11
.
84 in
.
if the record moves from rest to its
final angular speed in 4 s?
The conversion
between inches and meters is 0
.
0254 m
/
in.
Correct answer: 0
.
307057 m
/
s
2
.
Explanation:
Let :
ω
= 78 rpm
,
r
= (5
.
92 in) (0
.
0254 m
/
in) = 0
.
150368 m
t
= 4 s
.
ω
=
ω
0
+
α t
=
α t
α
=
Δ
ω
Δ
t
,
so
a
t
=
r α
=
r
Δ
ω
Δ
t
=
(0
.
150368 m) (78 rpm)
4 s
2
π
1 rev
1 min
60 s
=
0
.
307057 m
/
s
2
.
003 (part 2 of 3) 10.0 points
When the record is at its final speed, what is
the tangential velocity of the bug?
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 Fall '08
 moro
 Acceleration, Angular Momentum, Kinetic Energy, Moment Of Inertia, Rotation

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