homework 23 – PAPAGEORGE, MATT – Due: Mar 25 2008, 4:00 am
1
Question 1, chap 12, sect 2.
part 1 of 1
10 points
A large wheel is coupled to a wheel with
half the diameter as shown.
r
2
r
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 twice the rotational
speed and twice the 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 four times the ro
tational speed and the same 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.
The smaller wheel (with half the radius)
rotates twice as fast:
parenleftbigg
1
2
r
parenrightbigg
(2
ω
) =
r ω
=
v
Question 2, chap 12, sect 3.
part 1 of 1
10 points
A wheel rotating with a constant angular
acceleration turns through 17 rev during a 2 s
time interval. Its angular velocity at the end
of this interval is 17 rad
/
s.
What is the angular acceleration of the
wheel?
Note:
The initial angular velocity is
not
zero.
Correct answer:
−
36
.
4071
rad
/
s
2
(tolerance
±
1 %).
Explanation:
Let :
N
= 17 rev
,
t
= 2 s
,
and
ω
= 17 rad
/
s
.
Total angle through which the wheel rotates
for
N
revolutions is
Δ
θ
=
N
2
π
= 106
.
814 rad
Δ
θ
=
ω
0
t
+
1
2
α t
2
(1)
ω
−
ω
0
=
α t ,
or
ω
0
=
ω
−
α t ,
substituting into Eq
.
1
Δ
θ
= (
ω
−
α t
)
t
+
1
2
α t
2
,
so
α
= 2
ω t
−
Δ
θ
t
2
= 2
(17 rad
/
s) (2 s)
−
(106
.
814 rad)
(2 s)
2
=
−
36
.
4071 rad
/
s
2
.
Question 3, chap 12, sect 3.
part 1 of 3
10 points
As a result of friction, the angular speed of
a wheel changes with time according to
d θ
d t
=
ω
0
exp (
−
σ t
)
,
where
ω
0
and
σ
are constants.
The angular
speed changes from an initial angular speed
of 5
.
56 rad
/
s to 4
.
16 rad
/
s in 9
.
87 s
.
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homework 23 – PAPAGEORGE, MATT – Due: Mar 25 2008, 4:00 am
2
Hint:
Use this information to determine
σ
and
ω
0
.
Determine the magnitude of the angular
acceleration after 2
.
03 s.
Correct answer: 0
.
153946
rad
/
s
2
(tolerance
±
1 %).
Explanation:
Note:
The angular speed changes from
ω
1
= 5
.
56 rad
/
s at
t
= 0 to
ω
2
= 4
.
16 rad
/
s
at
t
2
= 9
.
87 s
.
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 Spring '09
 KLEINMAN
 Physics, Acceleration, Angular Momentum, Work, Moment Of Inertia, cm cm cm, MATT –

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